Question 12 Marks
Evaluate the following determinant:
$\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta \end{vmatrix}$
Answer$\triangle=\cos^2\theta-(-\sin^2\theta)$
$\triangle=\cos^2\theta+\sin^2\theta=1$
View full question & answer→Question 22 Marks
If A is symmetric matrix, write whether AT is symmetric or skew-symmetric.
AnswerFor any symmetric matrix, AT = A.
Hence, AT is also symmetric.
View full question & answer→Question 32 Marks
Prove that the determinant $\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$ is independent of θ.
Answer$\triangle=\begin{vmatrix}x&\sin\theta&\cos\theta\\-\sin\theta&-x&1\\\cos\theta&1&x\end{vmatrix}$
$= x(-x^2-1)-\sin\theta(-x\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+x\cos\theta)$
$=-x^3-x+x\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+x\cos^2\theta$
$=-x^3-x+x(\sin^2\theta+\cos^2\theta)$
$=-x^3-x+x$
$=-x^3$ (Independent of $\theta$)
Hence, $\triangle$ is independent of $\theta.$
View full question & answer→Question 42 Marks
Evaluate the determinants.
- $\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}$
- $\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}$
Answer -
$\begin{vmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{vmatrix}_{\ =\ \left(\cos\theta\right)\left(\cos\theta\right)\ -\ \left(-\sin\theta\right)\left(\sin\theta\right)\ =\ \cos^2\theta\ +\ \sin^2\theta\ =\ 1}$
-
$\begin{vmatrix}x^2-x+1&x-1\\x+1&x+1\end{vmatrix}$
$=\left(x^2-x+1\right)\left(x+1\right)-\left(x-1\right)\left(x+1\right)$
$=x^3-x^2+x+x^2-x+1-\left(x^2-1\right)$
$=x^3+1-x^2+1$
$=x^3-x^2+2$
View full question & answer→Question 52 Marks
If $\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix},$ find the value of $|\text{A}|+|\text{B}|.$
Answer $\text{A}=\begin{bmatrix}0&\text{i}\\\text{i}&1\end{bmatrix}$ $\Rightarrow|\text{A}|= 0-\text{i}^2$
$\Rightarrow|\text{A}|=-(-1)=1$
Also,
$\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\Rightarrow|\text{B}|=0-1=-1$
So,
$\Rightarrow|\text{A}|+|\text{B}|=1-1=0$
View full question & answer→Question 62 Marks
If A is a non-singular symmetric matrix, write whether A-1 is symmetric or skew-symmetric.
AnswerLet A be an invertible symmetric matrix. Then,
$|\text{A}|\neq0\text{ and }\text{A}^\text{T}=\text{A}$
Now, (A-1)T = (AT)-1
⇒ (A-1)T = A-1 [$\because$ AT = A]
Thus, A-1 is symmetric matrix.
View full question & answer→Question 72 Marks
Evaluate the determinant: $ \begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}$
Answer Let $\text{A}=\begin{vmatrix}3&-4&5\\1&1&-2\\2&3&1\end{vmatrix}.$ By expanding along the first row, we have: $|\text{A}|=3\begin{vmatrix}1&-2\\3&1\end{vmatrix}+4\begin{vmatrix}1&-2\\2&1\end{vmatrix}+5\begin{vmatrix}1&1\\2&3\end{vmatrix}$ $=3(1+6)+4(1+4)+5(3-2)$
$=3(7)+4(5)+5(1)$
$=21+20+5=46$
View full question & answer→Question 82 Marks
If A is a square matrix of order 3 such that adj (2A) = k adj (A), then write the value of k.
AnswerFor any matrix A of order n, adj $(\lambda\text{A})=\lambda^{\text{n}-1}=\lambda^{\text{n}-1}$ (adj A) where $\lambda$ is a constant.
