Question 1015 Marks
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
Answer$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$\text{L.H.S}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}$
$=\text{xyz}\begin{vmatrix}1&1&1\\\text{x}&\text{y}&\text{z}\\\text{x}^3&\text{y}^3&\text{z}^3\end{vmatrix}$
$=\text{xyz}\begin{vmatrix}0&1&0\\\text{x}-\text{y}&\text{y}&\text{z}-\text{y}\\\text{x}^3-\text{y}^3&\text{y}^3&\text{z}^3-\text{y}^3\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})\begin{vmatrix}0&1&0\\1&\text{y}&1\\\text{x}^2+\text{y}^2+\text{xy}&\text{y}^3&\text{z}^2+\text{y}^2+\text{zy}\end{vmatrix}$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[\text{z}^2+\text{y}^2+\text{zy}-\text{x}^2-\text{y}^2-\text{xy}]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[(\text{z}-\text{x})(\text{z}+\text{x})+\text{y}(\text{z}-\text{x})]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})(\text{z}-\text{x})[\text{z}+\text{x}+\text{y}]$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
View full question & answer→Question 1025 Marks
Prove that:
$\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
AnswerLet $\text{L.H.S}=\begin{vmatrix}1&\text{b}+\text{c}&\text{b}^2+\text{c}^2\\1&\text{c}+\text{a}&\text{c}^2+\text{a}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2 \end{vmatrix}$
$=\begin{vmatrix}0&(\text{b}+\text{c})-(\text{c}+\text{a})&(\text{b}^2+\text{c}^2)-(\text{c}^2+\text{a}^2)\\0&(\text{c}+\text{a})-(\text{a}+\text{b})&(\text{c}^2+\text{a}^2)-(\text{a}^2+\text{b}^2)\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix} [$Applying $R_1 → R_1 - R_2$ and $R_2 → R_2 - R_3]$
$=\begin{vmatrix}0&\text{b}-\text{a}&\text{b}^2-\text{a}^2\\0&\text{c}-\text{b}&\text{c}^2-\text{b}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(-1)^2\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{b}^2\\0&\text{b}-\text{c}&\text{b}^2-\text{c}^2\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix} [$Taking out $(-1)$ common from $R_1$ and $R_2]$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&\text{a}+\text{b}\\0&1&\text{b}+\text{c}\\1&\text{a}+\text{b}&\text{a}^2+\text{b}^2\end{vmatrix}$
$=(\text{a}+\text{b})(\text{b}-\text{c})\left\{1\times\begin{vmatrix} 1 & \text{a}+\text{b} \\ 1 & \text{b}+\text{c} \end{vmatrix}\right\}$
$=(\text{a}+\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
$=\text{R.H.S}$
View full question & answer→Question 1035 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c} \end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}=\text{bc}-\text{f}^2$
$\text{M}_{21}=\begin{vmatrix}\text{h}&\text{g}\\\text{f}&\text{c} \end{vmatrix}=\text{hc}-\text{fg}$
$\text{M}_{31}=\begin{vmatrix}\text{h}&\text{g}\\\text{b}&\text{f} \end{vmatrix}=\text{hf}-\text{gb}$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=\text{bc}-\text{f}^2$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{11}=-(\text{hc}-\text{fg})=\text{fg}-\text{hc}$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=\text{hf}-\text{gb}$
$\text{D}=\text{a}(\text{bc}-\text{f}^2)-\text{h}(\text{hc}-\text{fg})+\text{g}(\text{fh}-\text{bg})$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{bg}^2$
$=\text{abc}+2\text{hfg}-\text{af}^2-\text{bg}^2-\text{ch}^2$
View full question & answer→Question 1045 Marks
Prove the following identities:
$\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
$=\begin{vmatrix}\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{a}+\text{y}&\text{z}\\\text{a}+\text{x}+\text{y}+\text{z}&\text{y}&\text{a}+\text{z}\end{vmatrix} [$Applying $C_1 → C_1 + C_2 + C_3]$
$=(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\1&\text{a}+\text{y}&\text{z}\\1&\text{y}&\text{a}+\text{z}\end{vmatrix}$
$=(\text{a}+\text{x}+\text{y}+\text{z})\begin{vmatrix}1&\text{y}&\text{z}\\0&\text{a}&0\\0&0&\text{a}\end{vmatrix} [$Applying $R_2 → R_2 - R_1$ and $R_3 → R_2 - R_1]$
$=(\text{a}+\text{x}+\text{y}+\text{z})\text{a}^2$ [Expanding along first column]
$=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
$\therefore\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
View full question & answer→Question 1055 Marks
Prove that:
$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$\text{L.H.S}=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
