MCQ 512 Marks
If $A =\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]$ and $A ^2= I$, then $A ^{-1}$ is equal to
- ✓
$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- B
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
- C
$\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
- D
$\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
(A) $A ^2=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}x^2+1 & x \\ x & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow x=0$
$\therefore A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
$A ^2= I$
$\therefore \quad A^{-1} A \cdot A=A^{-1} I$
$\Rightarrow I \cdot A = A ^{-1} I$
$\Rightarrow A ^{-1}= A$
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If A is solution of $x^2-3 x-7=0$ and $A=\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]$, then $A^{-1}$ equals
- ✓
$\left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7} \\ \frac{-1}{7} & \frac{-5}{7}\end{array}\right]$
- B
$\left[\begin{array}{cc}2 & 3 \\ -1 & -5\end{array}\right]$
- C
$\left[\begin{array}{cc}\frac{1}{7} & \frac{1}{7} \\ \frac{-1}{7} & \frac{-5}{7}\end{array}\right]$
- D
$\left[\begin{array}{cc}3 & -1 \\ 1 & 2\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7} \\ \frac{-1}{7} & \frac{-5}{7}\end{array}\right]$
(A) $A^2-3 A-7 I=0$
$\Rightarrow A -3 I -7 A^{-1}=0 \Rightarrow A^{-1}=\frac{1}{7}(A-3 I )$
$\therefore \quad A ^{-1}=\frac{1}{7}\left\{\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]\right\}$
$=\left[\begin{array}{cc}\frac{2}{7} & \frac{3}{7} \\ -\frac{1}{7} & -\frac{5}{7}\end{array}\right]$
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If A and B are two square matrices such that $B=-A^{-1} B A$, then $(A+B)^2=$
- A
$0$
- ✓
$A ^2+ B ^2$
- C
$A^2+2 A B+B^2$
- D
$A+B$
AnswerCorrect option: B. $A ^2+ B ^2$
(B) Given, $B =- A ^{-1} BA$
$\therefore \quad AB =- AA ^{-1} BA$
$\Rightarrow AB =- I ( BA ) \Rightarrow AB =- BA$
Now $(A+B)^2=(A+B)(A+B)$
$= A ^2+ AB + BA + B ^2$
$= A ^2+ B ^2 \quad[\because BA =- AB ]$
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If $A ^2- A + I =0$, then $A ^{-1}=$
- A
$A^{-2}$
- B
$A + I$
- ✓
$I - A$
- D
$A - I$
AnswerCorrect option: C. $I - A$
(C) $A^2-A+I=0$
$\Rightarrow A \cdot A - A + I =0$
$\Rightarrow A ^{-1} \cdot A \cdot A - A ^{-1} \cdot A+ A ^{-1} \cdot I =0$
$\Rightarrow A - I + A ^{-1}=0$
$\Rightarrow A ^{-1}= I - A$
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If for the matrix $A, A^3=I$, then $A^{-1}=$
AnswerCorrect option: A. $A ^2$
(A) $A ^3= I$
$\Rightarrow A ^{-1} A^3= A ^{-1} \cdot I$
$\Rightarrow\left( A ^{-1} A\right) A ^2= A ^{-1}$
$\Rightarrow IA ^2= A ^{-1} \Rightarrow A^2= A ^{-1}$
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Let for any matrix $M , M ^{-1}$ exist, then which of the following is not true?