Thus, for matrix A of order 3, we have
adj (2A) = 23-1 (adj A)
⇒ adj (2A) = 22 (adj A)
⇒ adj (2A) = 4(adj) (A)
⇒ k adj (A) = 4 adj (A) [$\because$ adj (2A) = k adj (A)]
⇒ k = 4
View full question & answer→Question 92 Marks
If A is a non-singular square matrix such that |A| = 10, find |A-1|.
Answer$\big|\text{A}^{-1}\big|=\Big|\frac{1}{\text{A}}\Big|$
$=\Big|\frac{1}{\text{A}}\Big|$
$=\frac{1}{10}\ \big[\because|\text{A}|=10\text{ (Given})\big]$
Hence, $\big|\text{A}^{-1}\big|=\frac{1}{10}$
View full question & answer→Question 102 Marks
Show that points:
A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
AnswerArea of triangle ABC = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}a&b+c&1\\b&c+a&1\\c&a+b&1\end{bmatrix}\end{vmatrix}$$=\bigg|\frac{1}{2}\left[a(c+a-a-b)-(b+c)(b-c)+1\left\{b(a+b)-c(c+a)\right\}\right]\bigg|$
$=\bigg|\frac{1}{2}\left[a(c-b)-(b^2-c^2)+(ab+b^2-c^2-ac)\right]\bigg|$
$=\bigg|\frac{1}{2}(ac-ab-b^2+c^2+ab+b^2-c^2-ac)\bigg|$
$=\begin{vmatrix}\frac{1}{2}\times0\end{vmatrix}=0$
Therefore, points A, B and C are collinear.
View full question & answer→Question 112 Marks
If A is an invertible matrix such that |A-1| = 2, find the value of |A|.
Answer$\big|\text{A}^{-1}\big|=2$
$\Big|\frac{1}{\text{A}}\Big|=2$
$\frac{1}{|\text{A}|}=2$
$\therefore|\text{A}|=\frac{1}{2}$
Hence, $|\text{A}|=\frac{1}{2}$
View full question & answer→Question 122 Marks
If $\text{A}=\begin{bmatrix}2&4\\4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{n}\\1\end{bmatrix},\text{B}=\begin{bmatrix}8\\11\end{bmatrix}$and AX = B, then find n.
AnswerHere,
$\begin{bmatrix}2&4\\4&3\end{bmatrix}\begin{bmatrix}\text{n}\\1\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{n}+4\\4\text{n}+3\end{bmatrix}=\begin{bmatrix}8\\11\end{bmatrix}$
$\Rightarrow2\text{n}+4=8$
$\Rightarrow2\text{n}=4$
$\Rightarrow\text{n}=2$
View full question & answer→Question 132 Marks
If $\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0,$ find x.
Answer$\begin{vmatrix}2\text{x}+5&3\\5\text{x}+2&9\end{vmatrix}=0$
$\Rightarrow9(2\text{x}+5)-3(5\text{x}+2)=0$
$\Rightarrow18\text{x}+45-15\text{x}-6=0$
$\Rightarrow3\text{x}+39=0$
$\Rightarrow3\text{x}=-39$
$\Rightarrow\text{x}=\frac{-39}{3}$
$\Rightarrow\text{x}=-13$
View full question & answer→Question 142 Marks
If A is a square matrix of order 3 with determinant 4, then write the value of |-A|.
Answer|A| = 4
Here,
Order of the matrix (n) = 3
Using properties of matrices, we get
|kA| = kn|A| [For a square matrix of order n and constant k]
⇒ |-A| = (-1)3 |A| = (-1) × 4 = -4
View full question & answer→Question 152 Marks
Examine the consistency of the system of equations:
2x - y = 5
x + y = 4
AnswerMatrix form of given equations is AX = B $\Rightarrow \ \begin{bmatrix}2&-1\\1&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\4\end{bmatrix}$
$\therefore\ \text{A}=\begin{bmatrix}2&-1\\1&1\end{bmatrix}\text{and B}=\begin{bmatrix}5\\4\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}2&-1\\1&1\end{vmatrix}=2-(-1)=3\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 162 Marks
If A is a square matrix of order 3 such that |adj A| = 64, find |A|.
