Question 14 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}0&2&6\\1&5&0\\3&7&1 \end{vmatrix}$
AnswerLet Mij and Cij are respectively the minor and co-factor of the element aij.
Now,
$\text{M}_{11}=\begin{vmatrix}5&0\\7&1 \end{vmatrix}=5-0=5$
$\text{M}_{21}=\begin{vmatrix}2&6\\7&1 \end{vmatrix}=2-42=-40$
$\text{M}_{31}=\begin{vmatrix}2&6\\5&0 \end{vmatrix}=0-30=-30$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=5$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=(-1)(-40)=40$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=(-30)=-30$
Now, expanding the determinant along the first column.
$|\text{A}|=\text{a}_{11}\text{C}_{11}+\text{a}_{21}\text{C}_{21}+\text{a}_{31}\text{C}_{31}$
$=0\times5+1\times(40)+3\times(-30)$
$=40-90$
$=-50$
View full question & answer→Question 24 Marks
Evaluate:
$\begin{vmatrix}\text{a}&\text{b}+\text{c}&\text{a}^2\\\text{b}&\text{c}+\text{a}&\text{b}^2\\\text{c}&\text{a}+\text{b}&\text{c}^2\end{vmatrix}$
Answer$\begin{vmatrix}\text{a}&\text{b}+\text{c}&\text{a}^2\\\text{b}&\text{c}+\text{a}&\text{b}^2\\\text{c}&\text{a}+\text{b}&\text{c}^2\end{vmatrix}$
Apply: C2 → C2 + C1
$=\begin{vmatrix}\text{a}&\text{b}+\text{c}+\text{a}&\text{a}^2\\\text{b}&\text{c}+\text{a}+\text{b}&\text{b}^2\\\text{c}&\text{a}+\text{b}+\text{c}&\text{c}^2\end{vmatrix}$
Take (a + b + c) common from C2
$=(\text{b}+\text{c}+\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\\text{b}&1&\text{b}^2\\\text{c}&1&\text{c}^2\end{vmatrix}$
Apply: R2 → R2 - R1, R3 → R3 - R1
$=(\text{b}+\text{c}+\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\\text{b}-\text{a}&0&\text{b}^2-\text{a}^2\\\text{c}-\text{a}&0&\text{c}^2-\text{a}^2\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}\text{a}&1&\text{a}^2\\1&0&\text{b}+\text{a}\\1&0&\text{c}+\text{a}\end{vmatrix}$
$=(\text{b}+\text{c}+\text{a})(\text{b}-\text{a})(\text{c}-\text{a})(\text{b}-\text{c})$
View full question & answer→Question 34 Marks
Solve the following system of homogeneous linear equations:
x + y - 2z = 0,
2x + y - 3z = 0,
5x + 4y - 9z = 0
AnswerConsider,
x + y - 2z = 0
2x + y - 9z = 0
5x + 4y - 9z = 0
$\text{D}=\begin{vmatrix}1&1&-2\\2&1&-3\\5&4&-9 \end{vmatrix}$
$=1(-9+12)-1(-18+15)-2(8-5)=0$
So, the system has infinitely many solutions, putting z = k in the first two equations,
x + y = 2k
2x + y = 3k
Using cramer's rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{\begin{vmatrix}2\text{k}&1\\3\text{k}&1\end{vmatrix}}{\begin{vmatrix}1&1\\2&1\end{vmatrix}}=\frac{-\text{k}}{-1}=\text{k}$
$ \text{y}=\frac{\text{D}_2}{\text{D}}=\frac{\begin{vmatrix}1&2\text{k}\\1&3\text{k}\end{vmatrix}}{\begin{vmatrix}1&1\\2&1\end{vmatrix}}=\frac{-\text{k}}{-1}=\text{k}$
z = k
Clearly, these value satisfy the third equation.
Thus,
x = y = z - k $[\text{k}\in\text{R}]$
View full question & answer→Question 44 Marks
$\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
Answer$\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
$\text{L.H.S}=\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}$
Multiply R1, R2 and R3 by a, b and c respectively.
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{ab}^2+\text{ac}^2&\text{a}^2\text{b}&\text{a}^2\text{c}\\\text{b}^2\text{a}&\text{bc}^2+\text{ba}^2&\text{b}^2\text{c}\\\text{c}^2\text{a}&\text{c}^2\text{b}&\text{ca}^2+\text{cb}^2\end{vmatrix}$
Take a, b, and c common from C1, C2 and C3 respectively.
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}\text{b}^2+\text{c}^2&\text{a}^2&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
Now apply R1 → R1 + R2 + R3
$=\begin{vmatrix}2(\text{b}^2+\text{c}^2)&2(\text{c}^2+\text{a}^2)&2(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}(\text{b}^2+\text{c}^2)&(\text{c}^2+\text{a}^2)&(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}\text{c}^2&0&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\big[\text{c}^2\{(\text{c}^2+\text{a}^2)(\text{a}^2+\text{b}^2)-\text{b}^2\text{c}^2\}+\text{a}^2\{\text{b}^2\text{c}^2-(\text{c}^2+\text{a}^2)\text{c}^2\}\big]$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 54 Marks
Show that $\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}=0,$ where $\alpha,\beta,\gamma$ are in A.P.
