If $(AB)^{-1}=\begin{bmatrix}12 & -17\\-19 & 27\end{bmatrix}$ and $A^{-1}=\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}$, then $B^{-1}=$ (a) $\begin{bmatrix}8 & -5\\-3 & 2\end{bmatrix}$ (b) $\begin{bmatrix}2 & -5\\-3 & 8\end{bmatrix}$ (c) $\begin{bmatrix}8 & 5\\3 & 2\end{bmatrix}$ (d) $\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}$

> [!TLDR] > Using $(AB)^{-1}=B^{-1}A^{-1}\Rightarrow B^{-1}=(AB)^{-1}A$, and finding $A$ from $A^{-1}$, gives $B^{-1}=\begin{bmatrix}2&-5\\-3&8\end{bmatrix}$. > > 1. For invertible matrices, $(AB)^{-1}=B^{-1}A^{-1}$. > 2. Multiply both sides on the right by $A$: $(AB)^{-1}A=B^{-1}A^{-1}A=B^{-1}I=B^{-1}$. So $B^{-1}=(AB)^{-1}A$. > 3. We need $A$, but are given $A^{-1}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}$. Since $A=(A^{-1})^{-1}$, invert this matrix. > 4. For a $2\times2$ matrix $\begin{bmatrix}p&q\\r&s\end{bmatrix}$, the inverse is $\dfrac{1}{ps-qr}\begin{bmatrix}s&-q\\-r&p\end{bmatrix}$. Here $p=1,q=-1,r=-2,s=3$, so $\det(A^{-1})=1(3)-(-1)(-2)=3-2=1$. > 5. So $A=\dfrac{1}{1}\begin{bmatrix}3&1\\2&1\end{bmatrix}=\begin{bmatrix}3&1\\2&1\end{bmatrix}$. (Check: $A\,A^{-1}=\begin{bmatrix}3&1\\2&1\end{bmatrix}\begin{bmatrix}1&-1\\-2&3\end{bmatrix}=\begin{bmatrix}3-2&-3+3\\2-2&-2+3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$, correct.) …

If (AB)⁻¹= 12 -17; -19 27 and A⁻¹= 1 -1; -2 3 , then B⁻¹= (a) 8 -5; -3 2 (b) 2 -5; -3 8 (c) 8 5; 3 2 (d) 3 1;

Question 91 of 118

(AB)^{-1}=\begin{bmatrix}12 & -17\\-19 & 27\end{bmatrix}

(a)

\begin{bmatrix}8 & -5\\-3 & 2\end{bmatrix}

(b)

\begin{bmatrix}2 & -5\\-3 & 8\end{bmatrix}

(c)

\begin{bmatrix}8 & 5\\3 & 2\end{bmatrix}

(d)

\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2020MCQ· 1mImportance★★★★★

77% · 91/118 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Adjoint and Inverse of a Matrix

For a square matrix $A=[a_{ij}]$ of order $n$ , the cofactor of $a_{ij}$ is the signed minor $A_{ij}=(-1)^{i+j}M_{ij}$ (the minor $M_{ij}$ is the determinant left after deleting row $i$ and column $j$ ). Replace every entry of $A$ by its cofactor to get the cofactor matrix; its transpose is the adjoint, $\operatorname{adj}A$ .

Theorem (the central identity). For every square matrix $A$ of order $n$ ,

A(\operatorname{adj}A)=(\operatorname{adj}A)A=|A|\,I_n.

This follows from Laplace expansion: a row's entries dotted with their own cofactors reproduce $|A|$ , while a row's entries dotted with a different row's cofactors always give $0$ -- so the product matrix is $|A|$ on the diagonal and $0$ off it.

Definition of the inverse. A square matrix $B$ with $AB=BA=I_n$ is called the inverse of $A$ , written $A^{-1}$ . The inverse, when it exists, is unique. $A^{-1}$ exists if and only if $A$ is non-singular ( $|A|\ne0$ ): dividing the central identity by $|A|$ (possible exactly when $|A|\ne0$ ) gives the working formula

A^{-1}=\dfrac{1}{|A|}\operatorname{adj}A.

A singular matrix ( $|A|=0$ ) has no inverse.

Worked illustration (order 2). For $A=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$ , the cofactors are $A_{11}=d,\ A_{12}=-c,\ A_{21}=-b,\ A_{22}=a$ , so $\operatorname{adj}A=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}$ (swap the diagonal entries, negate the off-diagonal ones) and $A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}$ whenever $ad-bc\ne0$ .