AnswerFor any square matrix of order n,
|adj A| = |A|n-1
⇒ 64 = |A|2 [$\because$ |adj A| = 64]
$\Rightarrow|\text{A}|=\pm8$
View full question & answer→Question 172 Marks
If $\text{x}\in\text{N}$ and $\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8,$ then find the value of x.
Answer$\begin{vmatrix}\text{x}+3&-2\\-3\text{x}&2\text{x} \end{vmatrix}=8$
$\Rightarrow(\text{x}+3)2\text{x}-(-2)(-3\text{x})=8$
$\Rightarrow2\text{x}^2+6\text{x}-6\text{x}=8$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2-4=0$
$\Rightarrow\text{x}^2=4$
$\Rightarrow\text{x}=2$ $[\text{x}\neq-2\ \because\text{x}\in\text{N}]$
View full question & answer→Question 182 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}8&2&7\\12&3&5\\16&4&3 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}0&2&7\\12&3&5\\16&4&3 \end{vmatrix}$ [Applying C1 → C1 - 4C2]
$\Rightarrow\triangle=0$
View full question & answer→Question 192 Marks
Let A = [aij] be a square matrix of order 3 × 3 and Cij denote cofactor of aij in A. if |A| = 5, write the value of a13C13 + a23C23 + a33C33.
AnswerIf A = aij is a square matrix of order n and Cij is a cofactor of aij, then
$\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given, |A| = 5 and matrix A is of order 3 × 3
Since a13C13 + a23C23 + a33C33 represent expansion of A along third column, we get
⇒ a13C13 + a23C23 + a33C33 = |A| = 5
⇒ a13C13 + a23C23 + a33C33 = 5
View full question & answer→Question 202 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x + 2y = 5,
3x + 6y = 15
AnswerUsing the equations, we get
$\text{D}=\begin{vmatrix}1&2\\3&6\end{vmatrix}=6-6=0$
$\text{D}_1=\begin{vmatrix}5&2\\15&6\end{vmatrix}=30-30=0$
$\text{D}_2=\begin{vmatrix}1&5\\3&15\end{vmatrix}=15-15=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2$
Hence, the system of linear equation has infinitely many solutions.
View full question & answer→Question 212 Marks
If $\text{A}=\begin{bmatrix}1&2\\4&2\end{bmatrix}$, then show that $\left|2\text{A}\right|=4\left|\text{A}\right|$
AnswerThe given matrix is $\text{A}=\begin{bmatrix}1&2\\4&2\end{bmatrix}$
$\therefore2\text{A}=2\begin{bmatrix}1&2\\4&2\end{bmatrix}=\begin{bmatrix}2&4\\8&4\end{bmatrix}$
$\therefore\text{L.H.S.}=|2\text{A}|=\begin{vmatrix}2&4\\8&4\end{vmatrix}=2\times4-4\times8=8-32=-24$
$\text{Now},|\text{A}|=\begin{vmatrix}1&2\\4&2\end{vmatrix}=2-8=-6$
$\therefore\text{R.H.S.}=4|\text{A}|=4\times\left(-6\right)=-24$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
View full question & answer→Question 222 Marks
Write the adjoint of the matrix $\text{A}=\begin{bmatrix} -3 & 4 \\ 7 & -2 \end{bmatrix}$.
AnswerLet Cij be a cofactor of aij in A.