AnswerSince, $\alpha,\beta,\gamma$ are in A.P, $2\beta=\alpha+\gamma$
$\text{L.H.S}=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\frac{\text{R}_1}{2}-\frac{\text{R}_3}{2}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\$\text{x}-2)-\frac{\text{x}-3}{2}-\frac{\text{x}-1}{2}&(\text{x}-3)-\frac{\text{x}-4}{2}-\frac{\text{x}-2}{2}&(\text{x}-\beta)-\frac{\text{x}-\alpha}{2}-\frac{\text{x}-\gamma}{2}\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\0&0&0\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$ $[\because2\beta=\alpha+\gamma]$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 64 Marks
Prove the following identities:
$\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2\begin{vmatrix}\text{a}^3&1&\text{a}\\\text{b}^3&1&\text{b}\\\text{c}^3&1&\text{c}\end{vmatrix}$
$=2\{\text{a}^3(\text{c}-\text{d})-1(\text{b}^3\text{c}-\text{bc}^3)+\text{a}(\text{b}^3-\text{c}^3)\}$
$=2\{\text{a}^3(\text{c}-\text{b})-\text{bc}(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=(\text{b}-\text{c})\{-\text{a}^3-\text{bc}(\text{b}+\text{c})+\text{a}(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
$=\text{R.H.S}$
View full question & answer→Question 74 Marks
Prove that:
$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
Answer$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{b}+\text{c}+\text{a}&-\text{b}&\text{a}\\\text{c}+\text{a}+\text{b}&-\text{c}&\text{b}\\\text{a}+\text{b}+\text{c}&-\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\1&\text{c}&\text{b}\\1&\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\0&\text{c}-\text{b}&\text{b}-\text{a}\\0&\text{a}-\text{b}&\text{c}-\text{a}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})[(\text{c}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{a})(\text{a}-\text{b})]$
$=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$=\text{R.H.S}$
View full question & answer→Question 84 Marks
Prove the following identities:
$\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})^3$
Answer$\text{L.H.S}=\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$ [R1 = R1 + R2 + R3]
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&1&1\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&0&0\\2\text{z}&0&-\text{x}-\text{y}-\text{z}\\\text{x}-\text{y}-\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\end{vmatrix}$ [C2 = C2 - C1, C3 = C3 - C1]
$=(\text{x}+\text{y}+\text{z})\big[1\{0+(\text{x}+\text{y}+\text{z})(\text{x}+\text{y}+\text{z})\}\big]$
$=(\text{x}+\text{y}+\text{z})^3$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 94 Marks
Solve the following determinant equations:
$\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}15-2\text{x}-14+2\text{x}&11-3\text{x}&7-\text{x}\\11-28&17&14\\10-26&16&13\end{vmatrix}=0$ [Applying C1 → C1 - 2C3]
$\Rightarrow\begin{vmatrix}1&11-3\text{x}&7-\text{x}\\-17&17&14\\-16&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}12-3\text{x}&4-2\text{x}&7-\text{x}\\0&3&14\\0&3&13\end{vmatrix}=0$ [Applying C1 → C1 + C2 and C2 → C2 - C3]
$\Rightarrow(12-3\text{x})((3)\times13-(3\times14))=0$
$\Rightarrow(12-3\text{x})(-3)=0$
$\Rightarrow12-3\text{x}=0$
$\Rightarrow3\text{x}=12$
$\Rightarrow\text{x}=4$
View full question & answer→Question 104 Marks
Evaluate the following:
$\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}-1&1-\text{x}&0\\1&\text{x}&1\\0&1-\text{x}&\text{x}-1\end{vmatrix}$ [Applying R1 → R1 - R2 and R3 → R3 - R2]
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}&1\\0&-1&1\end{vmatrix}$
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}+1&1\\0&0&1\end{vmatrix}$ [Applying C2 → C2 + C3]
$=(\text{x}-1)^2(\text{x}+1+1)$ [Expanding along last row]
$=(\text{x}-1)^2(\text{x}+2)$
$\therefore\triangle=(\text{x}-1)^2(\text{x}+2)$
View full question & answer→Question 114 Marks
If the points (x, -2), (5, 2), (8, 8) are collinear, find x using determinants.
AnswerThe points (k, -2), (5, 2), (8, 8) are collinear.
$\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}=0$
$\triangle=\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8&8&1\end{vmatrix}$ [Applying R2 → R2 - R1]
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8-\text{x}&10&0\end{vmatrix}$ [Applying R3 → R3 - R1]
$=\begin{vmatrix}5-\text{x}&4\\8-\text{x}&10\end{vmatrix}$
$=50-10\text{x}-32+4\text{x}$
$=18-6\text{x}=0$
$\Rightarrow\text{x}=3$
View full question & answer→Question 124 Marks
Evaluate:
$\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&\lambda&\text{x}\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ [Applying C1 → C1 - C2, C2 → C2 - C3]
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&0&2\text{x}+\lambda\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ [Applying R1 → R2 + R3]
$=\lambda\begin{vmatrix}0&2\text{x}+\lambda\\-\lambda&\text{x}+\lambda\end{vmatrix}+\text{x}\begin{vmatrix}-\lambda&0\\0&-\lambda\end{vmatrix}$
$=\lambda\big[\lambda(2\text{x}+\lambda)\big]+\text{x}\lambda^2$
$=\lambda^2(2\text{x}+\lambda+\lambda^2\text{x})$
$=3\lambda^2\text{x}+\lambda^3$
$=\lambda^2(3\text{x}+\lambda)$
View full question & answer→Question 134 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}=0,\text{a}\neq\text{b}$
AnswerLet $\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$ [Applying R2 → R1 - R2]
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\0&\text{x}-\text{b}&\text{x}^2-\text{b}^2\end{vmatrix}$ [Applying R3 → R1 - R3]
$=(\text{x}-\text{a})(\text{x}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{x}+\text{a}\\0&1&\text{x}+\text{b}\end{vmatrix}$
$=(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}+\text{b}-\text{x}-\text{a})=0$
$\text{x}=\text{a},\text{b}$
View full question & answer→Question 144 Marks
$\begin{vmatrix}0&\text{b}^2\text{a}&\text{c}^2\text{a}\\\text{a}^2\text{b}&0&\text{c}^2\text{b}\\\text{a}^2\text{c}&\text{b}^2\text{c}&0\end{vmatrix}=2\text{a}^3\text{b}^3\text{c}^3$
Answer$\text{L.H.S}=\begin{vmatrix}0&\text{b}^2\text{a}&\text{c}^2\text{a}\\\text{a}^2\text{b}&0&\text{c}^2\text{b}\\\text{a}^2\text{c}&\text{b}^2\text{c}&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&\text{b}^3\text{a}&\text{c}^3\text{a}\\\text{a}^3\text{b}&0&\text{c}^3\text{b}\\\text{a}^3\text{c}&\text{b}^3\text{c}&0\end{vmatrix}$ [Multiplying the three columns by a, b, and c]
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}0&\text{b}^3&\text{c}^3\\\text{a}^3&0&\text{c}^3\\\text{a}^3&\text{b}^3&0\end{vmatrix}$ [Taking out a, b and c common from the three rows]
$=\text{b}^3\begin{vmatrix}\text{b}^3&\text{c}^3\\\text{a}^3&0\end{vmatrix}+\text{c}^3\begin{vmatrix}\text{a}^3&0\\\text{a}^3&\text{b}^3\end{vmatrix}$ [Expanding along R1]
$=2\text{a}^3\text{b}^3\text{c}^3$
View full question & answer→Question 154 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x + y - z = 0,
x - 2y + z = 0,
3x + 6y - 5z = 0
AnswerUsing the equations we get,
$\text{D}=\begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}$
$=1(10-6)-1(-5-3)-1(6+6)=0$
$\text{D}_1=\begin{vmatrix}1&1&-1\\0&-2&1\\0&6&-5\end{vmatrix}$
$=0(10-6)-1(0-0)-1(0+0)=0$
$\text{D}_2=\begin{vmatrix}1&0&-1\\1&0&1\\3&0&-5\end{vmatrix}$
$=1(0-0)-0(5-3)-1(0-0)=0$
$\text{D}_3=\begin{vmatrix}1&1&0\\1&-2&0\\3&6&0\end{vmatrix}$
$=1(0-0)-1(0-0)+0(6+6)=0$
$\text{D}=\text{D}_1=\text{D}_2$
Thus, the system has infinitely many solution.