Standing laws of inverses (for non-singular $A,B$ of the same order, $\lambda\ne0$ a scalar):

$|A^{-1}|=\dfrac{1}{|A|}$ .
$(A^T)^{-1}=(A^{-1})^T$ .
$(\lambda A)^{-1}=\dfrac1\lambda A^{-1}$ .
Left/right cancellation: $AB=AC\Rightarrow B=C$ ; $BA=CA\Rightarrow B=C$ (pre/post-multiply by $A^{-1}$ ) -- this fails when $A$ is singular.
Reversal law: $(AB)^{-1}=B^{-1}A^{-1}$ (note the order flips, exactly as for transposes).
Double inverse: $(A^{-1})^{-1}=A$ .

Six adjoint identities (non-singular $A$ , order $n$ ): (i) $\operatorname{adj}(A^{-1})=(\operatorname{adj}A)^{-1}=\dfrac{A}{|A|}$ ; (ii) $|\operatorname{adj}A|=|A|^{n-1}$ ; (iii) $\operatorname{adj}(\operatorname{adj}A)=|A|^{n-2}A$ ; (iv) $\operatorname{adj}(\lambda A)=\lambda^{n-1}\operatorname{adj}A$ ; (v) $|\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2}$ ; (vi) $(\operatorname{adj}A)^T=\operatorname{adj}(A^T)$ ; and for two non-singular matrices of the same order, $\operatorname{adj}(AB)=(\operatorname{adj}B)(\operatorname{adj}A)$ (order reverses, exactly like the inverse and the transpose). …

Start your 14-day free trial to unlock the full solution →

Using $(AB)^{-1}=B^{-1}A^{-1}\Rightarrow B^{-1}=(AB)^{-1}A$ , and finding $A$ from $A^{-1}$ , gives $B^{-1}=\begin{bmatrix}2&-5\\-3&8\end{bmatrix}$ .

For invertible matrices, $(AB)^{-1}=B^{-1}A^{-1}$ .
Multiply both sides on the right by $A$ : $(AB)^{-1}A=B^{-1}A^{-1}A=B^{-1}I=B^{-1}$ . So $B^{-1}=(AB)^{-1}A$ .
We need $A$ , but are given $A^{-1}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}$ . Since $A=(A^{-1})^{-1}$ , invert this matrix.
For a $2\times2$ matrix $\begin{bmatrix}p&q\\r&s\end{bmatrix}$ , the inverse is $\dfrac{1}{ps-qr}\begin{bmatrix}s&-q\\-r&p\end{bmatrix}$ . Here $p=1,q=-1,r=-2,s=3$ , so $\det(A^{-1})=1(3)-(-1)(-2)=3-2=1$ .
So $A=\dfrac{1}{1}\begin{bmatrix}3&1\\2&1\end{bmatrix}=\begin{bmatrix}3&1\\2&1\end{bmatrix}$ . (Check: $A\,A^{-1}=\begin{bmatrix}3&1\\2&1\end{bmatrix}\begin{bmatrix}1&-1\\-2&3\end{bmatrix}=\begin{bmatrix}3-2&-3+3\\2-2&-2+3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ , correct.) …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$3\times3$
(a) $I_3$
(b) $A$
(c) $B^T$
(d) $B$
›Reveal solutionSolution
Uses the given normal-matrix condition $AA^T=A^TA$ to collapse $BB^T$ down to the identity.

Given $B=A^{-1}A^T$ . Taking transpose: $B^T=(A^{-1}A^T)^T=(A^T)^TA^{-T}=A(A^T)^{-1}$ .

So $BB^T=\left(A^{-1}A^T\right)\left(A(A^T)^{-1}\right)=A^{-1}\left(A^TA\right)(A^T)^{-1}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2026Set SEM31 markMCQ
$A$
(a) $20$
(b) $15$
(c) $5$
(d) $25$
›Reveal solutionSolution

Use the identity $|\text{adj}\,A| = |A|^{n-1}$ with $n=3$ .

The adjoint–determinant relation is a CBSE/NCERT Class 12 determinants and adjoint/inverse result.

For a non-singular matrix of order $n$ , the standard identity is

$|\text{adj}\,A| = |A|^{\,n-1}.$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$3\times3$
(a) $5^2$
(b) $5$
(c) $\dfrac{1}{5^2}$
(d) $\dfrac15$
›Reveal solutionSolution
The standard determinant identity $|A^{-1}|=1/|A|$ gives the answer directly by substitution.

Since $AA^{-1}=I$ , taking determinants of both sides: $|A||A^{-1}|=|I|=1$ . …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix}$
(a) $19$
(b) $17$
(c) $21$
(d) $14$
›Reveal solutionSolution
Multiplying the given relation by $A$ turns it into $A^2=\lambda I$ , so $\lambda$ is found by directly computing $A^2$ .