- A
$\left( M ^{-1}\right)=( M )^{-1}$
- B
$\left(M^2\right)^{-1}=\left(M^{-1}\right)^2$
- ✓
$\left( M ^{-1}\right)^{-1}=\left( M ^{-1}\right)^1$
- D
$\left( M ^{-1}\right)^{-1}= M$
AnswerCorrect option: C. $\left( M ^{-1}\right)^{-1}=\left( M ^{-1}\right)^1$
(C) $\left(M^{-1}\right)^{-1} \neq\left(M^{-1}\right)^1$
$\therefore \quad\left( M ^{-1}\right)^{-1}=\left( M ^{-1}\right)^1$ is not true
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If $A=\left[\begin{array}{cc}x & -2 \\ 3 & 7\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right]$, then the value of $x$ is
Answer(D) Since $AA ^{-1}= I$,
$\left[\begin{array}{cc}x & -2 \\ 3 & 7\end{array}\right]\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}\frac{7 x+6}{34} & \frac{x-4}{17} \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
By equality of matrices,
$\frac{x-4}{17}=0 \Rightarrow x-4=0$
$\Rightarrow x=4$
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Let $A =\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$. If B is the inverse of matrix A , then $\alpha$ is
Answer(D) $10 A^{-1}=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] \quad \ldots\left[\because B = A ^{-1}\right]$
$\Rightarrow 10 A^{-1} A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] A$
$\Rightarrow 10 I =\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 0 \\ -5+\alpha & 5+\alpha & -5+\alpha \\ 0 & 0 & 10\end{array}\right]$
$\therefore \quad-5+\alpha=0 \Rightarrow \alpha=5$
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If the inverse of the matrix $A =\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ is $\frac{1}{5}\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & \alpha \\ 2 & 2 & -3\end{array}\right]$, then $\alpha=$ __________
Answer(C) $AA ^{-1}= I$
$\therefore \quad \frac{1}{5}\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & \alpha \\ 2 & 2 & -3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}1 & 0 & \frac{2 \alpha-4}{5} \\ 0 & 1 & \frac{\alpha-2}{5} \\ 0 & 0 & \frac{1+2 \alpha}{5}\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
∴ By equality of matrices,
$\frac{\alpha-2}{5}=0 \Rightarrow \alpha=2$
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From the matrix equation $AB = AC$ we can conclude that $B = C$, provided
Answer(B) $AB = AC$
$\Rightarrow A ^{-1} AB = A ^{-1} AC$
$\Rightarrow IB = IC$
$\Rightarrow B = C$
$\therefore \quad$ For $B = C , A ^{-1}$ must exist
$\Rightarrow A$ is non-singular
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$A =\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right] ; I =\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A^{-1}=\frac{1}{6}\left(A^2+c A+d I\right)$ where $c, d \in R$, then pair of values ( $c, d$ ) is
- A
$(6,11)$
- B
$(6,-11)$
- ✓
$(-6,11)$
- D
$(-6,-11)$
AnswerCorrect option: C. $(-6,11)$
(C) $|A|=\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right|=6 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{6}\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]$
$A^2=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4\end{array}\right]$
$=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14\end{array}\right]$
$\therefore \quad A ^2+ cA + dI$
$=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14\end{array}\right]+\left[\begin{array}{ccc} c & 0 & 0 \\ 0 & c & c \\ 0 & -2 c & 4 c \end{array}\right]+\left[\begin{array}{ccc} d & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & d\end{array}\right]$
$=\left[\begin{array}{ccc}1+ c + d & 0 & 0 \\ 0 & -1+ c + d & 5+ c \\ 0 & -10-2 c & 14+4 c + d \end{array}\right]$
Since $6 A^{-1}=A^2+c A+d I$