View full question & answer→Question 164 Marks
Solve the following systems of linear equations by cramer's rule:
x - 2y = 4,
-3x + 5y = -7
AnswerGiven, x - 2y = 4
-3x + 5y = -7
Using the properties of determinants, we get
$\text{D}=\begin{vmatrix}1&-2\\-3&5 \end{vmatrix}=5-6=-1\neq0$
$\text{D}_1=\begin{vmatrix}4&-2\\-7&5 \end{vmatrix}=20-14=6$
$\text{D}_2=\begin{vmatrix}1&4\\-3&-7 \end{vmatrix}=-7+12=5$
Using cramer's Rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6}{-1}=-6$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{5}{-1}=-5$
$\therefore\text{x}=-6$ and $\text{y}=-5$
View full question & answer→Question 174 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
Answer$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&3\text{a}+\text{b}\\2\text{a}+\text{b}&3\text{a}+\text{b}&4\text{a}+\text{b}\\4\text{a}+\text{b}&5\text{a}+\text{b}&6\text{a}+\text{b} \end{vmatrix}$
Apply: C3 → C3 - C2
$\begin{vmatrix}\text{a}+\text{b}&2\text{a}+\text{b}&\text{a}\\2\text{a}+\text{b}&3\text{a}+\text{b}&\text{a}\\4\text{a}+\text{b}&5\text{a}+\text{b}&\text{a} \end{vmatrix}$
Apply: C2 → C2 - C1
$\begin{vmatrix}\text{a}+\text{b}&\text{a}&\text{a}\\2\text{a}+\text{b}&\text{a}&\text{a}\\4\text{a}+\text{b}&\text{a}&\text{a} \end{vmatrix}$
$=0$
$\because\text{C}_3=\text{C}_2$
View full question & answer→Question 184 Marks
Prove that:
$\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}=16(3\text{x}+4)$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+4&3\text{x}+4&3\text{x}+4\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
[Applying R1 → R2 + R2 + R3]
$=(3\text{x}+4)\begin{vmatrix}1&1&1\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
[Taking out (3x + 4) common from R1]
$=(3\text{x}+4)\begin{vmatrix}1&0&0\\\text{x}&4&0\\\text{x}&0&4\end{vmatrix}$
[Applying C2 → C2 - C1 and C3 → C3 - C1]
$=(3\text{x}+4)(4)^2$ [Expanding along R1]
$=16(3\text{x}+4)$
$=\text{R.H.S}$
View full question & answer→Question 194 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = 17,
3x + 5y = 6
AnswerGiven, 2x - y = 17
3x + 5y = 6
Using cramers Rule, we get
$\text{D}=\begin{vmatrix}2&1\\3&5\end{vmatrix}=10+3=13$
$\text{D}_1=\begin{vmatrix}17&-1\\6&5\end{vmatrix}=85+6=91$
$\text{D}_2=\begin{vmatrix}2&17\\3&6\end{vmatrix}=12-51=-39$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{13}=7$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-39}{13}=-3$
$\therefore\text{x}=7$ and $\text{y}=-3$
View full question & answer→Question 204 Marks
A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission.
| Month | Sale of units | Total commission drawn (in Rs.) |
| | A | B | C | |
| Jan | 90 | 100 | 20 | 800 |
| Feb | 130 | 50 | 40 | 900 |
| March | 60 | 100 | 30 | 850 |
Find out the rates of commission on items A, B and C by using determinant method.
AnswerLet the rates of commissions on iteams A, B and C be x, y and z respectively.
Then we can express the given modal as system of linear equations
90x + 100y + 20z = 800
130x + 50y + 40z = 900
60x + 100y + 30z = 850
We will solve this using the Cramer's rule
Here,
$\text{D}=\begin{vmatrix}90&100&20\\130&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-200&0&-50\end{vmatrix}$
$=50(8500-12000)=-175000$
$\text{D}_1=\begin{vmatrix}800&100&20\\900&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-950&0&-50\end{vmatrix}$
$=50(50000-57000)=-350000$
$\text{D}_2=\begin{vmatrix}90&800&20\\130&900&40\\60&850&30\end{vmatrix}=\begin{vmatrix}90&800&20\\-50&-700&0\\-75&-350&0\end{vmatrix}$
$=20(17500-52500)=-700000$
$\text{D}_3=\begin{vmatrix}90&100&800\\130&50&900\\60&100&850\end{vmatrix}=\begin{vmatrix}-170&0&-1000\\130&50&900\\-200&0&-950\end{vmatrix}$
$=50(161500-200000)=-1925000$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-350000}{-175000}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-700000}{-175000}=4$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-1925000}{-175000}=11$
$\therefore$ The rates of commission of iteam A, B and C are 2%, 4% and 11% respectively.
View full question & answer→Question 214 Marks
Solve the following systems of linear equations by cramer's rule:
x + y + z + 1 = 0,
ax + by + cz + d = 0,
a2x + b2y + x2z + d2 = 0
AnswerThese equations can be written as
x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 0
$\text{D}=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2 \end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{a}&\text{a}-\text{b}&\text{b}-\text{c}\\\text{a}^2&\text{a}^2-\text{b}^2&\text{b}^2-\text{c}^2 \end{vmatrix} $ [Applying C2 → C1 - C2, C3 → C2 - C3]
Taking (b - a) and (c - a) common from C1 and C2, respectively, we get
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}1&0&0\\\text{a}&1&1\\\text{a}^2&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})\ ....(\text{i})$
$\text{D}_1=\begin{vmatrix}-1&1&1\\-\text{d}&\text{b}&\text{c}\\-\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{d}&\text{b}&\text{c}\\\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}$
$\text{D}_1=-(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})$ [Replacing a by d in eq. (i)]
$\text{D}_2=\begin{vmatrix}1&-1&1\\\text{a}&-\text{d}&\text{c}\\\text{a}^2&-\text{d}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{c}\\\text{a}^2&\text{d}^2&\text{c}^2\end{vmatrix}$
$\text{D}_2=-(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})$
$\text{D}_3=\begin{vmatrix}1&1&-1\\\text{a}&\text{b}&-\text{d}\\\text{a}^2&\text{b}^2&-\text{d}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{d}\\\text{a}^2&\text{b}^2&\text{d}^2\end{vmatrix}$
$\text{D}_3=-(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})$
Thus,
$\text{x}=\frac{\text{D}_1}{\text{D}}=-\frac{(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
View full question & answer→Question 224 Marks
Solve the following systems of linear equations by cramer's rule:
6x + y - 3z = 5,
x + 3y - 2z = 5,
2x + y + 4z = 8
AnswerLet $\text{D}=\begin{vmatrix}6&1&-3\\1&3&-2\\2&1&4\end{vmatrix}$
Expanding along R1
$=6(14)-1(8)-3(-5)$
$=84-8+15=91$
Also, $\text{D}_1=\begin{vmatrix}5&1&-3\\5&-3&-2\\8&1&4\end{vmatrix}$
Expanding along R1
$=5(14)-1(36)-3(-19)$
$=70-36+57=91$
Again $\text{D}_2=\begin{vmatrix}6&5&-3\\1&5&-2\\2&8&4\end{vmatrix}$
Expanding along R1
$=6(36)-5(8)-3(-2)$
$=216-40+6=182$
Also $\text{D}_3=\begin{vmatrix}6&1&5\\1&3&5\\2&1&8\end{vmatrix}$
Expanding along R1
$=6(19)-1(-2)+5(-5)$
