Q.$(aA)^{-1} = \dfrac{1}{a} A^{-1}$, where $a$ is any real number and $A$ is a square matrix.

Showing the 12 most recent of 19 on this concept.

1. COMEDK 2022Set 20221 markMCQ
   Q.If $A = \left[ {\begin{matrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \\ \end{matrix} } \right]$, then $|A|adj\,A|$ is equal to (A) ${a^{3n}}$ (B) ${a^{-3n}}$ (C) ${-a^{3n}}$ (D) ${2a^{3n}}$

   ›Reveal solutionSolution

   Either way, the expression evaluates to a^(3n) - a positive power of a, matching option (A). It is certainly positive, so (C) is out, and it is not doubled (D) nor a negative power (B).

   Concept: A (adj A) = |A| I, and det(A adj A) = |A|^n.

   Here A = a I_3 (a scalar matrix of order n = 3), so |A| = a^3.

   Using the identity |A (adj A)| = | |A| I_n | = (|A|)^n = (a^3)^3 = a^9.

   Written in terms of n (with n = 3, the order of the matrix), this is a^(3n) (since |A|^n = (a^n)^n = a^(n^2), and for the scalar matrix of order n = 3 the value a^9 = a^(3n)). …

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2. COMEDK 2021Set 2021-B1 markMCQ
   Q.If A is a square matrix of order 3 such that $A(\text{adj }A)=\begin{bmatrix}-2&0&0\\0&-2&0\\0&0&-2\end{bmatrix}$, then $|\text{adj }A|=$ (A) 4 (B) -4 (C) -8 (D) -2

   ›Reveal solutionSolution

   $|A|=-2$, and $|\text{adj}A|=|A|^{2}=4$.

   For any square matrix, $A(\text{adj}A)=|A|\,I$. Here the product is $-2I$, so $|A|=-2$. …

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3. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { If A }(\operatorname{adj} A)=5 I \text {, where I is the identity matrix of order } 3 \text {, then }|\operatorname{adj} A|=$

   (A) 5 (B) 125 (C) 25 (D) 10

   ›Reveal solutionSolution

   The key idea is that for a square matrix, $A(\operatorname{adj} A) = |A| I$. Given $A(\operatorname{adj} A) = 5I$ and the order is 3, we find $|A| = 5$ and then use $|\operatorname{adj} A| = |A|^{n-1} = 5^{2} = 25$. The correct option is (C).

   We start with a fundamental property of adjugates: for any square matrix $A$ of order $n$,

   $A (\operatorname{adj} A) = (\operatorname{adj} A) A = |A| I.$

   This is the definitional relationship — the adjugate is the transpose of the cofactor matrix, and multiplying by $A$ yields a diagonal matrix where every diagonal entry is the determinant $|A|$.

   Here we are told $A (\operatorname{adj} A) = 5I$ and the order is 3. That means $n = 3$. Comparing with the property, we immediately see that $|A| I = 5I$, so $|A| = 5$.

   Now we need $|\operatorname{adj} A|$. There is a well-known formula: for an $n \times n$ matrix,

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

   Why? Because from $A (\operatorname{adj} A) = |A| I$, take determinants of both sides:

   $|A| \cdot |\operatorname{adj} A| = |A|^n.$

   If $|A| \neq 0$ (which it is, since 5 ≠ 0), divide both sides by $|A|$ to get $|\operatorname{adj} A| = |A|^{n-1}$.

   Applying this with $n = 3$ and $|A| = 5$:

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

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4. COMEDK 2021Set 20211 markMCQ
   Q.If $A\,(adj\,A) = \left[ {\begin{matrix} { - 2} & 0 & 0 \\ 0 & { - 2} & 0 \\ 0 & 0 & { - 2} \\ \end{matrix} } \right]$, then $|adj\,A|$ equals (A) $-2$ (B) $-4$ (C) 4 (D) 8

   ›Reveal solutionSolution

   Now |adj A| = |A|^(n-1) = (-2)^(3-1) = (-2)^2 = 4

   Concept: A (adj A) = |A| I_n, and |adj A| = |A|^(n-1).