Now,
C11 = -2
C12 = -7
C21 = -4
C22 = -3
$\therefore\ \text{adj A}=\begin{bmatrix} -2 & -7 \\ -4 & -3 \end{bmatrix}^\text{T}=\begin{bmatrix} -2 & -4 \\ -7 & -3 \end{bmatrix}$
View full question & answer→Question 232 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&3&5\\2&6&10\\31&11&38\end{vmatrix}$
$=1\begin{vmatrix}6&10\\11&38\end{vmatrix}-3\begin{vmatrix}2&10\\31&38\end{vmatrix}+5\begin{vmatrix}2&6\\31&11\end{vmatrix}$
$=(228-110)-3(76-310)+5(22-186)$
$=1(118)-3(-234)+5(-164)$
$=118+702-820$
$=0$
View full question & answer→Question 242 Marks
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}1&bc&a(b+c)\\1&ca&b(c+a)\\1&ab&c(a+b)\end{vmatrix}=0$
Answer$\text{Given}:\ \begin{vmatrix}1&bc&a(b+c)\\1&ca&b(c+a)\\1&ab&c(a+b)\end{vmatrix}=\begin{vmatrix}1&bc&ab+ac\\1&ca&bc+ba\\1&ab&ca+cb\end{vmatrix}$
$\text{Operating}\ \text{C}_3\rightarrow\text{C}_3+\text{C}_2\ \begin{vmatrix}1&bc&ab+bc+ac\\1&ca&ab+bc+ca\\1&ab&ab+bc+ca\end{vmatrix}$
$=(ab+bc+ca)\begin{vmatrix}1&bc&1\\1&ca&1\\1&ab&1\end{vmatrix}$
$=(ab+bc+ca)(0)=0$ $\left[\because\text{two columns are identical Proved.}\right]$
View full question & answer→Question 252 Marks
If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.
AnswerSince A & B are square matrix of the same order, by the property of determinants we get
|AB| = |A| × |B|
|A| = 3, AB = I
⇒ |AB| = 1
⇒ |A| × |B| = 1
⇒ 3 × |B| = 1
$\Rightarrow|\text{B}|=\frac{1}{3}$
View full question & answer→Question 262 Marks
Write the value of a11C21 + a12C22 + a13C23.
AnswerWe know that in a square matrix of order n, the sum of the products of elements of a row (or a column) with the cofactors of the corresponding elements of some other row (or column) is zero. Therefore,
A = [aij] is a square matrix of order n.
$\Rightarrow\sum\limits_{\text{n}}^{\text{i}=1}\text{a}_{\text{ij}}\text{C}_\text{kj}=0$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ik}=0$
⇒ a11C21 + a12C22 + a13C23 = 0
[Since the elements are of first row and the cofactors are of elements of second row]
⇒ a11C21 + a12C22 + a13C23 = 0
View full question & answer→Question 272 Marks
Find the value of the determinant $\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}4200&1201\\4205&4203\end{vmatrix}$
$\triangle=\begin{vmatrix}4200&1\\4205&1\end{vmatrix}$ [Applying C2 → C2 - C1]
$\triangle=4200 - 4202$
$\triangle=-2$
View full question & answer→Question 282 Marks
If A = [Aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.
AnswerA scalar matrix is a digonal matrix, in which all the diagonal elements are equal to a given scalar number.
Given, A = [aij] is 3 × 3 matrix, where a11 = 2
$\Rightarrow\text{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$\Rightarrow\text{A}=\begin{vmatrix}2&0&0\\0&2&0\\0&0&2\end{vmatrix}$
$\Rightarrow|\text{A}|=2\times\begin{vmatrix}2&0\\0&2\end{vmatrix}$ [Expanding along C1]
$\Rightarrow|\text{A}|=2\times2\times2$
$\Rightarrow|\text{A}|=8$
View full question & answer→Question 292 Marks
Evaluate the following determinant:
$\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c}\end{vmatrix}$
$=\text{a}\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}-\text{h}\begin{vmatrix}\text{h}&\text{f}\\\text{g}&\text{c} \end{vmatrix}+\text{g}\begin{vmatrix}\text{h}&\text{b}\\\text{g}&\text{f} \end{vmatrix}$
$=\big(\text{bc}-\text{f}^2\big)-\text{h}\big(\text{hc}-\text{fg}\big)+\text{g}\big(\text{hf}-\text{gb}\big)$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{g}^2\text{b}$
$=\text{abc}+2\text{fgh}-\text{af}^2-\text{ch}^2-\text{bg}^2$
View full question & answer→Question 302 Marks
If A is a square matrix of order 3 such that |A| = 5, write the value of |adj A|.