$\therefore\left[\begin{array}{ccc}6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1\end{array}\right]=\left[\begin{array}{ccc}1+ c + d & 0 & 0 \\ 0 & -1+ c + d & 5+ c \\ 0 & -10-2 c & 14+4 c + d \end{array}\right]$
∴ by equality of matrices,
$1+c+d=6$ and $5+c=-1$
$\therefore \quad c=-6$ and $d=11$
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If the inverse of product of the matrix
$B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$ with a matrix $A$ is
$C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$, then $A^{-1}$ equals
- A
$\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6\end{array}\right]$
- B
$\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
- D
$\left[\begin{array}{ccc}-3 & -3 & 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
AnswerCorrect option: C. $\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
(C) $( BA )^{-1}= C$
$\Rightarrow A ^{-1} B^{-1}= C \Rightarrow A ^{-1}= CB$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$
$=\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
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If $\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right] A \left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then matrix A equals
- ✓
$\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]$
- B
$\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]$
- C
$\left[\begin{array}{cc}7 & 1 \\ 34 & 5\end{array}\right]$
- D
$\left[\begin{array}{cc}5 & 3 \\ 13 & 8\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]$
(A) If $XAY = I$, then $A = X ^{-1} Y ^{-1}=( YX )^{-1}$
Here, $YX =\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 7 & 4\end{array}\right]$
$=\left[\begin{array}{cc}8 & 5 \\ -11 & -7\end{array}\right]$
$\therefore \quad A=\left[\begin{array}{cc}8 & 5 \\ -11 & -7\end{array}\right]^{-1}$
$=\left[\begin{array}{cc}7 & 5 \\ -11 & -8\end{array}\right]$
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The matrix A satisfying $A \left[\begin{array}{ll}1 & 5 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}3 & -1 \\ 6 & 0\end{array}\right]$ is
- A
$\left[\begin{array}{cc}3 & 2 \\ 6 & -3\end{array}\right]$
- ✓
$\left[\begin{array}{ll}3 & -16 \\ 6 & -30\end{array}\right]$
- C
$\left[\begin{array}{cc}3 & -16 \\ 6 & 30\end{array}\right]$
- D
$\left[\begin{array}{cc}3 & -3 \\ 6 & 2\end{array}\right]$
AnswerCorrect option: B. $\left[\begin{array}{ll}3 & -16 \\ 6 & -30\end{array}\right]$
(B) If $AC = B$, then $A = BC ^{-1}$
$\therefore \quad A=\left[\begin{array}{cc}3 & -1 \\ 6 & 0\end{array}\right]\left[\begin{array}{cc}1 & 5 \\ 0 & 1\end{array}\right]^{-1}$
$=\left[\begin{array}{cc}3 & -1 \\ 6 & 0\end{array}\right]\left[\begin{array}{cc}1 & -5 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{ll}3 & -16 \\ 6 & -30\end{array}\right]$
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The element of second row and third column in the inverse of $\left[\begin{array}{ccc}1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$ is
Answer(B) Let $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1\end{array}\right] \Rightarrow|A|=-2 \neq 0$
Now, co-factor of element $a _{32}$ of $A = A _{32}$
$\therefore \quad A _{32}=(-1)^{3+2}\left|\begin{array}{ll}1 & 1 \\ 2 & 0\end{array}\right|=2$
$\therefore \quad$ Element $a _{23}$ of $A ^{-1}=\frac{ A _{32}}{|A|}=\frac{2}{-2}=-1$
Alternate method:
$|A|=-2 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}1 & -2 & -1 \\ -2 & 2 & 2 \\ 1 & -2 & -3\end{array}\right]$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc}-\frac{1}{2} & 1 & \frac{1}{2} \\ 1 & -1 & -1 \\ -\frac{1}{2} & 1 & \frac{3}{2}\end{array}\right]$
$\therefore \quad$ Element $a _{23}$ of $A ^{-1}=-1$.