$=114+2-25=91$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{91}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{182}{91}=2$
Also $\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{91}{91}=1$
Hence, x = 1, y = 2, z = 1
View full question & answer→Question 234 Marks
Solve the following systems of linear equations by cramer's rule:
5x - 7y + z = 11,
6x - 8y - z = 15,
3x + 2y - 6z = 7
Answer$\text{D}=\begin{vmatrix}5&-7&1\\6&-8&-1\\3&2&-6 \end{vmatrix}$
$=5(48+2)+7(-36+3)+1(12+24)$
$=5(50)+7(-33)+1(36)=55$
$\text{D}_1=\begin{vmatrix}11&-7&1\\15&-8&-1\\7&2&-6 \end{vmatrix}$
$=11(48+2)+7(-90+7)+1(30+36)$
$=11(50)+7(-83)+1(86)=55$
$\text{D}_2=\begin{vmatrix}5&11&1\\6&15&-1\\3&7&-6 \end{vmatrix}$
$=5(-90+7)-11(-36+3)+1(42-45)$
$=5(-83)-11(-33)+1(-3)=-55$
$\text{D}_3=\begin{vmatrix}5&-7&11\\6&-8&15\\3&2&7 \end{vmatrix}$
$=5(-56-30)+7(42-45)+11(12+24)$
$=5(-86)+7(-3)+11(36)=-55$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{55}{55}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-55}{55}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-55}{55}=-1$
$\therefore\text{x}=1,\text{ y}=-1$ and $\text{z}=-1$
View full question & answer→Question 244 Marks
If $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2,$ find $\frac{\text{dy}}{\text{dx}}$
AnswerWe have, $\sin(\text{xy})+\frac{\text{y}}{\text{x}}=\text{x}^2-\text{y}^2$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{ xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{d}}{\text{dx}}(\text{x}^2)-\frac{\text{d}}{\text{dx}}(\text{y}^2)$
$\Rightarrow \cos(\text{xy})\frac{\text{d}}{\text{dx}}(\text{xy})+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\frac{\text{d}}{\text{dx}}(\text{x})}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big\}+\Bigg\{\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}(1)}{\text{x}^2}\Bigg\}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \cos(\text{xy})\Big\{\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big\}+\frac{1}{\text{x}^2}\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})+\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}^2}=2\text{x}-2\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}\Big\{\text{x}\cos(\text{xy})+\frac{1}{\text{x}}+2\text{y}\Big\}=\frac{\text{y}}{\text{x}^2}-\text{y}\cos(\text{xy})+2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big\{\frac{\text{x}^2\cos(\text{xy})+1+2\text{xy}}{\text{x}}\Big\}=\frac{1}{\text{x}^2}\big(\text{y}-\text{x}^2\text{y}\cos(\text{xy})+2\text{x}^2\big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{2\text{x}^3+\text{y}-\text{x}^2\text{y}\cos(\text{xy})}{\text{x}\big(\text{x}^2\cos(\text{xy})+1+2\text{xy}\big)}$
View full question & answer→Question 254 Marks
Solve the following systems of linear equations by cramer's rule:
x + y = 5,
y + z = 3,
x + z = 4
AnswerLet $\text{D}=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}$
Expanding along R1
$=1(1)-1(-1)+0(-1)$
$=1+1+0=2$
Also $\text{D}_1=\begin{vmatrix}5&1&0\\3&1&1\\4&0&1\end{vmatrix}$
Expanding along R1
$=5(1)-1(-1)+0(-4)$
$=5+1+0=6$
Again $\text{D}_2=\begin{vmatrix}1&5&0\\0&3&1\\1&4&1\end{vmatrix}$
Expanding along R1
$=1(-1)-5(-1)+0(-3)$
$=-1+5+0=4$
Also $\text{D}_3=\begin{vmatrix}1&1&5\\0&1&3\\1&0&4\end{vmatrix}$
$=1(4)-1(-3)+5(-1)$
$=4+3-5=2$
Now $\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6}{2}=3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{4}{2}=2$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{2}{2}=1$
Hence, x = 3, y = 2, z = 1
View full question & answer→Question 264 Marks
Solve the following systems of linear equations by cramer's rule:
x + y = 1,
x + z = -6,
x - y - 2z = 3
AnswerThese equations can be written as
x + y + 0z = 1
x + 0y + z = -6
x - y - 2z = 3
$\text{D}=\begin{vmatrix}1&1&0\\1&0&1\\1&-1&-2 \end{vmatrix}$
$=1(0+1)-1(-2-1)+0(-1-0)=4$
$\text{D}_1=\begin{vmatrix}1&1&0\\-6&0&1\\3&-1&-2 \end{vmatrix}$
$=1(0+1)-1(12-3)+0(6-0)=-8$
$\text{D}_2=\begin{vmatrix}1&1&0\\1&-6&1\\1&3&-2 \end{vmatrix}$
$=1(12-3)-1(-2-1)+0(3+6)=12$
$\text{D}_3=\begin{vmatrix}1&1&0\\1&0&-6\\1&-1&3 \end{vmatrix}$
$=1(0-6)-1(3+6)+1(-1-0)=-16$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-8}{4}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{4}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-16}{4}=-4$
$\therefore\text{x}=-2,\text{y}=3$ and $\text{z}=-4$
View full question & answer→Question 274 Marks
Solve the following systems of linear equations by cramer's rule:
x + y + z + w = 2,
x - 2y + 2z + 2w = -6,
2x + y - 2z + 2w = -5,
3x - y + 3z - 3w = -3
AnswerHere, $\text{D}=\begin{vmatrix}1&1&1&1\\1&-2&2&2\\2&1&-2&2\\3&-1&3&-3\end{vmatrix}$
$\therefore\text{D}=\begin{vmatrix}1&0&0&0\\1&-3&1&1\\2&-1&-4&0\\3&-4&0&-6\end{vmatrix}=1\begin{vmatrix}-3&1&1\\-1&-4&0\\-4&0&-6\end{vmatrix}$
[C2 → C2 - C1, C3 → C3 - C1, C4 → C4 - C1]
$\therefore\text{D}=\begin{vmatrix}0&0&1\\-1&-4&0\\-22&6&-6\end{vmatrix}$ [C1 → C1 + 3C3, C2 → C2 - C3]
$=1(-6-88)=-94$
$\text{D}_1=\begin{vmatrix}2&1&1&1\\-6&-2&2&2\\-5&1&-2&2\\-3&-1&3&-3\end{vmatrix}=188$
$\text{D}_2=\begin{vmatrix}1&2&1&1\\1&-6&2&2\\2&-5&-2&2\\3&-3&3&-3\end{vmatrix}=-282$
$\text{D}_3=\begin{vmatrix}1&1&2&1\\1&-2&-6&2\\2&-1&-5&-2\\3&-1&-3&-3\end{vmatrix}=-141$
$\text{D}_4=\begin{vmatrix}1&1&1&2\\1&-2&2&-6\\2&1&-2&-5\\3&-1&3&-3\end{vmatrix}=47$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{188}{-94}=-2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-282}{-94}=3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-141}{94}=\frac{3}{2}$
$\text{w}=\frac{\text{D}_4}{\text{D}}=\frac{47}{-94}=-\frac{1}{2}$
Hence, $\text{x}=-2,\text{y}=3,\text{z}=\frac{3}{2},\text{w}=-\frac{1}{2}$
View full question & answer→Question 284 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x - y + z = 3,
2x + y - z = 2,
-x - 2y + 2z = 1
AnswerWe have,
$\text{D}=\begin{vmatrix}1&-1&1\\2&1&-1\\-1&-2&2\end{vmatrix}=\begin{vmatrix}0&-1&0\\3&1&0\\-3&-2&0\end{vmatrix}=0$
$\text{D}_1=\begin{vmatrix}3&-1&1\\2&1&-1\\1&-2&2\end{vmatrix}=\begin{vmatrix}3&-1&0\\2&1&0\\1&-2&0\end{vmatrix}=0$
$\text{D}_2=\begin{vmatrix}1&3&1\\2&2&-1\\-1&1&2\end{vmatrix}=\begin{vmatrix}1&0&0\\2&-4&-3\\-1&4&3\end{vmatrix}=1(-12+12)=0$
$\text{D}_3=\begin{vmatrix}1&-1&3\\2&1&2\\-1&-2&1\end{vmatrix}=\begin{vmatrix}1&0&0\\2&3&-4\\-1&-3&4\end{vmatrix}=1(12-12)=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
So, either the system is consistent with infinitely many solutions or it is inconsistent.
