Q.If $A=\begin{pmatrix}3 & 5\\ 1 & 2\end{pmatrix}$, $B=\operatorname{adj}A$ and $C=3A$, then $\dfrac{|\operatorname{adj}B|}{|C|}=$

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If A is a $3\times3$ non-singular matrix such that $AA^T=A^TA$ and $B=A^{-1}A^T$, then $BB^T=$

   (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.

   1. Given $B=A^{-1}A^T$. Taking transpose: $B^T=(A^{-1}A^T)^T=(A^T)^TA^{-T}=A(A^T)^{-1}$.
   2. So $BB^T=\left(A^{-1}A^T\right)\left(A(A^T)^{-1}\right)=A^{-1}\left(A^TA\right)(A^T)^{-1}$.
   3. Using the given condition $AA^T=A^TA$, substitute $A^TA=AA^T$: $BB^T=A^{-1}\left(AA^T\right)(A^T)^{-1}$.
   4. Regroup: $BB^T=(A^{-1}A)\left(A^T(A^T)^{-1}\right)=I_3\cdot I_3=I_3$.

   ✓Final answer

   (a) $I_3$

2. CBSE 2026Set SEM31 markMCQ
   Q.If the inverse of a matrix $A$ of order $3 \times 3$ exists and $|A| = 5$, then the value of $|adj\, A|$ is

   (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

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

   With $n=3$ and $|A|=5$:

   $|\text{adj}\,A| = 5^{3-1} = 5^2 = 25.$

   ✓Final answer

   $|\text{adj}\,A| = 25$ — option (d).

3. CBSE 2025Set ANNUAL1 markMCQ
   Q.If A is a non-singular matrix of order $3\times3$ and $|A|=5$ then $|A^{-1}|$ is :

   (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.

   1. Since $AA^{-1}=I$, taking determinants of both sides: $|A||A^{-1}|=|I|=1$.
   2. So $|A^{-1}|=\dfrac{1}{|A|}$.
   3. Given $|A|=5$: $|A^{-1}|=\dfrac15$.

   ✓Final answer

   (d) $\dfrac15$

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix}$ be such that $\lambda A^{-1}=A$, then $\lambda$ is :

   (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 $A^2$.

   1. Given $\lambda A^{-1}=A$. Multiply both sides on the left by $A$: $\lambda AA^{-1}=A\cdot A\Rightarrow\lambda I=A^2$.
   2. Compute $A^2$ for $A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}$.
   3. Row 1: $(2)(2)+(3)(5)=4+15=19$; $(2)(3)+(3)(-2)=6-6=0$.
   4. Row 2: $(5)(2)+(-2)(5)=10-10=0$; $(5)(3)+(-2)(-2)=15+4=19$.
   5. So $A^2=\begin{bmatrix}19&0\\0&19\end{bmatrix}=19I$.
   6. Comparing with $\lambda I=A^2$: $\lambda=19$.

   ✓Final answer

   (a) $19$

5. CBSE 2024Set ANNUAL1 markMCQ
   Q.If A is a non-singular matrix such that $A^{-1}=\begin{bmatrix}5 & 3\\-2 & -1\end{bmatrix}$, then $(A^T)^{-1}=$

   (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 $A^{-1}$.

   1. A standard matrix identity: $(A^T)^{-1}=(A^{-1})^T$ (the inverse of the transpose equals the transpose of the inverse).
   2. Given $A^{-1}=\begin{bmatrix}5&3\\-2&-1\end{bmatrix}$.
   3. $(A^{-1})^T=\begin{bmatrix}5&-2\\3&-1\end{bmatrix}$ (swap rows and columns).
   4. So $(A^T)^{-1}=\begin{bmatrix}5&-2\\3&-1\end{bmatrix}$.

   ✓Final answer

   (c) $\begin{bmatrix}5&-2\\3&-1\end{bmatrix}$

6. 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.

   1. (a) $\det A^{-1}=(\det A)^{-1}$: always true, since $\det(A)\det(A^{-1})=\det(AA^{-1})=\det I=1$.
   2. (b) $\text{adj}\,A=|A|A^{-1}$: always true — this is the defining relation $A\cdot\text{adj}\,A=|A|I$, rearranged using $A^{-1}$.
   3. (c) $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$: always true (reverse-order law for inverses of a product).
   4. (d) The correct reverse-order law for adjugates is $\text{adj}\,(AB)=(\text{adj}\,B)(\text{adj}\,A)$ — the stated $(\text{adj}\,A)(\text{adj}\,B)$ has the order un-reversed, and is not true in general (only when $A,B$ commute appropriately). So this is the one that is NOT true in general.

   ✓Final answer

   (d) $\text{adj}\,(AB)=(\text{adj}\,A)(\text{adj}\,B)$

7. 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.

   1. A square matrix $A$ of order $n$ has an inverse iff $A$ is non-singular, i.e. $|A|\ne 0$.
   2. $|A|\ne0$ for an $n\times n$ matrix happens exactly when all $n$ rows (columns) are linearly independent, i.e. when the rank $\rho(A)=n$.
   3. If $\rho(A)<n$, some row is a linear combination of others, $|A|=0$, and $A^{-1}$ does not exist.