AnswerFor any square matrix of order n,
|adj A| = |A|n-1
⇒ |adj A| = |A|2 = 52 = 25
View full question & answer→Question 312 Marks
If $\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix},$ find |AB|.
Answer$\Rightarrow\text{A}=\begin{bmatrix}1&2\\3&-1 \end{bmatrix}$
⇒ |A| = -1 - 6 = -7
$\Rightarrow\text{B}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
⇒ |B| = -2 + 12 = 10
If A and B are square matrix of the same order, then |AB| = |A| |B|.
⇒ |AB| = |A| |B|
⇒ |AB| = -7 × 10 = -70
View full question & answer→Question 322 Marks
If $\text{A}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},\text{B}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ find adj (AB).
Answer$\text{A}\times\text{B}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$
A × B is a non-singular matrix. Therefore, it is invertible.
Let Cij be a cofactor of aij in A.
The cofactors of element A are given by
C11 = d
C12 = -c
C21 = -b
C22 = a
$\therefore\ \text{adj A}=\begin{bmatrix} \text{d} & -\text{c} \\ -\text{b} & \text{a} \end{bmatrix}^\text{T}=\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
View full question & answer→Question 332 Marks
Find the maximum value of $\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}1&1&1\\1&1+\sin\theta&1\\1&1&1+\cos\theta \end{vmatrix}$
Applying R2 → R2 - R1 and R3 → R3 - R1 we get
$\triangle=\begin{vmatrix}1&1&1\\0&\sin\theta&0\\0&0&\cos\theta \end{vmatrix}$
$=\sin\theta\cos\theta$
$=\frac{\sin2\theta}{2}$
We know that $-1\leq\sin2\theta\leq1$
$\therefore$ Maximum value of $\triangle=\frac{1}{2}\times1=\frac{1}{2}$
View full question & answer→Question 342 Marks
Write the value of $\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
Answer$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
$=\text{a}^2-\text{iab}+\text{iab}-\text{i}^2\text{b}^2-(-\text{c}^2-\text{icd}+\text{icd}+\text{i}^2\text{d}^2)$
$=\text{a}^2-\text{i}^2\text{b}^2+\text{c}^2-\text{i}^2\text{d}^2$
Here, $\text{i}^2=-1$
$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}=\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2$
View full question & answer→Question 352 Marks
If w is an imaginary cube root of unity, find the value of $\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
Answer$\begin{vmatrix}1&\text{w}&\text{w}^2\\\text{w}&\text{w}^2&1\\\text{w}^2&1&\text{w}\end{vmatrix}$
$=\begin{vmatrix}1+\text{w}+\text{w}^2&\text{w}&\text{w}^2\\\text{w}+\text{w}^2+1&\text{w}^2&1\\\text{w}^2+1+\text{w}&1&\text{w}\end{vmatrix}$ [Applying C1 → C1 + C2 + C3]
$=\begin{vmatrix}0&\text{w}&\text{w}^2\\0&\text{w}^2&1\\0&1&\text{w}\end{vmatrix}$ $[\because$ 1 + w + w2 = 0, w is the imaginary cube root of unity$]$
View full question & answer→Question 362 Marks
Write the value of the determinant $\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
Answer$\begin{vmatrix}2&3&4\\5&6&8\\6\text{x}&9\text{x}&12\text{x}\end{vmatrix}$
$=\begin{vmatrix}2&3&4\\5&6&8\\2&3&4\end{vmatrix}$ [Taking 2x common from R3]
$=0$