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The inverse of matrix $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$ is
Answer(A) $|A|=\left|\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right|=-1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]$
$\therefore \quad A ^{-1}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]= A$
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If $P=\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], Q=\left[\begin{array}{ccc}1 & -2 & -3 \\ -3 & 1 & 9 \\ 0 & 0 & -5\end{array}\right]$ then $( PQ )^{-1}$ equals to
AnswerCorrect option: D. $-\frac{1}{5} I _3$
(D) Since $PQ =-5 I _3$,
$(P Q)^{-1}=-\frac{1}{5} I_3$
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If $A =\left[\begin{array}{ccc}\frac{ k }{2} & 0 & 0 \\ 0 & \frac{l}{3} & 0 \\ 0 & 0 & \frac{m}{4}\end{array}\right]$ and $A ^{-1}=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$ then $k +l+ m =$
Answer(D) Using Shortcut 1,
$A ^{-1}=\left[\begin{array}{ccc}\frac{2}{ k } & 0 & 0 \\ 0 & \frac{3}{l} & 0 \\ 0 & 0 & \frac{4}{m}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$
$\Rightarrow \frac{2}{k}=\frac{1}{2} \Rightarrow k=4$,
$\frac{3}{l}=\frac{1}{3} \Rightarrow l=9$ and
$\frac{4}{m}=\frac{1}{4} \Rightarrow m=16$
$\therefore \quad k +l+ m =4+9+16=29$
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The inverse of the matrix $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$ is
- A
$\frac{1}{24}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$
- B
$\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]$
- C
$\frac{1}{24}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$
(D) The inverse of the given diagonal matrix is
$A ^{-1}=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4}\end{array}\right]$
...[Using Shortcut 1]
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If $A=\left[\begin{array}{ccc}1 & -4 & -1 \\ 6 & 3 & 0 \\ 2 & 0 & 0\end{array}\right]$, then $6\left|A^{-1}\right|$ is equal to
Answer(A) $A=\left[\begin{array}{ccc}1 & -4 & -1 \\ 6 & 3 & 0 \\ 2 & 0 & 0\end{array}\right]$
$|A|=6 \neq 0$
$\therefore \quad\left| A ^{-1}\right|=\frac{1}{|A|}$
$=\frac{1}{6}$
$\therefore \quad 6\left|A^{-1}\right|=1$
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If $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$, then the sum of all the diagonal entries of $A ^{-1}$ is
Answer(D) $A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
$\therefore \quad$ sum of all the diagonal entries $=\frac{1}{2}+3+\frac{1}{2}=4$
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If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$, then $A^{-1}=$
AnswerCorrect option: C. $A ^3$
(C) $|A|=\left|\begin{array}{ccc}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right|=1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
$\therefore \quad A^{-1}=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
$A^2=\left[\begin{array}{ccc}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3\end{array}\right]$
$A^3=A^2 \cdot A=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
$= A ^{-1}$
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If $A=\left[\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]$, then $A^{-1}$ is
AnswerCorrect option: A. $A ^{ T }$
(A) $|A|=\left|\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right|=-1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{|A|}(\operatorname{adj} A )$
$=\left[\begin{array}{ccc}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right]= A ^{ T }$
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The inverse of the matrix $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$ is
- A
$\left[\begin{array}{ccc}\frac{3}{2} & \frac{6}{2} & \frac{-5}{2} \\ \frac{-15}{2} & \frac{-1}{2} & \frac{1}{2} \\ 5 & -1 & 1\end{array}\right]$
- B
$\left[\begin{array}{ccc}3 & 6 & 2 \\ -15 & -1 & 1 \\ 5 & -2 & -5\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
- D
$\left[\begin{array}{ccc}\frac{3}{2} & \frac{-1}{2} & \frac{1}{2} \\ \frac{-15}{2} & \frac{6}{2} & \frac{-5}{2} \\ 5 & -1 & 1\end{array}\right]$
AnswerCorrect option: C. $\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
(C) $|A|=\left|\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right|=1 \neq 0$
$\operatorname{adj} A=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
$\therefore \quad A ^{-1}=\frac{1}{|A|}(\operatorname{adj} A )$
$=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$
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If $A=\left[a_{i j}\right]_{2 \times 2}$, where $a_{i j}=\left\{\begin{array}{cc}i+j, & \text { if } i \neq j \\ i^2-2 j, & \text { if } i=j\end{array}\right.$ then $A ^{-1}$ is equal to
- A
$\frac{1}{9}\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]$
- B
$\frac{1}{9}\left[\begin{array}{cc}0 & -3 \\ -3 & -1\end{array}\right]$
- ✓
$\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$
- D
AnswerCorrect option: C. $\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$
(C) $A=\left[a_{i j}\right]_{2 \times 2} \Rightarrow A=\left[\begin{array}{cc}-1 & 3 \\ 3 & 0\end{array}\right]$
$|A|=-9 \neq 0$
$\therefore \quad A^{-1}=\frac{-1}{9}\left[\begin{array}{cc}0 & -3 \\ -3 & -1\end{array}\right]=\frac{1}{9}\left[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\right]$
...[Using Shortcut 2]
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Matrix $A=\left[\begin{array}{ccc}1 & 0 & -k \\ 2 & 1 & 3 \\ k & 0 & 1\end{array}\right]$ is invertible for
Answer(D) $| A |= k ^2+1$, which can be never zero.