   ✓Final answer

   (b) $\rho(A)=n$

8. CBSE 2023Set ANNUAL1 markMCQ
   Q.$|\text{adj}(\text{adj}A)|=|A|^{16}$, then the order of the square matrix A is :

   (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 $16$.

   1. For an $n\times n$ matrix $A$, the standard identity is $|\text{adj}\,A|=|A|^{\,n-1}$.
   2. Applying it to $\text{adj}\,A$ (also $n\times n$): $|\text{adj}(\text{adj}\,A)|=|\text{adj}\,A|^{\,n-1}=\left(|A|^{n-1}\right)^{n-1}=|A|^{(n-1)^2}$.
   3. Given $|\text{adj}(\text{adj}\,A)|=|A|^{16}$, matching exponents: $(n-1)^2=16\Rightarrow n-1=4\Rightarrow n=5$.

   ✓Final answer

   (c) $5$

9. CBSE 2022Set ANNUAL1 markMCQ
   Q.If $A=\begin{bmatrix}2 & 3\\5 & -2\end{bmatrix}$ be such that $\lambda A^{-1}=A$, then $\lambda$ is :

   (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 $\lambda=19$.

   1. We are given $\lambda A^{-1}=A$ for $A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}$.
   2. Post-multiplying both sides by $A$: $\lambda A^{-1}A=A\cdot A$, i.e. $\lambda I=A^2$.
   3. Compute $A^2=\begin{bmatrix}2&3\\5&-2\end{bmatrix}\begin{bmatrix}2&3\\5&-2\end{bmatrix}$.
   4. The $(1,1)$ entry is $2(2)+3(5)=4+15=19$; the $(1,2)$ entry is $2(3)+3(-2)=6-6=0$.
   5. The $(2,1)$ entry is $5(2)+(-2)(5)=10-10=0$; the $(2,2)$ entry is $5(3)+(-2)(-2)=15+4=19$.
   6. So $A^2=\begin{bmatrix}19&0\\0&19\end{bmatrix}=19I$.
   7. Comparing with $\lambda I=A^2=19I$, we get $\lambda=19$.

   ✓Final answer

   $\lambda=19$ — option (a).

10. 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.

    1. Each entry of $\text{Adj}(A)$ is a cofactor of $A$, obtained from an $(n-1)\times(n-1)$ minor.
    2. 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 $\lambda^{n-1}$.
    3. 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.
    4. Statement (b) is a standard true property: the adjoint (transpose of the cofactor matrix) of a symmetric matrix is symmetric.
    5. Statement (c) is the fundamental adjoint identity $A(\text{Adj }A)=(\text{Adj }A)A=|A|I$, which always holds.
    6. Statement (d) is true: cofactors of a diagonal matrix off the leading diagonal vanish, so its adjoint is also diagonal.
    7. Since (b), (c), (d) are all true and (a) misstates the exponent, (a) is the incorrect statement.

    ✓Final answer

    Statement (a) is incorrect — the correct law is $\text{Adj}(\lambda A)=\lambda^{n-1}(\text{Adj }A)$.

11. CBSE 2020Set ANNUAL1 markMCQ
    Q.If $(AB)^{-1}=\begin{bmatrix}12 & -17\\-19 & 27\end{bmatrix}$ and $A^{-1}=\begin{bmatrix}1 & -1\\-2 & 3\end{bmatrix}$, then $B^{-1}=$

    (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 $B^{-1}=\begin{bmatrix}2&-5\\-3&8\end{bmatrix}$.

    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.)
    6. Now compute $B^{-1}=(AB)^{-1}A=\begin{bmatrix}12&-17\\-19&27\end{bmatrix}\begin{bmatrix}3&1\\2&1\end{bmatrix}$.
    7. Row 1: $\left[12(3)+(-17)(2),\ 12(1)+(-17)(1)\right]=\left[36-34,\ 12-17\right]=[2,-5]$.
    8. Row 2: $\left[-19(3)+27(2),\ -19(1)+27(1)\right]=\left[-57+54,\ -19+27\right]=[-3,8]$.
    9. So $B^{-1}=\begin{bmatrix}2&-5\\-3&8\end{bmatrix}$.

    ✓Final answer

    $B^{-1}=\begin{bmatrix}2&-5\\-3&8\end{bmatrix}$ — option (b).

12. CBSE 2019Set ANNUAL1 markMCQ
    Q.If $A$ is a scalar matrix with scalar $k \neq 0$, of order $3$, then $A^{-1}$ is :

    (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 $A^{-1}=\dfrac{1}{k}I$.

    1. A scalar matrix of order $3$ with scalar $k$ is $A=kI$, where $I$ is the $3\times3$ identity matrix.
    2. To find $A^{-1}$, we need a matrix $B$ such that $AB=I$.
    3. Try $B=\dfrac{1}{k}I$: then $AB=(kI)\left(\dfrac{1}{k}I\right)=\left(k\cdot\dfrac{1}{k}\right)I=I$.
    4. Since $k\neq0$, $\dfrac{1}{k}$ is defined, so this $B$ is a valid two-sided inverse.
    5. Hence $A^{-1}=\dfrac{1}{k}I$.

    ✓Final answer

    $A^{-1}=\dfrac{1}{k}I$ — option (a).