MCQ 11 Mark
If $A$ is a matrix of order $3$ and $|A| = 8$, then $\text{|adj A| =}$
Answer$|A| = d$
$|adj\ A| = |A|^{n-1}$
Here, $n = 3, |A| = 8$
$|adj\ A| = 8^2$
$|\text{adj A}|={(2^3)}^2=2^6$
View full question & answer→MCQ 21 Mark
If A is a singular matrix, then adj A is:
AnswerIf A is singular matrix then adjoint of A is also singular.
View full question & answer→MCQ 31 Mark
For non$-$singular square matrix $A, B$ and $C$ of the same order $(AB^{-1} C) =$
- A
$A^{-1} BC^{-1}$
- B
$C^{-1} B^{-1} A^{-1}$
- C
$CBA^{-1}$
- ✓
$C^{-1} BA^{-1}$
AnswerCorrect option: D. $C^{-1} BA^{-1}$
We know that $(AB)^{-1} = B^{-1} A^{-1}$
Hence, $(AB^{-1}C)^{-1} = C^{-1}BA^{-1}$
View full question & answer→MCQ 41 Mark
Let $\text{A}=\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$ and $X$ be a matrix such that $A = BX,$ then $X$ is equal to:
- ✓
$\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
- B
$\frac{1}{2}\begin{bmatrix} -2 & 4 \\ 3 & 5 \end{bmatrix}$
- C
$\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
- D
AnswerCorrect option: A. $\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
$A = BX$
$B^{-1}A = B^{-1}BX$
$X = B^{-1}A$
Using adjoint method of inverse
$\text{B}^{-1}=\frac{1}{2}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$
$\text{X}=\text{B}^{-1}\text{A}$
$\text{X}=\frac{1}{2}\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$
$\text{x}=\frac{1}{2}\begin{bmatrix} 2 & 4 \\ 3 & -5 \end{bmatrix}$
View full question & answer→MCQ 51 Mark
Let $\text{A}=\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ be such that $A^{-1} = kA$, then $k$ equals:
- A
$19$
- ✓
$\frac{1}{19}$
- C
$-19$
- D
$-\frac{1}{19}$
AnswerCorrect option: B. $\frac{1}{19}$
$\text{adj A}=\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
$|\text{A}|=-19$
$\therefore\ \text{A}^{-1}=-\frac{1}{|\text{A}|}\text{ adj A}$
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$
Now,
$A^{-1} = kA$
$\Rightarrow-\frac{1}{19}\begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}=\text{kA}$
$\Rightarrow\frac{1}{19}\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}=\text{kA}$
$\Rightarrow\frac{1}{19}\text{A}=\text{kA}$
$\Rightarrow\text{k}=\frac{1}{19}$
View full question & answer→MCQ 61 Mark
If a matrix $A$ is such that $3A^3 + 2A^2 + 5A + I = 0,$ then $A^{-1}$ equal to:
- A
$-(3A^2 + 2A + 5)$
- B
$3A^2 + 2A + 5$
- C
$3A^2 - 2A - 5$
- ✓
Answer$3A^3 + 2A^2 + 5A + I = 0$
$\Rightarrow 3A^{-1} A^3 + 2A^{-1}A^2 + 5A^{-1}A + A^{-1}I = A^{-1}0$
$\Rightarrow 3A^2 + 2A + 5I + A^{-1} = 0$
$\Rightarrow A^{-1} = -(3A^2 + 2A + 5I)$
View full question & answer→MCQ 71 Mark
If $A$ and $B$ are invertible matrices, which of the following statement is not correct.
- A
$\operatorname{adj} A=|A| A^{-1}$
- B
$\operatorname{det}\left(A^{-1}\right)=(\operatorname{det} A)^{-1}$
- ✓
$(A+B)^{-1}=A^{-1}+B^{-1}$
- D
$(A B)^{-1}=B^{-1} A^{-1}$
AnswerCorrect option: C. $(A+B)^{-1}=A^{-1}+B^{-1}$
We have$, \text{adj} A=|A| A^{-1}, \operatorname{det}\left(A^{-1}\right)=(\operatorname{det} A)^{-1}$ and $(A B)^{-1}=B^{-1} A^{-1}$ all are the properites of inverse of a matrix.