Hence matrix A is invertible for all real k .
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Let A be a square matrix of order 3 whose all entries are 1 and let $I_3$ be the identity matrix of order 3 . Then the matrix $A -3 I _3$ is
- A
- B
- ✓
- D
real skew symmetric matrix
Answer(C) $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right], I_3=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$A-3 I_3=\left[\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right]$
$\left|A-3 I_3\right|=\left|\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right|$
$=0$
$\therefore \quad$ the matrix $A -3 I _3$ is non-invertible.
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The matrix $\left[\begin{array}{ccc}\lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{array}\right]$ is invertible, if
- A
$\lambda \neq-15$
- ✓
$\lambda \neq-17$
- C
$\lambda \neq-16$
- D
$\lambda \neq-18$
AnswerCorrect option: B. $\lambda \neq-17$
(B) The given matrix will be invertible, if
$\left|\begin{array}{ccc}\lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2\end{array}\right| \neq 0$
$\begin{array}{l}\Rightarrow \lambda(0-1)+1(-6+1)+4(-3) \neq 0 \\ \Rightarrow-\lambda-5-12 \neq 0 \\ \Rightarrow \lambda \neq-17\end{array}$
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The matrix $\left[\begin{array}{lll}1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1\end{array}\right]$ is not invertible, if ' $a$ ' has the value
Answer(B) The matrix is not invertible if $\left|\begin{array}{lll}1 & \text { a } & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1\end{array}\right|=0$
$\Rightarrow 1(2-5)-a(1-10)+2(1-4)=0$
$\Rightarrow-3+9 a-6=0$
$\Rightarrow a=1$
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If A is a square matrix such that $A(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$, then $\frac{|\operatorname{adj}(\operatorname{adj} A)|}{|\operatorname{adj} A|}$ is equal to
Answer(B) A. $(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]$ . . .(i)
$=4\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$=4 . I$
Using Shortcut 4(i),
$A (\operatorname{adj} A )=| A | . I$,
$|A|=4$
From (i), $| A | \cdot|\operatorname{adj} A |=64$
$\Rightarrow|\operatorname{adj} A|=\frac{64}{4}=16$
Using Shortcut 4(xiv),
$|\operatorname{adj}(\operatorname{adj} A )|=| A |^{( n -1)^2}$
$=| A |^{(3-1)^2}$
$=(4)^4=256$
$\therefore \frac{|\operatorname{adj}(\operatorname{adj} A)|}{|\operatorname{adj} A|}=\frac{256}{16}=16$
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Let A be a $2 \times 2$ matrix
Statement-1: $\operatorname{adj}(\operatorname{adj} A )= A$
Statement-2 : $|\operatorname{adj} A |=| A |$
- A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
- ✓
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
- C
Statement- 1 is true, Statement- 2 is false
- D
Statement-1 is false, Statement-2 is true
AnswerCorrect option: B. Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
(B) $|\operatorname{adj} A |=| A |^{ n -1}=| A |^{2-1}=| A |$
…[Using Shortcut 4(viii)]
$\operatorname{adj}(\operatorname{adj} A )=| A |^{ n -2} A=| A |^0 A= A$
...[Using Shortcut 4(xiii)]
$\therefore$ option (B) is the correct answer.
