Q.Which of the following is/are correct?

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 $BB^T=\left(A^{-1}A^T\right)\left(A(A^T)^{-1}\right)=A^{-1}\left(A^TA\right)(A^T)^{-1}$. …

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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

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

   …

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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: $|A||A^{-1}|=|I|=1$. …

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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: $(2)(2)+(3)(5)=4+15=19$; $(2)(3)+(3)(-2)=6-6=0$. …

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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 $A^{-1}=\begin{bmatrix}5&3\\-2&-1\end{bmatrix}$. …

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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). …

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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. $|A|\ne 0$. …

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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 $|\text{adj}\,A|=|A|^{\,n-1}$. …

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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 $2(3)+3(-2)=6-6=0$. …

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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. …

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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.) …

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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 $AB=(kI)\left(\dfrac{1}{k}I\right)=\left(k\cdot\dfrac{1}{k}\right)I=I$. …

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