Given $\lambda A^{-1}=A$ . Multiply both sides on the left by $A$ : $\lambda AA^{-1}=A\cdot A\Rightarrow\lambda I=A^2$ .

Compute $A^2$ for $A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}$ .

Row 1: $(2)(2)+(3)(5)=4+15=19$ ; $(2)(3)+(3)(-2)=6-6=0$ . …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$A^{-1}=\begin{bmatrix}5 & 3\\-2 & -1\end{bmatrix}$
(a) $\begin{bmatrix}-1 & -3\\2 & 5\end{bmatrix}$
(b) $\begin{bmatrix}-5 & 3\\2 & 1\end{bmatrix}$
(c) $\begin{bmatrix}5 & -2\\3 & -1\end{bmatrix}$
(d) $\begin{bmatrix}5 & 3\\-2 & -1\end{bmatrix}$
›Reveal solutionSolution
Uses the identity $(A^T)^{-1}=(A^{-1})^T$ and simply transposes the given $A^{-1}$ .

A standard matrix identity: $(A^T)^{-1}=(A^{-1})^T$ (the inverse of the transpose equals the transpose of the inverse).

Given $A^{-1}=\begin{bmatrix}5&3\\-2&-1\end{bmatrix}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. If A, B and C are invertible matrices of some order, then which one of the following is not true ?$
(a) $\det A^{-1}=(\det A)^{-1}$
(b) $\text{adj}\,A=|A|A^{-1}$
(c) $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$
(d) $\text{adj}\,(AB)=(\text{adj}\,A)(\text{adj}\,B)$
›Reveal solutionSolution
Checking each identity against the standard matrix-algebra facts, only the adjugate-of-a-product rule has its order reversed from what's stated.

(a) $\det A^{-1}=(\det A)^{-1}$ : always true, since $\det(A)\det(A^{-1})=\det(AA^{-1})=\det I=1$ .

(b) $\text{adj}\,A=|A|A^{-1}$ : always true — this is the defining relation $A\cdot\text{adj}\,A=|A|I$ , rearranged using $A^{-1}$ .

(c) $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$ : always true (reverse-order law for inverses of a product). …
Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. A square matrix A of order n has inverse if and only if :$
(a) $\rho(A) > n$
(b) $\rho(A) = n$
(c) $\rho(A) \ne n$
(d) $\rho(A) < n$
›Reveal solutionSolution
A square matrix has an inverse exactly when it is non-singular, i.e. when its rank equals its order.

A square matrix $A$ of order $n$ has an inverse iff $A$ is non-singular, i.e. $|A|\ne 0$ . …
Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$|\text{adj}(\text{adj}A)|=|A|^{16}$
(a) $2$
(b) $3$
(c) $5$
(d) $4$
›Reveal solutionSolution
Using the standard identity $|\text{adj}\,A|=|A|^{n-1}$ twice gives $|\text{adj}(\text{adj}A)|=|A|^{(n-1)^2}$ , matched against the exponent $16$ .

For an $n\times n$ matrix $A$ , the standard identity is $|\text{adj}\,A|=|A|^{\,n-1}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix}$
(a) $19$
(b) $17$
(c) $21$
(d) $14$
›Reveal solutionSolution
Multiplying both sides of $\lambda A^{-1}=A$ by $A$ gives $\lambda I=A^2$ ; computing $A^2$ shows it equals $19I$ , so $\lambda=19$ .

We are given $\lambda A^{-1}=A$ for $A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}$ .

Post-multiplying both sides by $A$ : $\lambda A^{-1}A=A\cdot A$ , i.e. $\lambda I=A^2$ .

Compute $A^2=\begin{bmatrix}2&3\\5&-2\end{bmatrix}\begin{bmatrix}2&3\\5&-2\end{bmatrix}$ .

The $(1,1)$ entry is $2(2)+3(5)=4+15=19$ ; the $(1,2)$ entry is $2(3)+3(-2)=6-6=0$ . …
Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. Which one of the following is incorrect ?$
(a) If A is a square matrix of order n, and $\lambda$ is a scalar, then Adj $(\lambda A)=\lambda^n$ (Adj A).
(b) Adjoint of a symmetric matrix is also a symmetric matrix.
(c) A(Adj A) = (Adj A)A = |A|I.
(d) Adjoint of a diagonal matrix is also a diagonal matrix.
›Reveal solutionSolution
The adjoint scaling law is $\text{Adj}(\lambda A)=\lambda^{n-1}\text{Adj}(A)$ , not $\lambda^{n}\text{Adj}(A)$ , so statement (a) is the incorrect one.