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If A is a matrix of order 3, such that $A(\operatorname{adj} A)=10 I$, then $|\operatorname{adj} A |=$
Answer(C) Using Shortcut 4(i),
$A (\operatorname{Adj} A )=| A | I$
$\therefore|A|=10$
Using Shortcut 4(viii),
$\begin{aligned}|\operatorname{Adj} A | & =| A |^{ n -1} \\ \therefore \quad|\operatorname{Adj} A| & =| A |^{3-1}=| A |^2=10^2=100\end{aligned}$
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If $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then $\operatorname{adj}(\operatorname{adj} A)$ is equal to
- A
$\operatorname{adj} A$
- ✓
- C
$A ^{ T }$
- D
$-A$
Answer(B) $\operatorname{adj} A=\left[\begin{array}{cc} d & - b \\ - c & a \end{array}\right]$
$\therefore \quad \operatorname{adj}(\operatorname{adj} A )=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]= A$
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If $A$ is a singular matrix of order $n$, then $A \cdot(\operatorname{adj} A )$ is
Answer(A) A is a singular matrix.
$\therefore|A|=0$ and $A \cdot(\operatorname{adj} A)=|A| . I=0 . I=0$
$\Rightarrow A (\operatorname{adj} A )$ is a zero matrix.
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If $A=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$, then $A(\operatorname{adj} A)=$
- ✓
$\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$
- B
$\left[\begin{array}{cc}0 & 10 \\ 10 & 0\end{array}\right]$
- C
$\left[\begin{array}{cc}10 & 1 \\ 1 & 10\end{array}\right]$
- D
AnswerCorrect option: A. $\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$
(A) Using Shortcut 4(i),
$A (\operatorname{adj} A )=| A | \cdot I _{ n }$
$\therefore \quad A (\operatorname{adj} A )=\left|\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right|\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]$
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If the adjoint of a $3 \times 3$ matrix $P$ is $\left[\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right]$, then the possible values of the determinant of $P$ are
- ✓
$\pm 2$
- B
$\pm 1$
- C
$\pm 3$
- D
$\pm 4$
AnswerCorrect option: A. $\pm 2$
(A) $\operatorname{adj} P=\left[\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right]$
Using Shortcut 4(viii),
$|\operatorname{adj} P |=| P |^2$
$\Rightarrow\left|\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right|=| P |^2$
$\begin{array}{l}\Rightarrow|P|^2=1(-4)-4(-1)+4(1) \\ \Rightarrow|P|^2=4 \Rightarrow|P|= \pm 2\end{array}$
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If $A$ is a singular matrix, then $\operatorname{adj} A$ is
Answer(A) Given, A is a singular matrix.
$\therefore \quad| A |=0$
Using Shortcut 4(viii),
$|\operatorname{adj} A |=| A |^{ n -1}$
$\Rightarrow|\operatorname{adj} A|=0$
$\Rightarrow \operatorname{adj} A$ is also singular.
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If X is a square matrix of order $3 \times 3$ and $\lambda$ is a scalar, then adj $(\lambda X)$ is equal to
- A
$\lambda \operatorname{adj} X$
- B
$\lambda^3 \operatorname{adj} X$
- ✓
$\lambda^2 \operatorname{adj} X$
- D
$\lambda^4 \operatorname{adj} X$
AnswerCorrect option: C. $\lambda^2 \operatorname{adj} X$
(C) $\operatorname{adj}(\lambda X )=\lambda^{3-1}(\operatorname{adj} X )$
…[Using Shortcut 4(vi)]
$=\lambda^2 \operatorname{adj} X$
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If the value of a third order determinant is 16, then the value of the determinant formed by replacing each of its elements by its cofactor is
Answer(D) If A is a square matrix of order 3 , then
$|\operatorname{adj} A |=| A |^2$
...[Using Shortcut 4(x)]
$\begin{array}{l}=(16)^2 \\ =256\end{array}$
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If $M$ is any square matrix of order 3 over $R$ and if $M^{\prime}$ be the transpose of $M$, then $\operatorname{adj}\left( M ^{\prime}\right)-(\operatorname{adj} M )^{\prime}$ is equal to