View full question & answer→Question 372 Marks
Find area of the triangle with vertices at the point given: (2, 7), (1, 1), (10, 8)
AnswerArea of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}2&7&1\\1&1&1\\10&8&1\end{bmatrix}\end{vmatrix}$ $=\bigg|\frac{1}{2}\left[2(1-8)-7(1-10)+1(8-10)\right]\bigg|$
$=\bigg|\frac{1}{2}\left[2(-7)-7(-9)-2\right]\bigg|$
$=\bigg|\frac{1}{2}(-14+63-2)\bigg|=\bigg|\frac{1}{2}(63-16)\bigg|$
$=\begin{vmatrix}\frac{47}{2}\end{vmatrix}=\frac{47}{2}\text{sq.units}$
View full question & answer→Question 382 Marks
Find the value of x from the following: $\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
Answer$\begin{vmatrix}\text{x}&4\\2&2\text{x}\end{vmatrix}=0$
$\Rightarrow2\text{x}^2-8=0$
$\Rightarrow2\text{x}^2=8$
$\Rightarrow\text{x}^2=\frac{8}{2}=4$
$\Rightarrow\text{x}=\sqrt{4}=\pm2$
View full question & answer→Question 392 Marks
If A is a square matrix, then write the matrix adj (AT) − (adj A)T.
AnswerIn a non-singular matrix, adj AT = (adj A)T.
⇒ (adj AT) - (adj A)T = Null matrix
View full question & answer→Question 402 Marks
Evaluate: $\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
Answer$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}$
$=\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ$
$=\cos(15^\circ+75^\circ)$ $\big[\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}=\cos(\text{A}+\text{B})\big]$
$=\cos90^\circ$
$=0$
$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ\end{vmatrix}=0$
View full question & answer→Question 412 Marks
Find the value of x, if:
if $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10,$ find the value of x.
AnswerGiven, $\begin{vmatrix}3\text{x}&7\\2&4\end{vmatrix}=10$
$\Rightarrow12\text{x}-14=10$
$\Rightarrow12\text{x}=24$
$\Rightarrow\text{x}=2$
View full question & answer→Question 422 Marks
If the matrix $\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$ is a singular, find the value of x.
AnswerA matrix is said to be singular if its determinant is zero. since the given matrix is singular, we get
$\text{A}=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=\begin{bmatrix}5\text{x}&2\\-10&1\end{bmatrix}$
$\Rightarrow|\text{A}|=0$
$\Rightarrow5\text{x}+20=0$ [Expanding]
$\Rightarrow\text{x}=-\frac{20}{5}$
$\Rightarrow\text{x}=-4$
View full question & answer→Question 432 Marks
Examine the consistency of the system of equations:
x + 3y = 5
2x + 6y = 8
AnswerMatrix from of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&3\\2&6\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}5\\8\end{bmatrix}$
$\therefore\ \text{A}=\begin{bmatrix}1&3\\2&6\end{bmatrix}\text{and B}=\begin{bmatrix}5\\8\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&3\\2&6\end{vmatrix}=6-6=0$
$\text{Now},\ \text{(adj. A)B}=\begin{bmatrix}6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8\end{bmatrix}=\begin{bmatrix}33-24\\-10+8\end{bmatrix}=\begin{bmatrix}6\\-2\end{bmatrix}\neq0$
Therefore, given equations are inconsistent, i.e., have no common solution.
