Q.If $A$ is an invertible matrix of order 2, then $\det(A^{-1})$ is equal to (A) $\det(A)$ (B) $\dfrac{1}{\det(A)}$ (C) $1$ (D) $0$

Showing the 12 most recent of 19 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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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 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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8. COMEDK 2025Set 2025-A1 markMCQ
   Q.If $A(\operatorname{adj} A)=\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]$, then the value of $|A|+|\operatorname{adj} A|$ is equal to : (A) 5 (B) 25 (C) 125 (D) 30

   ›Reveal solutionSolution

   The key idea is that for any square matrix $A$, $A(\operatorname{adj} A) = |A| I$. Here that gives $|A| = 5$, and since $|\operatorname{adj} A| = |A|^{n-1}$ for an $n \times n$ matrix, we get $|\operatorname{adj} A| = 5^2 = 25$. Their sum is $5 + 25 = 30$, so the answer is (D).

   ---

   The problem gives us $A(\operatorname{adj} A) = 5I$, where $I$ is the $3 \times 3$ identity matrix. This is a classic property: for any square matrix $A$, the product $A$ times its adjugate equals the determinant times the identity. So the scalar on the diagonal is exactly $|A|$. That means $|A| = 5$ immediately.

   Now, we also need $|\operatorname{adj} A|$. There's a neat formula: for an $n \times n$ matrix, $|\operatorname{adj} A| = |A|^{n-1}$. Here $n = 3$, so $|\operatorname{adj} A| = 5^{2} = 25$.

   Thus the sum is $5 + 25 = 30$.

   Let's walk through it step by step.

   1. Recall the defining property of the adjugate. For any square matrix $A$, we have

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

   This is the fundamental relation. The problem gives us the left-hand side explicitly as $5I$, so we can directly compare:

   $|A| I = 5I \quad \Rightarrow \quad |A| = 5.$

   2. Find $|\operatorname{adj} A|$ using the determinant of both sides. Take determinants of the equation $A (\operatorname{adj} A) = |A| I$:

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

   The right-hand side is a scalar matrix: $|A| I$ is $|A|$ times the identity, so its determinant is $(|A|)^n$ for an $n \times n$ matrix. Here $n = 3$, so

   $\big| |A| I \big| = (|A|)^3.$

   Thus

   $$$$

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9. COMEDK 2025Set 2025-E1 markMCQ
   Q.Value of the determinant of a matrix $A$ of order $3 \times 3$ is 7 . Then the value of the determinant formed by the cofactors of matrix A is (A) 7 (B) 49 (C) 14 (D) 343

   ›Reveal solutionSolution

   The determinant of the cofactor matrix (the adjugate) of a $3\times3$ matrix $A$ equals $(\det A)^{n-1} = (\det A)^2$. Given $\det A = 7$, the answer is $7^2 = 49$, so option (B).

   ---

   The key idea is that the matrix of cofactors is intimately linked to the inverse of $A$. For any square matrix $A$, the product $A \cdot (\text{adj } A) = (\det A) I$, where $\text{adj } A$ is the transpose of the cofactor matrix. Taking determinants on both sides gives a direct relationship between $\det(\text{cofactor matrix})$ and $\det A$.

   For an $n \times n$ matrix, the determinant of the cofactor matrix (strictly, of the adjugate) is $(\det A)^{n-1}$. Here $n=3$, so the exponent is $2$.

   Let’s walk through it carefully.

   1. Define the cofactor matrix and adjugate. For a $3 \times 3$ matrix $A$, let $C_{ij}$ be the cofactor of entry $a_{ij}$. The cofactor matrix is $C = [C_{ij}]$. The adjugate (or classical adjoint) is the transpose: $\operatorname{adj}(A) = C^T$. The fundamental property is:

   $A \cdot \operatorname{adj}(A) = \operatorname{adj}(A) \cdot A = (\det A) I_3.$

   This holds for any square matrix.

   2. Take determinants of both sides. From $A \cdot \operatorname{adj}(A) = (\det A) I_3$, we have:

   $\det(A \cdot \operatorname{adj}(A)) = \det((\det A) I_3).$

   The left side, by the product rule, is $\det A \cdot \det(\operatorname{adj}(A))$.

   The right side: multiplying a $3\times3$ identity matrix by the scalar $\det A$ gives a diagonal matrix with $\det A$ on each diagonal entry. Its determinant is $(\det A)^3$.

   3. Set up the equation.

   $\det A \cdot \det(\operatorname{adj}(A)) = (\det A)^3.$

   Since $\det A = 7 \neq 0$, we can divide both sides by $\det A$:

   $$$$

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10. 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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11. 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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12. 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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