Answer(C) $\operatorname{adj}\left(M^{\prime}\right)-(\operatorname{adj} M)^{\prime}$ is a null matrix as $\operatorname{adj}\left( M ^{\prime}\right)=\left(\operatorname{adj} M )^{\prime}\right.$
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If $A =\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right], B =\left[\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right]$, then $\operatorname{adj}( AB )$ is equal to
- ✓
$\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]$
- B
$\left[\begin{array}{ll}94 & -39 \\ 82 & -34\end{array}\right]$
- C
$\left[\begin{array}{cc}94 & -82 \\ -39 & 34\end{array}\right]$
- D
$\left[\begin{array}{cc}-94 & -39 \\ 82 & 34\end{array}\right]$
AnswerCorrect option: A. $\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]$
(A) $AB =\left[\begin{array}{ll}34 & 39 \\ 82 & 94\end{array}\right] \Rightarrow \operatorname{adj}( AB )=\left[\begin{array}{cc}94 & -39 \\ -82 & 34\end{array}\right]$
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Adjoint of the matrix $N=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]$ is
Answer(A) Matrix of co-factors,
$\left[ A _{ ij }\right]_{3 \times 3}=\left[\begin{array}{ccc}-4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3\end{array}\right]$
$\therefore \quad$ adj $N=\left[A_{i j}\right]_{3 \times 3}^T=\left[\begin{array}{ccc}-4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{array}\right]=N$
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If $A=\left[\begin{array}{lll}1 & 2 & 1 \\ 3 & 2 & 3 \\ 2 & 1 & 2\end{array}\right]$, then $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}=$
Answer(C) $a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}$
$=1(4-3)+3[-(4-1)]+2(6-2)=0$
and $|A|=1(4-3)-2(6-6)+1(3-4)=0$
$\therefore \quad a_{11} A_{11}+a_{21} A_{21}+a_{31} A_{31}=|A|$
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If $A=\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$, then $C_2+2 C_1$ and then $R_1+R_3$ gives
- A
$\left[\begin{array}{lll}2 & 1 & 3 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$
- B
$\left[\begin{array}{ccc}2 & -1 & 3 \\ 2 & 4 & 3 \\ 3 & 1 & 2\end{array}\right]$
- C
$\left[\begin{array}{lll}5 & 5 & 5 \\ 3 & 4 & 2 \\ 2 & 6 & 3\end{array}\right]$
- ✓
$\left[\begin{array}{lll}5 & 5 & 5 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{lll}5 & 5 & 5 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$
(D) $A=\left[\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right]$
Applying $C _2 \rightarrow C _2+2 C _1$,
$A \sim\left[\begin{array}{lll}2 & 1 & 3 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$
Applying $R _1 \rightarrow R _1+ R _3$,
$A \sim\left[\begin{array}{lll}5 & 5 & 5 \\ 2 & 6 & 3 \\ 3 & 4 & 2\end{array}\right]$
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If $\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]= A \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
then $C _2 \rightarrow C _2-3 C _1$ and $C _3 \rightarrow C _3+2 C _1$ gives
- A
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & -9 & 11 \\ -2 & 1 & 4\end{array}\right]=A\left[\begin{array}{ccc}-1 & 3 & 2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
- B
$\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & -1 & -4 \\ 2 & -9 & 11\end{array}\right]-A\left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
- C
$\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 1 & 4 \\ -2 & 9 & -11\end{array}\right]=A\left[\begin{array}{ccc}-1 & 3 & 2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
- ✓
$\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 9 & -11 \\ 2 & -1 & 4\end{array}\right]=A\left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
AnswerCorrect option: D. $\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 9 & -11 \\ 2 & -1 & 4\end{array}\right]=A\left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