View full question & answer→Question 442 Marks
Find area of the triangle with vertices at the point given:
(1, 0), (6, 0), (4, 3)
AnswerArea of triangle = Modulus of $\frac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}\frac{1}{2}\begin{bmatrix}1&0&1\\6&0&1\\4&3&1\end{bmatrix}\end{vmatrix}$ $=\bigg|\frac{1}{2}\left[1(0-3)-0(6-4)+1(18-0)\right]\bigg|$
$=\bigg|\frac{1}{2}(-3+18)\bigg|=\bigg|\frac{15}{2}\bigg|=\frac{15}{2}\text{sq.units}$
View full question & answer→Question 452 Marks
On expanding by first row, the value of the determinant of 3 × 3 square matrix A = [aij] is a11 C11 + a12 C12 + a13 C13, where [Cij] is the cofactor of aij in A. Write the expression for its value on expanding by second column.
AnswerIf A = aij is a square matrix of order n, then the sum of the products of elements of a row (or a column) with their cofactors is always equal to det (A). Therefore,
$\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$ and $\sum\limits_{\text{i}=1}^{\text{n}}\text{a}_{\text{ij}}\text{C}_\text{ij}=|\text{A}|$
Given, |A| = a11C11 + a12C12 + a13C13 [Expanding along R1]
Now,
|A| = a12C12 + a22C22 + a32C32 [Expanding along R2] [a12, a22 and a32 are elements of C3]
View full question & answer→Question 462 Marks
State whether the matrix $\begin{vmatrix}2&3\\6&4\end{vmatrix}$ is singular or non-singular.
AnswerLet $\triangle=\begin{vmatrix}2&3\\6&4\end{vmatrix}$
= 2 × 4 - 6 × 3
= 18 - 18 = -10
A matrix is said to be singular if its determinant is equal to zero. Since $\triangle=-10\neq0,$ the given matrix is non-singular.
View full question & answer→Question 472 Marks
If $\text{A}=\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix},$ find $\left|\text{A}\right|.$
Answer$\text{Let A}=\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix}.$
By expanding along the first row, we have:
$\left|\text{A}\right|=1\begin{vmatrix}1&-3\\4&-9\end{vmatrix}-1\begin{vmatrix}2&-3\\5&-9\end{vmatrix}-2\begin{vmatrix}2&1\\5&4\end{vmatrix}$
$=1(-9+12)-1(-18+15)-2(8-5)$
$ =1(3)-1(-3)-2(3)$
$ =3+3-6$
$=6-6$
$=0$
View full question & answer→Question 482 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
AnswerLet $\triangle$ be the determinant.
$\triangle=\begin{vmatrix}1&4&9\\4&9&16\\9&16&25 \end{vmatrix}$
Applying R3 → R3 - R2, we get
$\Rightarrow\triangle=\begin{vmatrix}1&4&9-4\\4&9&16-9\\9&16&25-16 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&4&5\\4&9&7\\9&16&9\end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}1&5&5\\4&13&7\\9&25&9 \end{vmatrix}$ [Applying C2 → C1 + C2]
$\Rightarrow\triangle=\begin{vmatrix}1&0&0\\4&-7&-13\\9&-20&-36 \end{vmatrix}$ [Applying C2 → 5C1 - C2 and C3 → 5C1 - C3]
$\Rightarrow\triangle=1(7\times36-13\times20)$
$\Rightarrow\triangle=252-260=-8$
View full question & answer→Question 492 Marks
A matrix A of order 3 × 3 has determinant 5. What is the value of |3A|?
AnswerIf A = [aij] is a square matrix of order n and k is a constant, then
|kA| = kn|A|
Here,
Number of rows = n
k is a common factor from each row of k
|3A| = 33|A| = 27 × 5 = 135 [Given matrix is 3 × 3 such that |A| = 5]
Thus, |3A| = 135
View full question & answer→Question 502 Marks
Examine the consistency of the system of equations:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\4\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}1&1&1\\2&3&2\\a&a&2a\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}1&1&1\\2&3&2\\a&a&2a\end{vmatrix}$
$\Rightarrow\ \text{|A|}=1(6a-2a)-1(4a-2a)+1(2a-3a)=4a-2a-a=a\neq0$
Therefore, Unique solution and hence equations are consistent.
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