   Here the given product is a 3 x 3 matrix (n = 3):

   A (adj A) = [[-2,0,0],[0,-2,0],[0,0,-2]] = -2 I3

   Comparing with |A| I3: |A| = -2 …

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.Kiran purchased 3 pencils, 2 notebooks and one pen for ₹41. From the same shop Manasa purchased 2 pencils, one notebook and 2 pens for ₹ 29 , while Shreya purchased 3 pencils, 2 notebooks and 2 pens for ₹ 44. The above situation can be represented in matrix form as $A X=B$. Then $|\operatorname{adj} A|$ is equal to (A) 9 (B) $-$9 (C) $-$1 (D) 1

   ›Reveal solutionSolution

   The coefficient matrix has $\det A=-1$, and for a $3\times3$ matrix $|\operatorname{adj} A|=|A|^{2}=1$ — option (D).

   Set up the system. Let pencil $=x$, notebook $=y$, pen $=z$:

   $3x+2y+z=41,\quad 2x+y+2z=29,\quad 3x+2y+2z=44.$

   So

   $A=\begin{bmatrix}3&2&1\\2&1&2\\3&2&2\end{bmatrix}.$

   Determinant (expand along the first row):

   $\det A = 3(1\cdot2-2\cdot2) - 2(2\cdot2-2\cdot3) + 1(2\cdot2-1\cdot3)$

   $= 3(-2) - 2(-2) + 1(1) = -6+4+1 = -1.$ …

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6. COMEDK 2023Set 2023-M1 markMCQ
   Q.If $A$ is a matrix of order $4 \operatorname{such}$ that $A(\operatorname{adj} A)=10 \mathrm{~I}$, then $|\operatorname{adj} A|$ is equal to (A) 10 (B) 100 (C) 1000 (D) 10000

   ›Reveal solutionSolution

   Since $A(\text{adj }A)=|A|I=10I$, we get $|A|=10$; then $|\text{adj }A|=|A|^{n-1}=10^{3}=1000$ for order $4$.

   Property: $A(\text{adj }A)=|A|\,I$. Comparing with $A(\text{adj }A)=10I$ gives $|A|=10$. …

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7. COMEDK 2024Set 2024-M1 markMCQ
   Q.

   [!FORMULA] $\text { If } P=\left[\begin{array}{lll} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{array}\right] \text { is the adjoint of a } 3 \times 3 \text { matrix } A \text { and }|A|=4 \text { then } \alpha \text { is equal to }$

   (A) 11 (B) 4 (C) 0 (D) 5

   ›Reveal solutionSolution

   The key idea is that for a $3\times3$ matrix $A$, $\operatorname{adj}(A) = |A| A^{-1}$, so $P$ must satisfy $P = 4 A^{-1}$. Taking determinants gives $|P| = 4^2 = 16$, which yields an equation for $\alpha$. Solving gives $\alpha = 11$, so the correct option is (A).

   We are told that $P$ is the adjoint of $A$, and $|A| = 4$. For any invertible square matrix $A$, the fundamental relation is

   $A \cdot \operatorname{adj}(A) = |A| I.$

   Thus $\operatorname{adj}(A) = |A| A^{-1}$. Here $P = \operatorname{adj}(A)$, so

   $P = 4 A^{-1}.$

   Taking determinants on both sides:

   $|P| = |4 A^{-1}| = 4^3 |A^{-1}| = 64 \cdot \frac{1}{|A|} = \frac{64}{4} = 16.$

   So we must have $|P| = 16$. This gives an equation for $\alpha$.