Consider the first two equations, written as
x - y = 3 - z
2x + y = 2 + z
to solve these equation, written as
Here,
$\text{D}=\begin{vmatrix}1&-1\\2&1\end{vmatrix}=1+2=3$
$\text{D}_1=\begin{vmatrix}3-\text{z}&-1\\2+\text{z}&1\end{vmatrix}=(3-\text{z})+(2+\text{z})=5$
$\text{D}_2=\begin{vmatrix}1&3-\text{z}\\1&2+\text{z}\end{vmatrix}=(2+\text{z})-(6-2\text{z})=-4+3\text{z}$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{5}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4+3\text{z}}{3}$
Let z = k, then the equations have the solution.
$\text{x}=\frac{5}{3},\text{ y}=\frac{-4+3\text{k}}{3},\text{ z}=\text{ k}$
View full question & answer→Question 294 Marks
Solve the following systems of linear equations by cramer's rule:
2x - 3y - 4z = 29,
-2x + 5y - z = -15,
3x - y + 5z = -11
AnswerGiven, 2x - 3y - 4z = 29
-2x + 5y - z = -15
3x - y + 5z = -11
$\text{D}=\begin{vmatrix}2&-3&-4\\-2&5&-1\\3&-1&5 \end{vmatrix}$
$=2(25-1)+3(-10+3)-4(2-15)$
$=2(24)+3(-7)-4(-13)=79$
$\text{D}_1=\begin{vmatrix}29&-3&-4\\-15&5&-1\\-11&-1&5 \end{vmatrix}$
$=29(25-1)+3(-72-11)-4(15+55)$
$=29(24)+3(-86)-4(70)=158$
$\text{D}_2=\begin{vmatrix}2&29&-4\\-2&-15&-1\\3&-11&5 \end{vmatrix}$
$=2(-75-11)-29(-10+3)-4(22+45)$
$=2(-86)-29(-7)-4(67)=-237$
$\text{D}_3=\begin{vmatrix}2&-3&29\\-2&5&-15\\3&-1&-11 \end{vmatrix}$
$=2(-55-15)+3(22+45)+29(2-15)$
$=2(-70)+3(67)+29(-13)=-316$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{158}{79}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-237}{79}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-316}{79}=-4$
$\therefore\text{x}=2,\text{y}=-3$ and $\text{z}=-4$
View full question & answer→Question 304 Marks
Solve the following systems of linear equations by cramer's rule:
x - 4y - z = 11,
2x - 5y + 2z = 39,
-3x + 2y + z = 1
AnswerGiven, x - 4y - z = 11
2x - 5y + 2z = 39
-3x + 2y + z = 1
$\text{D}=\begin{vmatrix}1&-4&-1\\2&-5&2\\-3&2&1\end{vmatrix}$
$=1(-5-4)-(-4)(2+6)+(-1)(4-15)$
$=1(-9)-(-4)(8)+(-1)(-11)=34$
$\text{D}_1=\begin{vmatrix}11&-4&-1\\39&-5&2\\1&2&1\end{vmatrix}$
$=11(-5-4)-(-4)(39-2)+(-1)(78+5)$
$=11(-9)-(-4)(37)+(-1)(83)=-34$
$\text{D}_2=\begin{vmatrix}1&11&-1\\2&39&2\\-3&1&1\end{vmatrix}$
$=1(39-2)-11(2+6)+(-1)(2+117)$
$=1(37)-11(8)+(-1)(119)=-170$
$\text{D}_3=\begin{vmatrix}1&-4&11\\2&-5&39\\-3&2&1\end{vmatrix}$
$=1(-5-78)-(-4)(2+117)+11(4-15)$
$=1(-83)-(-4)(119)+11(-11)=272$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-34}{34}=-1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-170}{34}=-5$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{272}{34}=8$
$\therefore\text{x}=-1,\text{ y}=-5$ and $\text{z}=8$
View full question & answer→Question 314 Marks
An automobile company uses three types of steel S
1, S
2 and S
3 for producing three types of cars C
1, C
2 and C
3. Steel requirements (in tons) for each type of cars are given below:
| Steel | Cars |
| | C1 | C2 | C3 |
| S1 | 2 | 3 | 4 |
| S2 | 1 | 1 | 2 |
| S3 | 3 | 2 | 1 |
Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.
AnswerExpressing the given information as a system of linear equations we get,
2x +3y + 4z = 29
x + y + 2z = 13
3x + 2y + z = 16
Where x, y, z is the number of cars C1, C2 and C3 produced.
We use Cramer's rule to solve this system.
Here,
$\text{D}=\begin{vmatrix}2&3&4\\1&1&2\\3&2&1\end{vmatrix}=\begin{vmatrix}-10&-5&0\\-5&-3&0\\3&2&1\end{vmatrix}=1(30-25)=5$
$\text{D}_1=\begin{vmatrix}29&3&4\\13&1&2\\16&2&1\end{vmatrix}=\begin{vmatrix}-35&-5&0\\-19&-3&0\\16&2&1\end{vmatrix}=1(105-95)=10$
$\text{D}_2=\begin{vmatrix}0&29&4\\1&13&2\\3&16&1\end{vmatrix}=\begin{vmatrix}-10&-35&0\\-5&-19&0\\3&16&1\end{vmatrix}=(190-175)=15$
$\text{D}_3=\text{D}=\begin{vmatrix}2&3&29\\1&1&13\\3&2&16\end{vmatrix}=\begin{vmatrix}-2&0&0\\1&1&13\\3&2&16\end{vmatrix}=-2(16-26)=20$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{10}{5}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{15}{5}=3$
and $\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{20}{5}=4$
Hence, the number of cars produced of the type C1, C2 and C3 are 2, 3 and 4 respectively.