Apply: $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&\text{b}^2+\text{ca}-\text{a}^2-\text{bc}&\text{b}^3-\text{a}^3\\0&\text{c}^2+\text{abb}+\text{ca}-\text{a}^2-\text{bc}&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}^2-\text{a}^2)-\text{c}(\text{b}-\text{a})&\text{b}^3-\text{a}^3\\0&(\text{c}^2-\text{a}^2)-\text{b}(\text{c}-\text{a})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}-\text{a})(\text{b}+\text{a}-\text{c})&\text{b}^3-\text{a}^3\\0&(\text{c}-\text{a})(\text{c}+\text{a}-\text{b})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}+\text{a}-\text{c})&\text{b}^2+\text{a}^2+\text{ab}\\0&(\text{c}+\text{a}-\text{b})&\text{c}^2+\text{a}^2+\text{ac} \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\big[(\text{b}+\text{a}-\text{c})(\text{c}^2+\text{a}^2+\text{ac})\\-(\text{b}^2+\text{a}^2+\text{ab})(\text{c}^2+\text{a}^2+\text{ac})\big]$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
View full question & answer→Question 1065 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
$=\begin{vmatrix}1&1&\text{x}\\\text{p}&\text{p}&\text{p}\\3&\text{x}+1&\text{x}+2\end{vmatrix} [$Applying $R_2→ R_2 - R_1]$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\3&\text{x}+1&\text{x}+2\end{vmatrix}$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\2&\text{x}&2\end{vmatrix}$
$=\text{p}\begin{vmatrix}0&1&\text{x}\\0&1&1\\2-\text{x}&\text{x}&2\end{vmatrix} [$Applying $C_1 → C_2 - C_1]$
$=\text{p}\left\{(2-\text{x})\times\begin{vmatrix}1&\text{x}\\1&1 \end{vmatrix}\right\}[$ Expanding along $C_1]$
$=\text{p}(2-\text{x})(1-\text{x})=0$
$\text{x}=1,2$
View full question & answer→Question 1075 Marks
Solve the following systems of linear equations by cramer's rule:
2x + 3y = 10,
x + 6y = 4
AnswerGiven, 2x + 3y = 10
x + 6y = 4
Using Cramer's Rule, we get
$\text{D}=\begin{vmatrix}2&3\\1&6\end{vmatrix}=12-3=9$
$\text{D}_1=\begin{vmatrix}10&3\\4&6\end{vmatrix}=60-12=48$
$\text{D}_2=\begin{vmatrix}2&10\\1&4\end{vmatrix}=8-10=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{48}{9}=\frac{16}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{9}$
$\therefore\text{x}=\frac{16}{3}$ and $\text{y}=\frac{-2}{9}$
View full question & answer→Question 1085 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^3&\text{abc}\\1&\text{b}^3&\text{abc}\\1&\text{c}^3&\text{abc}\end{vmatrix}$ [Applying $R_1 \rightarrow a R_1, R_2 \rightarrow b R_2$ and $R_3 \rightarrow c R_3$ ]
$=\text{abc}\begin{vmatrix}1&\text{a}^3&1\\1&\text{b}^3&1\\1&\text{c}^3&1 \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 1095 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 $R_1, R_2$ and $R_3$ 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 $C_1, C_2$ and $C_3$ 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 $R_1 → R_1 + R_2 + R_3$
$=\begin{vmatrix}2(\text{b}^2+\text{c}^2)&2(\text{c}^2+\text{a}^2)&2(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}(\text{b}^2+\text{c}^2)&(\text{c}^2+\text{a}^2)&(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}\text{c}^2&0&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\big[\text{c}^2\{(\text{c}^2+\text{a}^2)(\text{a}^2+\text{b}^2)-\text{b}^2\text{c}^2\}+\text{a}^2\{\text{b}^2\text{c}^2-(\text{c}^2+\text{a}^2)\text{c}^2\}\big]$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 1105 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 1115 Marks
Show that the following systems of linear equations is inconsistent:
3x - y + 2z = 6,
2x - y + z = 2,
3x + 6y + 5z = 20
Answer$\text{D}=\begin{vmatrix}3&-1&2\\2&-1&1\\3&6&5\end{vmatrix}$
$=3(-5-6)+1(10-3)+2(12+3)=4$
Since D is non zero,
$\text{D}_1=\begin{vmatrix}6&-1&2\\2&-1&1\\20&6&5\end{vmatrix}$
$=6(-5-6)+1(10-20)+2(12+20)$
$=-66-10+64=-12$
$\text{D}_2=\begin{vmatrix}3&6&2\\2&2&1\\3&20&5\end{vmatrix}$
$=3(10-20)-6(10-3)+2(40-6)$
$=-30+42+68=-4$
$\text{D}_3=\begin{vmatrix}3&-1&6\\2&-1&2\\3&6&20\end{vmatrix}$
$=3(-20-12)+1(40-6)+6(12+3)$
$=-96+34+90=28$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-12}{4}=-3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4}{4}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{28}{4}=7$
$\therefore\text{x}=-3,\text{ y}=-1$ and $\text{z}=7$
View full question & answer→Question 1125 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 1135 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
$=\frac{\text{xyz}}{\text{xyz}}\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
$=\frac{1}{\text{xyz}}\begin{vmatrix}0&\text{xz}&\text{yz}\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}$
$=\frac{1}{\text{xyz}}\begin{vmatrix}-2\text{xy}&0&2\text{yz}\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}$ $[$Applying $R_1 → R_1 + R_2 + R_3]$