View full question & answer→MCQ 81 Mark
If $\text{A}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ \text{a} & \text{b} & 2 \end{bmatrix},$ then $aI + bA + 2 A^2$ equals:
Answer$\text{A}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 1 \\ \text{a} & \text{b} & 2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix} 1+\text{a} & \text{b} & 3 \\ \text{a} & \text{b} & 2 \\ 3\text{a} & 2\text{b} & \text{a}+\text{b}+4 \end{bmatrix}$
$\Rightarrow\text{aI}+\text{bA}+2\text{A}^2$
$=\begin{bmatrix} 3\text{a}+2+\text{b} & 2\text{b} & 6+\text{b} \\ 2\text{a} & \text{a}+2\text{b} & \text{b}+4 \\ \text{ab}6\text{a} & 6\text{b}+\text{b}^2 & 3\text{a}+4\text{b}+8 \end{bmatrix}$
View full question & answer→MCQ 91 Mark
If x, y, z are non-zero real numbers, then the inverse, then the inverse of the matrix $\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$, is:
- ✓
$\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
- B
$\text{xyz}\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
- C
$\frac{1}{\text{xyz}}\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}$
- D
$\frac{1}{\text{xyz}}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
AnswerCorrect option: A. $\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
A = IA
$\Rightarrow\begin{bmatrix}\text{x} & 0 & 0\\ 0 & \text{y} & 0 \\ 0 & 0 & \text{z}\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix}\text{x}^1 & 0 & 0\\ 0 & \text{y}^1 & 0 \\ 0 & 0 & \text{z}^1\end{bmatrix}\text{A}$
$\Big[\text{Applying R}_1=\frac{1}{\text{x}}\text{R}_1,\text{R}_2=\frac{1}{\text{y}}\text{R}_2\text{ and R}_3=\frac{1}{\text{z}}\text{R}_3\Big]$
$\Rightarrow\text{A}^{-1}=\begin{bmatrix}\text{x}^{-1} & 0 & 0\\ 0 & \text{y}^{-1} & 0 \\ 0 & 0 & \text{z}^{-1}\end{bmatrix}$
View full question & answer→MCQ 101 Mark
If $\text{A}=\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix},$ then $A^n =$
- ✓
$\text{A}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ if $n$ is an even natural number
- B
$\text{A}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ if $n$ is an odd natural number
- C
$\text{A}=\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix},$if $\text{n}\in\text{N}$
- D
AnswerCorrect option: A. $\text{A}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$ if $n$ is an even natural number
$\text{A}=\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=1$
If $n$ is an natural number.
View full question & answer→MCQ 111 Mark
If $A$ and $B$ are square matrices such that $B = -A^{-1} BA,$ then $(A + B)^2 =$
- A
$O$
- ✓
$A^2 + B^2$
- C
$A^2 + 2AB + B^2$
- D
$A + B$
AnswerCorrect option: B. $A^2 + B^2$
$B=-A^{-1} B A$
$\Rightarrow A B=-A A^{-1} B A$
$\Rightarrow A b=-I B A$
$\Rightarrow A B=-B A$
$\Rightarrow A B+B A=0 \ldots . .(i)$
Consider,
$(A+B)^2=A^2+A B+B A+B^2$
$(\because A B \neq B A)$
$(A+B)^2=A^2+O+B^2$
$(A+B)^2=A^2+B^2$
View full question & answer→MCQ 121 Mark
If $\text{A}=\begin{bmatrix} 3 & 4 \\ 2 & 4 \end{bmatrix},\text{B}=\begin{bmatrix} -2 & -2 \\ 0 & -1 \end{bmatrix}$ then $(A + B)^{-1} =$
AnswerWe have
$(\text{A}+\text{B})=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
$\therefore|\text{A}+\text{B}| = -1\neq0$
Thus, $(A + B)^{-1}$ exists.
Now,
$(\text{A}+\text{B})^\text{T}=\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$
Here,
$(\text{A}+\text{B})^\text{T}\neq-(\text{A}+\text{B})$
Hence, it is not a akew symmetric matrix.
We also know that $A^{-1} + B^{-1}$ is not the same as $(A + B)^{-1}$.