Each entry of $\text{Adj}(A)$ is a cofactor of $A$ , obtained from an $(n-1)\times(n-1)$ minor.

If every entry of $A$ is scaled by $\lambda$ to form $\lambda A$ , each $(n-1)\times(n-1)$ minor (a determinant of $n-1$ rows/columns, each scaled by $\lambda$ ) scales by $\lambda^{n-1}$ .

Hence the correct identity is $\text{Adj}(\lambda A)=\lambda^{n-1}\text{Adj}(A)$ , so the claim in (a) that it equals $\lambda^{n}\text{Adj}(A)$ is false.

Statement (b) is a standard true property: the adjoint (transpose of the cofactor matrix) of a symmetric matrix is symmetric. …
Start your 14-day free trial to see the full solution →
CBSE 2020Set ANNUAL1 markMCQ
$(AB)^{-1}=\begin{bmatrix}12 & -17\\-19 & 27\end{bmatrix}$
(a) $\begin{bmatrix}8 & -5\\-3 & 2\end{bmatrix}$
(b) $\begin{bmatrix}2 & -5\\-3 & 8\end{bmatrix}$
(c) $\begin{bmatrix}8 & 5\\3 & 2\end{bmatrix}$
(d) $\begin{bmatrix}3 & 1\\2 & 1\end{bmatrix}$
›Reveal solutionSolution
Using $(AB)^{-1}=B^{-1}A^{-1}\Rightarrow B^{-1}=(AB)^{-1}A$ , and finding $A$ from $A^{-1}$ , gives $B^{-1}=\begin{bmatrix}2&-5\\-3&8\end{bmatrix}$ .

For invertible matrices, $(AB)^{-1}=B^{-1}A^{-1}$ .

Multiply both sides on the right by $A$ : $(AB)^{-1}A=B^{-1}A^{-1}A=B^{-1}I=B^{-1}$ . So $B^{-1}=(AB)^{-1}A$ .

We need $A$ , but are given $A^{-1}=\begin{bmatrix}1&-1\\-2&3\end{bmatrix}$ . Since $A=(A^{-1})^{-1}$ , invert this matrix.

For a $2\times2$ matrix $\begin{bmatrix}p&q\\r&s\end{bmatrix}$ , the inverse is $\dfrac{1}{ps-qr}\begin{bmatrix}s&-q\\-r&p\end{bmatrix}$ . Here $p=1,q=-1,r=-2,s=3$ , so $\det(A^{-1})=1(3)-(-1)(-2)=3-2=1$ .

So $A=\dfrac{1}{1}\begin{bmatrix}3&1\\2&1\end{bmatrix}=\begin{bmatrix}3&1\\2&1\end{bmatrix}$ . (Check: $A\,A^{-1}=\begin{bmatrix}3&1\\2&1\end{bmatrix}\begin{bmatrix}1&-1\\-2&3\end{bmatrix}=\begin{bmatrix}3-2&-3+3\\2-2&-2+3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ , correct.) …
Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$A$
(a) $\dfrac{1}{k} I$
(b) $kI$
(c) $\dfrac{1}{k^2} I$
(d) $\dfrac{1}{k^3} I$
›Reveal solutionSolution
The inverse of the scalar matrix $A=kI$ (order $3$ , $k\neq0$ ) is $A^{-1}=\dfrac{1}{k}I$ .

A scalar matrix of order $3$ with scalar $k$ is $A=kI$ , where $I$ is the $3\times3$ identity matrix.

To find $A^{-1}$ , we need a matrix $B$ such that $AB=I$ .

Try $B=\dfrac{1}{k}I$ : then $AB=(kI)\left(\dfrac{1}{k}I\right)=\left(k\cdot\dfrac{1}{k}\right)I=I$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ27 — If ρ(A) = ρ([A mid B]), then the system AX = B of linear equations is …→Next questionQ29 — If A= 2 9; 1 7 then prove that (AT)⁻¹=(A⁻¹)T.

Q.If (AB)−1=[12−17−1927](AB)^{-1}=\begin{bmatrix}12 & -17\\-19 & 27\end{bmatrix}(AB)−1=[12−19​−1727​] and A−1=[1−1−23]A^{-1}=\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}A−1=[1−2​−13​], then B−1=B^{-1}=B−1=

🎓Unlock everything free for 14 days

Q.If $(AB)^{-1}=\begin{bmatrix}12 & -17\\-19 & 27\end{bmatrix}$ and $A^{-1}=\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}$ , then $B^{-1}=$

Unlock everything free for 14 days