View full question & answer→Question 324 Marks
Solve the following systems of linear equations by cramer's rule:
2x - 3z + w = 1,
x - y + 2w = 1,
-3y + z + w = 1,
x + y + z = 1
Answer$\text{D}=\begin{vmatrix}2&0&-3&1\\1&-1&0&2\\0&-3&1&1\\1&1&1&0\end{vmatrix}$
$=2\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-0-3\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 0(0 - 1) + 2(-3 - 1)] - 3[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 0(0 + 3)]
= -21
$\text{D}_1=\begin{vmatrix}1&0&-3&1\\1&-1&0&2\\1&-3&1&1\\1&1&1&0\end{vmatrix}$
$=\begin{vmatrix}-1&0&2\\-3&1&1\\1&1&0\end{vmatrix}-0-3\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 0(0 - 1) + 2(-3 - 1)] - 3[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 0(0 + 3)]
= -21
$\text{D}_2=\begin{vmatrix}2&1&3&1\\1&1&0&2\\0&1&1&1\\1&1&1&0\end{vmatrix}$
$=2\begin{vmatrix}1&0&2\\1&1&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&0&2\\0&1&1\\1&1&0\end{vmatrix}+(-3)\begin{vmatrix}1&1&2\\0&1&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&1&0\\0&1&1\\1&1&1\end{vmatrix}$
= 2[1(0 - 1) + 2(1 - 1)] - 1[1(0 - 1) + 2(0 - 1)] - 3[1(0 - 1) - 1(0 - 1) + 2(0 - 1)] - 1[1(1 - 1) - 1(0 - 1)]
= 6
$\text{D}_3=2\begin{vmatrix}-1&1&2\\-3&1&1\\1&1&0\end{vmatrix}-0+1\begin{vmatrix}1&-1&2\\0&-3&1\\1&1&0\end{vmatrix}-1\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(0 - 1) - 1(0 - 1) + 2(-3 - 1)] + 1[1(0 - 1) + 1(0 - 1) + 2(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1) + 1(0 + 3)]
= -6
$\text{D}_4=\begin{vmatrix}2&0&-3&1\\1&-1&0&1\\0&-3&1&1\\1&1&1&1\end{vmatrix}$
$=2\begin{vmatrix}-1&0&1\\-3&1&1\\1&1&1\end{vmatrix}-0-3\begin{vmatrix}1&-1&1\\0&-3&1\\1&1&1\end{vmatrix}-1\begin{vmatrix}1&-1&0\\0&-3&1\\1&1&1\end{vmatrix}$
= 2[-1(1 - 1) + 1(-3 - 1)] - 3[1(-3 - 1) + 1(0 - 1) + 1(0 + 3)] - 1[1(-3 - 1) + 1(0 - 1)]
= 3
So, by Cramer's rule, we obtain
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{21}{21}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{6}{-21}=-\frac{2}{7}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{3}{-21}=-\frac{1}{7}$
$\text{w}=\frac{\text{D}_4}{\text{D}}=\frac{3}{-21}=-\frac{1}{7}$
Hence, $\text{x}=1,\text{ y}=-\frac{2}{7},\text{ z}=\frac{2}{7},\text{ w}=-\frac{1}{7}$
View full question & answer→Question 334 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
2x + y - 2z = 0,
x - 2y + z = -2,
5x - 5y + z = -2
Answer$=\begin{vmatrix}2&1&-2\\1&-2&1\\5&5&1\end{vmatrix}=\begin{vmatrix}12&9&-12\\-4&-3&1\\0&0&1\end{vmatrix}=1(-36+36)=0$
$\text{D}_1=\begin{vmatrix}4&1&-2\\-2&-2&1\\-2&-5&1\end{vmatrix}=\begin{vmatrix}0&1&-2\\0&-2&1\\0&-5&1\end{vmatrix}=0$
$\text{D}_2=\begin{vmatrix}2&4&-2\\1&-2&1\\5&-2&1\end{vmatrix}=\begin{vmatrix}2&0&-2\\1&0&1\\5&0&1\end{vmatrix}=0$
$\text{D}_3=\begin{vmatrix}2&1&4\\1&-2&-2\\5&-5&-2\end{vmatrix}=\begin{vmatrix}4&-3&0\\1&-2&-2\\4&-3&0\end{vmatrix}=2(-12+12)=0$
So, $\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
So, the given system is either inconsistent or has infinite solutions.
Consider the 2nd and 3rd equations, written as
x - 2y = -2 - z
5x - 5y = -2 - z
Then,
$\text{D}=\begin{vmatrix}1&-2\\5&-5\end{vmatrix}=-5-(-10)=5$
$\text{D}_2=\begin{vmatrix}-2-\text{z}&-2\\-2-\text{z}&-5\end{vmatrix}$
$=(2+\text{z})+(5)-2(2+\text{z})=3(2+\text{z})=6+3\text{z}$
$\text{D}_2=\begin{vmatrix}1&-(2+\text{z})\\5&-(2+\text{z})\end{vmatrix}$
$=-(2+\text{z})+5(2+\text{z})=4(2+\text{z})=8+4\text{z}$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6+3\text{z}}{5}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{8+4\text{z}}{5}$
Let z = k, then
$\text{x}=\frac{6+3\text{k}}{5},\text{ y}=\frac{8+4\text{k}}{5},\text{ z}=\text{k}$ are the infinite solution of the given system of equations.