View full question & answer→MCQ 131 Mark
If $A$ is an invertible matrix of order $3,$ then which of the following is not true:
- A
$|\text{adj A}|=|\text{A}|^2$
- B
$(\text{A}^{-1})^{-1}=\text{A}$
- ✓
If $BA = CA,$ than $\text{B}\neq\text{C},$ where $B$ and $C$ are square matrices of order $3$
- D
$(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1},$ where $\text{B}\neq\big[\text{b}_{\text{ij}}\big]_{3\times3}$ and $|B|\neq0$
AnswerCorrect option: C. If $BA = CA,$ than $\text{B}\neq\text{C},$ where $B$ and $C$ are square matrices of order $3$
$BA = CA$
$\Rightarrow \text{BAA}^{-1} = \text{CAA}^{-1}$
$\Rightarrow BI = CI$
$\Rightarrow B = C$
Hence$, (c)$ is not correct.
View full question & answer→MCQ 141 Mark
If $A$ is a square matrix such that $A^2 = I,$ then $A^{-1}$ is equal to:
Answer$A^2 = I$
$A^{-1}A^2 = A^{-1}I$
$A = A^{-1}$
View full question & answer→MCQ 151 Mark
If $A$ is an invertible matrix, then which of the following is not true:
- ✓
$(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$
- B
$|\text{A}^{-1}|=|\text{A}|^{-1}$
- C
$(\text{A}^\text{T})^\text{-1}=(\text{A}^{-1})^\text{T}$
- D
$|\text{A}|\neq0$
AnswerCorrect option: A. $(\text{A}^2)^\text{-1}=(\text{A}^{-1})^2$
We have, $|A^{-1}| = |A|^{-1}, (AT)^{-1} = (A^{-1})^T$ and $|\text{A}|\neq0$ all are the properties of the inverse of a matrix $A.$
View full question & answer→MCQ 161 Mark
If $A$ is an invertible matrix, then det $(A^{-1})$ is equal to:
AnswerCorrect option: B. $\frac{1}{\text{det(A)}}$
We know that $\big|\text{A}^{-1}\big|=\frac{1}{|\text{A}|}$
View full question & answer→MCQ 171 Mark
If $\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix},$ then the value of $\text{|adj A|}$ is:
- A
$a^{27}$
- B
$a^9$
- ✓
$a^6$
- D
$a^2$
Answer$\text{A}=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}$
$\therefore|\text{A}|=\begin{bmatrix} \text{a} & 0 & 0 \\ 0 & \text{a} & 0 \\ 0 & 0 &\text{a} \end{bmatrix}=\text{a}^3\neq0$
and
$n = 3$
Thus, we have
$\text{|adj A|} = |A|^{n-1} = (a^3)^2 = a^6$.
View full question & answer→MCQ 181 Mark
If $A^5 = 0$ Such that $\text{A}^{\text{n}}\neq\text{I for }1\leq\text{n}\leq4,\text{ then}(\text{I}-\text{A})^{-1}$ equals:
Answer$A^5 = 0$
Using $a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$
$I - A^5 = (I - A)(I + A + A^2 + A^3 + A^4)$
$I = (I - A)(I + A + A^2 + A^3 + A^4)$
$(I - A)^{-1}I = (I - A)^{-1}(I - A)(I + A + A^2 + A^3 + A^4)$
$(I - A)-1 = I + A + A^2 + A^3 + A^4$
View full question & answer→MCQ 191 Mark
If $\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}-1=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix},$ then:
- A
$\text{a}=1,\text{b}=1$
- ✓
$\text{a}=\cos2\theta,\text{b}=\sin2\theta$
- C
$\text{a}=\sin2\theta,\text{b}=\cos2\theta$
- D
AnswerCorrect option: B. $\text{a}=\cos2\theta,\text{b}=\sin2\theta$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}^{-1}=\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{c} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} \frac{1-\tan^2\theta}{\sec^2\theta} & \frac{-2\tan\theta}{\sec^2\theta} \\ \frac{2\tan\theta}{\sec^2\theta} & \frac{1-\tan^2\theta}{\sec^2\theta} \end{bmatrix}=\begin{bmatrix}\text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
On comparing, we get