$=\frac{1}{\text{xyz}}\begin{vmatrix}0&0&0\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}=0 [$Applying $R_1 → R_1 - 2R_2]$
View full question & answer→Question 1145 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$\text{L.H.S}=\begin{vmatrix}2\text{y}&\text{y}-\text{z}-\text{x}&2\text{y}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$ $[R_1 = R_1 + R_2 + R_3]$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&1&1\\2\text{z}&2\text{z}&\text{z}-\text{x}-\text{y}\\\text{x}-\text{y}-\text{z}&2\text{x}&2\text{x}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&0&0\\2\text{z}&0&-\text{x}-\text{y}-\text{z}\\\text{x}-\text{y}-\text{z}&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\end{vmatrix}$ $[C_2 = C_2 - C_1, C_3 = C_3 - C_1]$
$=(\text{x}+\text{y}+\text{z})\big[1\{0+(\text{x}+\text{y}+\text{z})(\text{x}+\text{y}+\text{z})\}\big]$
$=(\text{x}+\text{y}+\text{z})^3$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 1155 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 1165 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Answer$\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Applying: $C_3 \rightarrow C_3-C_2, C_4 \rightarrow C_4-C_1$
$=\begin{vmatrix}1^1&2^2&3^2-2^2&4^2-1^2\\2^2&3^2&4^2-3^2&5^2-2^2\\3^3&4^2&5^2-4^2&6^2-3^2\\4^2&5^2&6^2-5^2&7^2-4^2 \end{vmatrix}$
$=\begin{vmatrix}1^1&2^2&5&15\\2^2&3^2&7&21\\3^3&4^2&9&27\\4^2&5^2&11&33 \end{vmatrix}$
Take 3 common from $C_4$
$=0$
$\because\text{C}_3=\text{C}_4$
View full question & answer→Question 1175 Marks
Evaluate the following:
$\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&\text{x}&\text{x}\\\text{y}&0&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$
$[$Taking $x^2$ common from from $C_1, y^2$ common from $C_2$ and $z^2$ common from $C_3]$
$=\text{x}^3\text{y}^3\text{z}^3\begin{vmatrix}0&0&1\\1&-1&1\\1&1&0\end{vmatrix}$
$[$Applying $C_2 → C_2 - C_3]$
$=\text{x}^3\text{y}^3\text{z}^3(1+1)$ [Expanding along first row]
$=2\text{x}^3\text{y}^3\text{z}^3$
$\therefore\ \triangle=2\text{x}^3\text{y}^3\text{z}^3$
View full question & answer→Question 1185 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
Answer$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\begin{vmatrix}\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}&5&\sqrt{10}\\3&\sqrt{15}&5\end{vmatrix}+\begin{vmatrix}\sqrt{23}&\sqrt{5}&\sqrt{5}\\\sqrt{46}&5&\sqrt{10}\\\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{15}&5\end{vmatrix}+\sqrt{23}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{2}&5&\sqrt{10}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{3}&5\end{vmatrix}+\sqrt{23}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&1\\\sqrt{2}&5&\sqrt{2}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=0+0$
$=0$
View full question & answer→Question 1195 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$
$=\begin{vmatrix}3\text{a}+3\text{b}&3\text{a}+3\text{b}&3\text{a}+3\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix} [$Applying $R_1 → R_2 + R_2 + R_3]$
$=3(\text{a}+\text{b})\begin{vmatrix}1&1&1\\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix} [$Taking out $3(a + b)$ common from $R_1]$
$=3(\text{a}+\text{b})\begin{vmatrix}0&0&1\\2\text{b}&-\text{b}&\text{a}+\text{b}\\-\text{b}&2\text{b}&\text{a} \end{vmatrix}$ $[$Applying $C_1 → C_1 - C_2$ and $C_2 → C_2 - C_3]$
$=3(\text{a}+\text{b})\text{b}^2\begin{vmatrix}0&0&1\\2&-1&\text{a}+\text{b}\\-1&2&\text{a} \end{vmatrix} [$Taking out $b$ common from $C_1$ and $C_2]$
$=3(\text{a}+\text{b})\text{b}^2\times3$
$=9(\text{a}+\text{b})\text{b}^2$
$=\text{R.H.S}$
View full question & answer→Question 1205 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 $R_1$]
$=2\text{a}^3\text{b}^3\text{c}^3$
View full question & answer→Question 1215 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, $2\beta=\alpha+\gamma$
$\text{L.H.S}=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\frac{\text{R}_1}{2}-\frac{\text{R}_3}{2}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\$\text{x}-2)-\frac{\text{x}-3}{2}-\frac{\text{x}-1}{2}&(\text{x}-3)-\frac{\text{x}-4}{2}-\frac{\text{x}-2}{2}&(\text{x}-\beta)-\frac{\text{x}-\alpha}{2}-\frac{\text{x}-\gamma}{2}\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\0&0&0\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$ $[\because2\beta=\alpha+\gamma]$
$=0$
$=\text{R.H.S}$
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