View full question & answer→Question 344 Marks
Solve the following systems of linear equations by cramer's rule:
2y - 3z = 0,
x + 3y = -4,
3x + 4y = 3
AnswerThese equations can be written as
0x + 2y - 3z = 0
x + 3y + 0z = -4
3x + 4y + 0z= 3
$\text{D}=\begin{vmatrix}0&2&-3\\1&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-2(0-0)-3(4-9)$
$=15$
$\text{D}_1=\begin{vmatrix}0&2&-3\\-4&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-0(0-0)-3(3+12)$
$=-45$
$\text{D}_3=\begin{vmatrix}0&2&0\\1&3&-4\\3&4&3 \end{vmatrix}$
$=0(9+16)-2(3+12)-0(4-9)$
$=-30$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{75}{15}=5$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-45}{15}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-30}{12}=-2$
$\therefore\text{x}=5,\text{y}=-3$ and $\text{z}=-2$
View full question & answer→Question 354 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+1&3&5\\2&\text{x}+2&5\\2&3&\text{x}+4\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}\text{x}+1&3&5\\2&\text{x}+2&5\\2&3&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}\text{x}+9&3&5\\\text{x}+9&\text{x}+2&5\\\text{x}+9&3&\text{x}+4\end{vmatrix}$ [Applying C1 = C1 + C2 + C3]
$=(\text{x}+9)\begin{vmatrix}1&3&5\\1&\text{x}+2&5\\1&3&\text{x}+4\end{vmatrix}$
$=(\text{x}+9)\begin{vmatrix}1&3&5\\0&\text{x}-1&0\\1&3&\text{x}+4\end{vmatrix}=0$ [Applying R2 → R2 - R1]
$=(\text{x}+9)\begin{vmatrix}1&3&5\\0&\text{x}-1&0\\1&0&\text{x}-1\end{vmatrix}$ [Applying R3 → R3 - R1]
$=(\text{x}+9)(\text{x}-1)^2=0$
$\text{x}=-9,1,1$
View full question & answer→Question 364 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}+2\text{c}&\text{a}&\text{b}\\\text{c}&\text{b}+\text{c}+2\text{a}&\text{b}\\\text{c}&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}=2(\text{a}+\text{b}+\text{c})^3$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}+2\text{c}&\text{a}&\text{b}\\\text{c}&\text{b}+\text{c}+2\text{a}&\text{b}\\\text{c}&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\1&\text{b}+\text{c}+2\text{a}&\text{b}\\1&\text{a}&\text{c}+\text{a}+2\text{b} \end{vmatrix}$ [Taking out 2(a + b + c) common from C1]
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\1&\text{b}+\text{c}+\text{a}&0\\0&-\text{b}-\text{c}-\text{a}&\text{c}+\text{a}+\text{b} \end{vmatrix}$ [Applying R2 → R2 - R1 and R2 → R2 - R3]
$=2(\text{a}+\text{b}+\text{c})(\text{a}+\text{b}+\text{c})(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{a}&\text{b}\\0&1&0\\0&-1&0\end{vmatrix}$ [Taking out (a + b + c) common from R2 and R3]
$=2(\text{a}+\text{b}+\text{c})^3\{1(1-0)\}$ [Expanding along C1]
$=2(\text{a}+\text{b}+\text{c})^3$
$=\text{R.H.S}$
View full question & answer→Question 374 Marks
Solve the following systems of linear equations by cramer's rule:
3x + y + z = 2,
2x - 4y + 3z = -1,
4x + y - 3z = -11
AnswerLet $\text{D}=\begin{vmatrix}3&1&1\\2&-4&3\\4&1&-3\end{vmatrix}$
Expanding along R1
$=3(9)+(-1)(-18)+1(18)$
$=27+18+18=63$
Again $\text{D}_2=\begin{vmatrix}3&2&1\\2&-1&3\\4&-11&-3\end{vmatrix}$
Expanding along R1
$=3(3+33)-2(-18)+1(-22+4)$
$=108+36-18=126$
Also $\text{D}_3=\begin{vmatrix}3&1&2\\2&-4&-1\\4&1&-11\end{vmatrix}$
Expanding along R1
$=3(45)-1(-18)+2(18)$
$=135+18+36=189$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-63}{63}=-1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{126}{63}=2$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{189}{63}=3$
View full question & answer→Question 384 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
Answer$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\sin^2(90-23)^{\circ}&-1\\-\sin^2(90-23)^{\circ}&-\sin^223^{\circ}&1\\1&\sin^223^{\circ}&\sin^2(90-23)^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}&-\sin^223^{\circ}&1\\-1&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}+\cos^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}-\sin^223^{\circ}&-\sin^223^{\circ}&1\\-1+\sin^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$ [Applying C1 → C1 + C2]
$=\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=(-1)\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=0$
View full question & answer→Question 394 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{a}^2-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&\text{b}^2-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&\text{c}^2-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2&-(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&-(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&-(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$ [Applying C2 → C2 - C1]
$=(-1)\begin{vmatrix}\text{a}^2&(\text{b}-\text{c})^2&\text{bc}\\\text{b}^2&(\text{c}-\text{a})^2&\text{ca}\\\text{c}^2&(\text{a}-\text{b})^2&\text{ab}\end{vmatrix}$
$=-\begin{vmatrix}\text{a}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$ [Applying C2 → C2 - 2C1]
$=-\begin{vmatrix}\text{a}^2+\text{b}^2+\text{c}^2&\text{b}^2+\text{c}^2&\text{bc}\\\text{b}^2+\text{c}^2+\text{a}^2&\text{c}^2+\text{a}^2&\text{ca}\\\text{c}^2+\text{a}^2+\text{b}&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$ [Applying C1 → C1 + C2]
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\1&\text{c}^2+\text{a}^2&\text{ca}\\1&\text{a}^2+\text{b}^2&\text{ab}\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}^2-\text{b}^2&\text{c}(\text{a}-\text{b})\\0&\text{a}^2-\text{c}^2&\text{b}(\text{a}-\text{c})\end{vmatrix}$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{a}-\text{c})\begin{vmatrix}1&\text{b}^2+\text{c}^2&\text{bc}\\0&\text{a}+\text{b}&\text{c}\\0&\text{a}+\text{c}&\text{b}\end{vmatrix} $
[Taking (a - b) common from R2 and (a - c) common from R3]
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\times\left\{1\times\begin{vmatrix}\text{a}+\text{b}&\text{c}\\\text{a}+\text{c}&\text{b}\end{vmatrix}\right\}$
$[\because(\text{c}-\text{a})=-(\text{a}-\text{c})]$ [Expanding along C1]
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})(\text{ab}+\text{b}^2-\text{ac}-\text{c}^2)$
$=-(\text{a}^2+\text{b}^2+\text{c}^2)(\text{a}-\text{b})(\text{c}-\text{a})\{\text{a}(\text{b}-\text{c})+(\text{b}+\text{c})(\text{b}-\text{c})\}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
Hence prove.
View full question & answer→Question 404 Marks
Evaluate:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{c}&\text{a}&\text{b}\\\text{b}&\text{c}&\text{a}\end{vmatrix}$
Applying C1 → C1 + C2 + C3 we get,
$\triangle=\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{b}&\text{c}\\\text{a}+\text{b}+\text{c}&\text{a}&\text{b}\\\text{a}+\text{b}+\text{c}&\text{c}&\text{a}\end{vmatrix}$
Taking (a + b + c) common, we have
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\1&\text{a}&\text{b}\\1&\text{c}&\text{a}\end{vmatrix}$
Applying R2 → R2 - R1, R3 - R1, we get
$\triangle=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&\text{b}&\text{c}\\0&\text{a}-\text{b}&\text{b}-\text{c}\\0&\text{c}-\text{b}&\text{a}-\text{c}\end{vmatrix}$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})[(\text{a}-\text{b})(\text{a}-\text{c})-(\text{b}-\text{c})(\text{c}-\text{b})]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2-\text{ac}-\text{ab}+\text{bc}+\text{b}^2+\text{c}^2-2\text{ab}\big]$
$\Rightarrow\triangle=(\text{a}+\text{b}+\text{c})\big[\text{a}^2+\text{b}^2+\text{c}^2-\text{ac}-\text{ab}-\text{bc}\big]$
View full question & answer→Question 414 Marks
Prove that:
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
Answer$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
$\text{L.H.S}=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}$
Apply R3 → R3 - R2
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)2&1&0\end{vmatrix}$
Apply R2 → R2 - R1
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)2&1&0\$\text{a}+3)2&1&0\end{vmatrix}$
$=[(2\text{a}+4)(1)-(1)(2\text{a}+6)]$
$=-2$
$=\text{R.H.S}$
View full question & answer→Question 424 Marks
Evaluate:
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
When a = b, the first two rows become identical. Hence, a - b is a factor. Similarly, when b = c and c = a, the second and third and third and first rows become indetical. Hence, b - c and c - a are also factors. The degree of product of the diagonal elements is 3. Hence, there are no other factors.