$\text{a}=\frac{1-\tan^2\theta}{\sec^2\theta}\text{ and b}=\frac{2\tan\theta}{\sec^2\theta}$
$\Rightarrow\text{a}=\cos^2\theta-\sin^2\theta\text{ and b}=2\sin\theta\cos\theta$
$\Rightarrow\text{a}=\cos2\theta\text{ and b}=\sin2\theta$
View full question & answer→MCQ 201 Mark
If $\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix},$ then $A^5 =$
Answer$\text{A}=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}=2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\text{A}=2\text{I}$
$\Rightarrow\text{A}^5=(2\text{I})^5$
$\Rightarrow\text{A}^5=16\times2\text{I}$
$\Rightarrow\text{A}^5=16\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\Rightarrow\text{A}^5=16\text{A}$
View full question & answer→MCQ 211 Mark
If for the matrix $A, A^3 = I$, than $A^{-1} =$
Answer$A^3 = I$
$\Rightarrow A^{-1}A^3 = A^{-1}I$
$\Rightarrow IA^2 = A^{-1}I$
$\Rightarrow A^2 = A^{-1}$
View full question & answer→MCQ 221 Mark
For any 2 × 2 matrix, if $\text{A(adj A)}=\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix},$ then |A| is equal to:
Answer$\text{A(adj A)}\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$
By definition, we have
A(adj A) = |A|I = (adj A)A (Where I is the identity matrix)
⇒ |A|I = A(adj A)
$\Rightarrow|\text{A}|\text{I}=10\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow|\text{A}|=10$
View full question & answer→MCQ 231 Mark
If $A, B$ are two $n \times n$ non $-$ singular matrices, then
- ✓
$AB$ is non $-$ singular.
- B
$AB$ is singular.
- C
$(AB)^{-1} A^{-1} B^{-1}$.
- D
$(AB)^{-1}$ does not exist.
AnswerCorrect option: A. $AB$ is non $-$ singular.
$A$ and $B$ are non $-$ singular matrices of order $n \times n.$
$\therefore|\text{A}|\neq0$ and $|\text{B}|\neq0\ .....(\text{i})$
$A$ and $B$ are of the same order, so $AB$ is defined and is of the same order.
Thus,
$|AB| = |A\|B|$
$\Rightarrow|\text{AB}|\neq0\ \big[\text{Using (1)}\big]$
Thus, $AB$ is non $-$ singular.
View full question & answer→MCQ 241 Mark
If $A^2 - A + I = 0,$ then the inverse of $A$ is :
- A
$A^{-2}$
- B
$A + I$
- ✓
$I - A$
- D
$A - I$
AnswerCorrect option: C. $I - A$
$A^2 - A + I = 0$
$\Rightarrow A^{-1}A^2 - A^{-1}A + A^{-1}I = A^{-1}0$
$\Rightarrow A - I + A^{-1} = 0$
$\Rightarrow A^{-1} = I - A$
View full question & answer→MCQ 251 Mark
The matrix $\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is a singular matrix, if the value of b is:
Answer$\begin{bmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{bmatrix}$ is singular matrix.
$\Rightarrow\begin{vmatrix} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & \text{b} \end{vmatrix}=0$
⇒ 5(-4b + 12) - 10(-2b + 6) + 3(4 - 4) = 0
⇒ -20b + 60 + 20b - 60 = 0
b does not exist.
View full question & answer→MCQ 261 Mark
If $A$ satisfies the equation $\text{x}^2-5\text{x}^2+4\text{x}+\lambda=0$ then $A^{-1}$ exists if :
- A
$\lambda\neq1$
- B
$\lambda\neq2$
- C
$\lambda\neq-1$
- ✓
$\lambda\neq0$
AnswerCorrect option: D. $\lambda\neq0$
A satisfies $\text{x}^3-5\text{x}^2+4\text{x}+\lambda=0$
$\Rightarrow\text{A}^3-5\text{A}^2+4\text{A}=-\lambda$
Assuming $A^{-1}$ exists, we get
$\text{A}^{-1}(\text{A}^3-5\text{A}^2+4\text{A})=-\lambda\text{A}^{-1}$
$\Rightarrow\text{A}^2-5\text{A}+4=-\text{A}^{-1}\lambda$
$\Rightarrow\text{A}-1=\frac{-(\text{A}^2-5\text{A}+4)}{\lambda}$
Thus, $A^{-1}$ exists if $\lambda\neq0$.