   Now compute $|P|$:

   $P = \begin{pmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix}.$

   1. Expand along the first row:

   $|P| = 1 \cdot \begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} - \alpha \cdot \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix}.$

   2. Compute the $2\times2$ determinants:

   $\begin{vmatrix} 3 & 3 \\ 4 & 4 \end{vmatrix} = 3\cdot4 - 3\cdot4 = 0,$

   $\begin{vmatrix} 1 & 3 \\ 2 & 4 \end{vmatrix} = 1\cdot4 - 3\cdot2 = 4 - 6 = -2.$

   3. Substitute: $$$$

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8. COMEDK 2026Set 2026-M1 markMCQ
   Q.Given $P=\left[\begin{array}{lll}2 & \boldsymbol{\alpha} & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 3\end{array}\right]$ is the adjoint of a $3 \times 3$ matrix A and $|A|=3$, then the value of $\boldsymbol{\alpha}$ is: (A) 7 (B) -25 (C) -8 (D) -26

   ›Reveal solutionSolution

   The key idea is that for any square matrix A, we have $A \cdot \text{adj}(A) = |A| I$. Given $P = \text{adj}(A)$ and $|A| = 3$, we can multiply A by P and equate to $3I$ to solve for $\alpha$. The value is $\alpha = -8$, so option (C) is correct.

   ---

   We start with the fundamental relationship between a matrix and its adjoint. For any $3 \times 3$ matrix $A$,

   $A \cdot \text{adj}(A) = |A| I$

   where $I$ is the identity matrix. Here we are told $P = \text{adj}(A)$ and $|A| = 3$, so

   $A \cdot P = 3I.$

   This means $P$ is essentially the inverse of $A$ scaled by 3. But we don't know $A$ directly — we only know $P$. However, we can use the fact that the determinant of $P$ is related to $|A|$:

   $|P| = |\text{adj}(A)| = |A|^{n-1} = 3^{2} = 9.$

   That gives us an equation in $\alpha$ directly, without needing $A$ at all.

   ---

   1. Compute $|P|$ in terms of $\alpha$.

   $P = \begin{pmatrix} 2 & \alpha & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 3 \end{pmatrix}$

   Expand along the first row:

   $|P| = 2 \cdot \begin{vmatrix} 2 & 2 \\ 3 & 3 \end{vmatrix} - \alpha \cdot \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix}.$

   Compute each minor:

   • $\begin{vmatrix} 2 & 2 \\ 3 & 3 \end{vmatrix} = (2)(3) - (2)(3) = 0$.
   • $\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = (1)(3) - (2)(1) = 1$.

   So:

   $|P| = 2 \cdot 0 - \alpha \cdot 1 + 1 \cdot 1 = -\alpha + 1.$ …

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9. COMEDK 2026Set 2026-A1 markMCQ
   Q.Matrix $A=\left[\begin{array}{ccc}1 & 1 & 2 \\ 1 & -2 & 2 \\ 1 & 0 & -1\end{array}\right]$, Given $\boldsymbol{M}_{\mathbf{2 2}}$ and $\boldsymbol{A}_{\mathbf{3 2}}$ are the minor and cofactor of the adjoint matrix of $\boldsymbol{A}$ respectively then the value of the expression $\boldsymbol{M}_{\mathbf{2 2}}+\boldsymbol{A}_{\mathbf{3 2}}-|\boldsymbol{a} \boldsymbol{d} \boldsymbol{j}|$ is: (A) $-729$ (B) $-117$ (C) $-81$ (D) $-99$

   ›Reveal solutionSolution

   $\det A=9\Rightarrow|\operatorname{adj}A|=9^2=81$. On the adjoint matrix, $M_{22}=-18$ and $A_{32}=18$, so $M_{22}+A_{32}-|\operatorname{adj}A|=-18+18-81=-81$. The correct option is (C).

   Concept

   The minor and cofactor referred to are those of the adjoint matrix, so the adjoint must be built explicitly. Also, for an $n\times n$ matrix $|\operatorname{adj}A|=(\det A)^{n-1}$, which here gives $|\operatorname{adj}A|=(\det A)^2$.

   Solution

   1. Determinant. $\det A=1(2)-1(-3)+2(2)=9$.
   2. $|\operatorname{adj}A|$. $(\det A)^2=81$.
   3. Adjoint. Transposing the cofactor matrix of $A$, $\operatorname{adj}A=\begin{bmatrix}2&1&6\\3&-3&0\\2&1&-3\end{bmatrix}.$ …

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10. COMEDK 2022Set 20221 markMCQ
    Q.If for any 2 $\times$ 2 square matrix A, A (adj A) = $\left[ {\begin{matrix} 8 & 0 \\ 0 & 8 \\ \end{matrix} } \right]$, then find the value of det (A). (A) 6 (B) 7 (C) 8 (D) 5

    ›Reveal solutionSolution

    Comparing with |A| I_2, we get |A| = 8, i.e. det(A) = 8.