(D) $\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0\end{array}\right]= A \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Applying $C _2 \rightarrow C _2-3 C _1$ and $C _3 \rightarrow C _3+2 C _1$,
$\left[\begin{array}{ccc}1 & 3-3 & -2+2 \\ -3 & 0+9 & -5-6 \\ 2 & 5-6 & 0+4\end{array}\right]= A \left[\begin{array}{lll}1 & 0-3 & 0+2 \\ 0 & 1-0 & 0+0 \\ 0 & 0-0 & 1+0\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ -3 & 9 & -11 \\ 2 & -1 & 4\end{array}\right]= A \left[\begin{array}{ccc}1 & -3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
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If $3 x-4 y+2 z=-1,2 x+3 y+5 z=7$, $x+z=2$, then $x=$
Answer(A) $A X=B$
$\therefore \quad\left[\begin{array}{ccc}3 & -4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-1 \\ 7 \\ 2\end{array}\right]$
$R _2-5 R _3 \Rightarrow\left[\begin{array}{ccc}3 & -4 & 2 \\ -3 & 3 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-1 \\ -3 \\ 2\end{array}\right]$
$R _1-2 R _3 \Rightarrow\left[\begin{array}{ccc}1 & -4 & 0 \\ -3 & 3 & 0 \\ 1 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-5 \\ -3 \\ 2\end{array}\right]$
$\Rightarrow x-4 y=-5$, and …(i)
$-3 x+3 y=-3$ …(ii)
Solving (i) and (ii), we get $x=3$
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If $x+2 y=3$ and $2 x+3 y=4$, then the values of $x$ and $y$ respectively are
- A
$1,-2$
- B
$-2,1$
- ✓
$-1,2$
- D
$2,-1$
AnswerCorrect option: C. $-1,2$
(C) The given system of equations can be written in matrix form AX = B, where
$A =\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], X =\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B =\left[\begin{array}{l}3 \\ 4\end{array}\right]$
$\therefore \quad$ Now, $\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}3 \\ 4\end{array}\right]$
Applying $R _2 \rightarrow R _2-2 R _{ 1 }$,
$\left[\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -2\end{array}\right]$
$\Rightarrow x+2 y=3$, and $\ldots$(i)
$- y = - 2$ $\ldots$(ii)
$\Rightarrow y=2$
putting y = 2 in (i), we get
x + 2(2) = 3
$\Rightarrow x=-1$
Alternate method:
$\begin{array}{l}A X=B \Rightarrow X=A^{-1} B \\ |A|=-1 \neq 0 \\ A^{-1}=\frac{1}{-1}\left[\begin{array}{cc}3 & -2 \\ -2 & 1\end{array}\right]\end{array}$
$\begin{aligned} A^{-1} & =\frac{1}{-1}\left[\begin{array}{cc}3 & -2 \\ -2 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & 2 \\ 2 & -1\end{array}\right] \end{aligned}$
$\begin{aligned} X & =\left[\begin{array}{cc}-3 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{l}3 \\ 4\end{array}\right] \\ & =\left[\begin{array}{c}-1 \\ 2\end{array}\right]\end{aligned}$
$\therefore x=-1, y=2$
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If $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 4\end{array}\right]$, then the values of $x$ and $y$ respectively are
AnswerCorrect option: D. $-1,3$
(D) $\left[\begin{array}{cc}1 & 1 \\ -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 4\end{array}\right]$
$\begin{array}{l}\Rightarrow x+y=2 \text { and }-x+y=4 \\ \Rightarrow x=-1, y=3\end{array}$
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If A and B are non-singular matrices, then
- A
$( AB )^{-1}= A ^{-1} B^{-1}$
- B
$AB = BA$
- C
$( AB )^{\prime}= A ^{\prime} B ^{\prime}$
- ✓
$( AB )^{-1}= B ^{-1} A^{-1}$
AnswerCorrect option: D. $( AB )^{-1}= B ^{-1} A^{-1}$
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Let A be an invertible matrix. Which of the following is not true?
- A
$\left(A^T\right)^{-1}=\left(A^{-1}\right)^T$
- ✓
$A ^{-1}=| A |^{-1}$
- C
$\left(A^2\right)^{-1}=\left(A^{-1}\right)^2$
- D
$\left| A ^{-1}\right|=| A |^{-1}$
AnswerCorrect option: B. $A ^{-1}=| A |^{-1}$
(B) Consider option (B),
$A ^{-1}$ is a matrix and $| A |^{-1}$ is a number.
∴ option (B) is not true.
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