View full question & answer→MCQ 271 Mark
If $B$ is a non $-$ singular matrix and $A$ is a square matrix, then det $(B^{-1} AB)$ is equal to :
- A
Det $(A^{-1})$
- B
Det $(B^{-1})$
- ✓
Det $(A)$
- D
Det $(B)$
AnswerCorrect option: C. Det $(A)$
$B$ is non $-$ singular.
This implies that $|\text{B}|\neq0,$ that $B$ is invertible and that $B^{-1}$ exists.
Here $B$ is invertible.
$\therefore|\text{B}^{-1}|=|\text{B}|^{-1}=\frac{1}{|\text{B}|}$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{B}^{-1}\|\text{AB}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{B}|^{-1}|\text{A}\|\text{B}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=\frac{1}{|\text{B}|}|\text{A}\|\text{B}|$
$\Rightarrow|\text{B}^{-1}\text{AB}|=|\text{A}|$
View full question & answer→MCQ 281 Mark
If $\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix},$ then det $\text{(adj (adj A))}$ is :
- ✓
$14^4$
- B
$14^3$
- C
$14^2$
- D
$14$
AnswerCorrect option: A. $14^4$
$\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}$
$|\text{A}|=14$
det $\text{(adj (adj A))}=|\text{A}|^{{\text{n}-1}^{2}}$
det $\text{(adj (adj A))}=|14|^{{3-1}^{2}}=14^4$
View full question & answer→MCQ 291 Mark
If $d$ is the determinant of a square matrix $A$ of order $n,$ then the determinant of its adjoint is :
- A
$d^n$
- ✓
$d^{n-1}$
- C
$d^{n+1}$
- D
$d$
AnswerCorrect option: B. $d^{n-1}$
We know,
$|\text{adj A}| = |A|^{n-1}$
$\Rightarrow |\text{adj A}| = d^{n-1}$
View full question & answer→MCQ 301 Mark
If $\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$ is prthogonal, than x + y =
Answer$\text{A}=\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 & \text{y} \end{bmatrix}$
$\text{A}^\text{T}\text{A}=\text{I}$
$\frac{1}{3}\begin{bmatrix} 1 & 2 & \text{x} \\ 1 & 1 & 2 \\ 2 & -2 & \text{y} \end{bmatrix}\frac{1}{3}\begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & -2 \\ \text{x} & 2 &\text{y} \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\frac{1}{9}\begin{bmatrix} 1+4+\text{x}^2 & 1+2+2\text{x} & \text{xy}-2 \\ 1+2+2\text{x} & 1+1+4 & 2-2+2\text{y} \\ 2-4+\text{xy} & 2-2+2\text{y} & 4+4+\text{y}^2 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 5+\text{x}^2 & 3+2\text{x} & \text{xy}-2 \\ 3+2\text{x} & 6 & 2\text{y} \\ -6+\text{xy} & 2\text{y} & 8+\text{y}^2 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Equality of two matrices does not hold Matrix A is not orthogonal.
Hence, the given question is incorrect.
View full question & answer→MCQ 311 Mark
If $\text{S}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix},$ then adj A is:
- A
$\begin{bmatrix} -\text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
- ✓
$\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
- C
$\begin{bmatrix} \text{d} & \text{b} \\ \text{c} & \text{a} \end{bmatrix}$
- D
$\begin{bmatrix} \text{d} & \text{c} \\ \text{b} & \text{a} \end{bmatrix}$
AnswerCorrect option: B. $\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
Adjoint of a square matrix of order 2 is obtained by interchancing the diagoinal elements and changing the signs of off-diagonal elements.
Here,
$\text{A}=\begin{bmatrix}\text{a} & \text{bc} & \text{d} \end{bmatrix}$
$\Rightarrow\text{adj A}=\begin{bmatrix}\text{d} & -\text{b}-\text{c} & \text{a} \end{bmatrix}$
View full question & answer→