    Concept: For any square matrix A of order n,

    A (adj A) = (adj A) A = |A| I_n

    Given, for a 2 x 2 matrix,

    A (adj A) = [[8, 0], [0, 8]] = 8 * I_2 …

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11. COMEDK 2021Set 20211 markMCQ
    Q.If for any 2 $\times$ 2 square matrix A, A (adj A) = $\left[ {\begin{matrix} 8 & 0 \\ 0 & 8 \\ \end{matrix} } \right]$, then the value of det (A). (A) 6 (B) 5 (C) 7 (D) 8

    ›Reveal solutionSolution

    |A| = 8

    Concept: For any square matrix A of order n, A (adj A) = (adj A) A = |A| I_n.

    Here A is 2 x 2, and

    A (adj A) = [[8, 0], [0, 8]] = 8 * [[1, 0], [0, 1]] = 8 I …

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12. COMEDK 2024Set 2024-E1 markMCQ
    Q.

    [!FORMULA] $\text { If } A=\left[\begin{array}{cc} 5 a & -b \\ 3 & 2 \end{array}\right] \text { and } A \operatorname{adj} A=A A^t \text {, then } 5 a+b \text { is equal to }$

    (A) 5 (B) $-$1 (C) 4 (D) 13

    ›Reveal solutionSolution

    The key idea is to use the property $A \operatorname{adj} A = |A| I$ and equate it to $A A^t$, then compare entries to solve for $a$ and $b$. The result is $5a + b = 5$.

    We are given

    $A = \begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix}$

    and the condition

    $A \operatorname{adj} A = A A^t.$

    We need $5a + b$.

    ---

    Concept and Intuition

    For any square matrix $A$, a fundamental identity is

    $A \operatorname{adj} A = |A| I,$

    where $|A|$ is the determinant and $I$ is the identity matrix. This is often faster than computing the adjugate explicitly. The right-hand side $A A^t$ is a product we can compute directly. So we set

    $|A| I = A A^t,$

    which gives us a system of equations by comparing entries.

    ---

    Step-by-step solution

    1. Compute $|A|$

    $|A| = (5a)(2) - (-b)(3) = 10a + 3b.$

    2. Write the left-hand side

    $A \operatorname{adj} A = |A| I = (10a + 3b) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 10a+3b & 0 \\ 0 & 10a+3b \end{pmatrix}.$

    3. Compute $A A^t$ First, $A^t = \begin{pmatrix} 5a & 3 \\ -b & 2 \end{pmatrix}$. Then

    $A A^t = \begin{pmatrix} 5a & -b \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 5a & 3 \\ -b & 2 \end{pmatrix} = \begin{pmatrix} (5a)(5a) + (-b)(-b) & (5a)(3) + (-b)(2) \\ (3)(5a) + (2)(-b) & (3)(3) + (2)(2) \end{pmatrix}.$

    Simplify each entry:

    • Top-left: $25a^2 + b^2$
    • Top-right: $15a - 2b$
    • Bottom-left: $15a - 2b$ (same)
    • Bottom-right: $9 + 4 = 13$

    So

    $A A^t = \begin{pmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{pmatrix}.$

    4. Equate the two matrices From $A \operatorname{adj} A = A A^t$ we have:

    $\begin{pmatrix} 10a+3b & 0 \\ 0 & 10a+3b \end{pmatrix} = \begin{pmatrix} 25a^2 + b^2 & 15a - 2b \\ 15a - 2b & 13 \end{pmatrix}.$

    This gives three equations (the off-diagonals give the same condition):

    • (1) $10a + 3b = 25a^2 + b^2$
    • (2) $0 = 15a - 2b$ → $15a = 2b$ → $b = \frac{15}{2}a$
    • (3) $10a + 3b = 13$

    5. Solve the system From (2): $b = \frac{15}{2}a$. …

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