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=\lambda(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$ [Where $\lambda$ is a constant]
$\begin{vmatrix}1&0&2\\1&1&0\\1&2&0\end{vmatrix}=2\lambda$ $[$Putting a = 0, b = 1 and c = 2 to find $\lambda]$
$\Rightarrow2=2\lambda$
$\Rightarrow\lambda=1$
Hence,
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
View full question & answer→Question 434 Marks
Prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+2\text{p}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
Answer$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+\begin{vmatrix}1&1&\text{p}\\2&3&3\text{p}\\3&6&6\text{p}\end{vmatrix}+(\text{pq})\begin{vmatrix}1&1&1\\2&2&2\\3&3&3\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+(\text{p})\begin{vmatrix}1&1&\text{p}\\2&3&3\\3&6&6\end{vmatrix}+0$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+0$
$[\because$ Value of determinant with two identical columns is zero$]$
$=\begin{vmatrix}1&0&0\\2&1&2\\3&3&7\end{vmatrix}$ [Applying C2 → C2 - C1 and C3 → C3 - C1]
$=\left\{1\times\begin{vmatrix}1&2\\3&7\end{vmatrix}\right\}$ [Expanding along R1]
$=7-6$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 444 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
Answer$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos\alpha\cos\delta&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos\beta\cos\delta&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos\gamma\cos\delta&\cos(\gamma+\delta)\end{vmatrix}$$[\text{Applying} \text{ C}_1\rightarrow\sin\delta\text{ C}_1\text{ and}\text{ C}_2\rightarrow\cos\delta\text{ C}_2]$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos(\alpha+\delta)&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos(\beta+\delta)&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos(\gamma+\delta)&\cos(\gamma+\delta)\end{vmatrix}$ [Applying C2 → C2 - C1]
$=0$
View full question & answer→Question 454 Marks
Prove the following identities:
$\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(5\text{x}+\lambda)(\lambda-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}-\text{x}-\lambda&\text{x}+\lambda-2\text{x}&0\\2\text{x}-\text{x}-\lambda&0&\text{x}+\lambda-2\text{x}\end{vmatrix}$ [Applying R2 → R2 - R1 and R3 → R3 - R1]
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-(\lambda-\text{x})&\lambda-\text{x}&0\\-(\lambda-\text{x})&0&\lambda-\text{x}\end{vmatrix}$
$=(\lambda-\text{x})^2\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-1&1&0\\-1&0&1\end{vmatrix}$ [Taking $(\lambda-\text{x})$ common from R2 and $(\lambda-\text{x})$ common from R3]
$=(\lambda-\text{x})^2[-1(-2\text{x})+1(\text{x}+\lambda+2\text{x})]$ [Expanding along last row]
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
$=\text{R.H.S}$
$\because\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
View full question & answer→Question 464 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}=2\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)\begin{vmatrix}\text{a}&\text{c}&\text{b}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{b}&\text{a} \end{vmatrix}$
$[$Applying $\text{C}_1\leftrightarrow\text{C}_3$ in second determinant to get negative value of the deteminant$]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)(-1)\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$ $[$Applying $\text{C}_2\leftrightarrow\text{C}_3]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}=\text{R.H.S}$
View full question & answer→Question 474 Marks
Solve the following systems of linear equations by cramer's rule:
3x + ay = 4,
2x + ay = 2, $\text{a}\neq0$
AnswerGiven, 3x + ay = 4
2x + ay = 2
Using Cramer's rule, we get
$\text{D}=\begin{vmatrix}3&\text{a}\\2&\text{a}\end{vmatrix}=3\text{a}-2\text{a}=\text{a}$
$\text{D}_1=\begin{vmatrix}4&\text{a}\\2&\text{a}\end{vmatrix}=4\text{a}-2\text{a}=2\text{a}$
$\text{D}_2=\begin{vmatrix}3&4\\2&2\end{vmatrix}=6-8=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{2\text{a}}{\text{a}}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{\text{a}}=-\frac{2}{\text{a}}$
$\therefore\text{x}=2$ and $\text{y}=-\frac{2}{\text{a}}$
View full question & answer→Question 484 Marks
Using properties of determinants prove that:
$\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=(5\text{x}+4)(4-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+4&2\text{x}&2\text{x}\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}5\text{x}+4&5\text{x}+4&5\text{x}+4\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Applying R1 → R1 + R2 + R3]
$=5\text{x}+4\begin{vmatrix}1&1&1\\2\text{x}&\text{x}+4&2\text{x}\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Take out 5x + 4 common from R1]
$=5\text{x}+4\begin{vmatrix}1&0&0\\2\text{x}&4-\text{x}&0\\2\text{x}&2\text{x}&\text{x}+4\end{vmatrix}$ [Applying C2 → C2 - C1 and C3 → C3 - C1]
$=5\text{x}+4(4-\text{x})^2$
View full question & answer→Question 494 Marks
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
AnswerGiven,
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
$\Rightarrow\triangle=(-1)^{1+1}\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)\\+(-1)^{1+2}\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\+(-1)^{1+3}(-\sin\alpha)(-\sin^2\beta\sin\alpha-\sin\alpha\cos^2\beta)$ [Expanding along R1]
$=\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)-\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\-\sin\alpha(\sin^2\beta\sin\alpha-\sin\alpha-\sin\alpha\cos^2\beta)$
$=\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha\sin^2\beta+\sin^2\alpha\cos^2\beta$
$=\cos^2\alpha(\cos^2\beta+\sin^\beta)+\sin^2\alpha(\sin^2\beta+\cos^2\beta)$
$\Rightarrow\triangle=\cos^2\alpha+\sin^2\alpha$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\triangle=1$
View full question & answer→Question 504 Marks
$\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}+\text{c}&-\text{c}&-\text{b}\\-\text{c}&\text{a}+\text{b}+\text{c}&-\text{a}\\-\text{b}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&-\text{c}&-\text{b}\\\text{b}&\text{a}+\text{b}+\text{c}&-\text{a}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying C1 → C1 + C2 + C3]
$=\begin{vmatrix}\text{a}+\text{b}&\text{a}+\text{b}&-(\text{a}+\text{b})\\\text{b}+\text{c}&\text{b}+\text{c}&\text{b}+\text{c}\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying R1 → R1 + R2 and R2 → R2 + R3]
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}1&1&-1\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$
[Taking out common factor from R1 and R2]
$=(\text{a}+\text{b})(\text{b}+\text{c})\begin{vmatrix}0&0&-2\\1&1&1\\\text{c}&-\text{a}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying R1 → R1 - R2]
$=(\text{a}+\text{b})(\text{b}+\text{c})\{(-2)(-\text{a}-\text{c})\}$ [Expanding along R1]
$=2(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
$=\text{R.H.S}$
Hence proved.
View full question & answer→