Q.$\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$ Verify $A (\text{adj } A) = (\text{adj } A) A = |A| I$ in Exercises 3 and 4

Showing the 12 most recent of 31 on this concept.

1. CBSE 2025Set 65/1/11 markMCQ
   Q.If A is a square matrix of order 2 such that $\det (A) = 4$, then $\det(4 \text{ adj } A)$ is equal to : (A) $16$ (B) $64$ (C) $256$ (D) $512$

   ›Reveal solutionSolution

   The key idea is to use the property $\det(\text{adj } A) = (\det A)^{n-1}$ for an $n \times n$ matrix, then combine with the scalar multiplication rule $\det(kB) = k^n \det B$. For a $2 \times 2$ matrix with $\det A = 4$, we get $\det(4 \text{ adj } A) = 4^2 \cdot 4^{2-1} = 16 \cdot 4 = 64$. The answer is (B).

   The problem asks for $\det(4 \text{ adj } A)$ given that $A$ is a $2 \times 2$ matrix with $\det A = 4$. This is a classic exam question that tests two fundamental determinant properties together: how the determinant behaves when you multiply a matrix by a scalar, and the relationship between a matrix and its adjoint.

   Let’s unpack the intuition first. The adjoint (or adjugate) of a matrix is the transpose of its cofactor matrix. For a $2 \times 2$ matrix, the adjoint has a simple form: if $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then $\text{adj } A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Notice that $\det(\text{adj } A) = ad - bc = \det A$. That’s not a coincidence — it’s a special case of a general rule.

   For any $n \times n$ matrix $A$, $\det(\text{adj } A) = (\det A)^{n-1}$.

   For $n = 2$, this gives $\det(\text{adj } A) = (\det A)^{1} = \det A$. So here, $\det(\text{adj } A) = 4$.

   Now we need $\det(4 \text{ adj } A)$. The scalar multiplication rule says: if you multiply an $n \times n$ matrix by a scalar $k$, the determinant gets multiplied by $k^n$. Why? Because each of the $n$ rows gets a factor of $k$, and pulling out $k$ from each row gives $k^n$ times the original determinant.

   Watch out

   A common mistake is to forget the exponent $n$ and write $\det(kB) = k \det B$. That’s only true for a $1 \times 1$ matrix. For a $2 \times 2$ matrix, it’s $k^2$.

   So here $n = 2$ and $k = 4$, so $\det(4 \text{ adj } A) = 4^2 \cdot \det(\text{adj } A) = 16 \cdot 4 = 64$.

   Let’s walk through it step by step.

   1. Identify the order. $A$ is $2 \times 2$, so $n = 2$. This matters for both the adjoint property and the scalar multiplication rule.

   2. Find $\det(\text{adj } A)$. Using the formula $\det(\text{adj } A) = (\det A)^{n-1}$, we get $(\det A)^{1} = 4$. So $\det(\text{adj } A) = 4$.

   3. Apply the scalar multiplication rule. We have $B = \text{adj } A$, and we want $\det(4B)$. For a $2 \times 2$ matrix, $\det(4B) = 4^2 \det B = 16 \det B$.

   4. Combine the results. $\det(4 \text{ adj } A) = 16 \times 4 = 64$.

   Tip

   You can also verify with a concrete example. Take $A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ (so $\det A = 4$). Then $\text{adj } A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$, so $4 \text{ adj } A = \begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix}$, whose determinant is $8 \times 8 = 64$. This confirms the result.

   ✓Final answer

   The value is $\boxed{64}$, which corresponds to option (B).

2. CBSE 2023Set 65/1/11 markMCQ
   Q.Let A be a $3 \times 3$ matrix such that $|\text{adj A}| = 64$. Then $|A|$ is equal to : (A) Only $8$ (B) Only $-8$ (C) $64$ (D) $8$ or $-8$

   ›Reveal solutionSolution

   For a $3 \times 3$ matrix, the determinant of its adjugate is $|\text{adj } A| = |A|^{n-1} = |A|^2$. Given $|A|^2 = 64$, the possible values are $|A| = 8$ or $|A| = -8$, so the correct option is (D).

   The key here is the adjugate matrix property — a beautiful and often-tested result in linear algebra. For any square matrix $A$ of order $n$, the adjugate (or classical adjoint) satisfies:

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

   Taking determinants on both sides gives:

   $|A| \cdot |\text{adj } A| = |A|^n$

   which simplifies (for $|A| \neq 0$) to:

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

   This formula holds even when $|A| = 0$ (both sides are zero), so it’s universally true.

   Now, let’s apply it step by step.

   1. Identify the order of the matrix. Here $A$ is $3 \times 3$, so $n = 3$. Therefore $n-1 = 2$, and the formula becomes:

   $|\text{adj } A| = |A|^{2}$

   2. Plug in the given value. We are told $|\text{adj } A| = 64$. So:

   $|A|^2 = 64$

   3. Solve for $|A|$. Taking square roots:

   $|A| = \pm 8$

   Both $8$ and $-8$ satisfy the equation, because squaring eliminates the sign.

   Watch out

   A common mistake is to forget that $|A|$ can be negative. Determinants are real numbers — they can be positive, negative, or zero. The square root of $64$ gives two possibilities, not just the positive one.

   4. Check if both are valid. The adjugate formula $|\text{adj } A| = |A|^{n-1}$ works for any $|A|$, positive or negative. So both $8$ and $-8$ are possible values of $|A|$ for some $3 \times 3$ matrix $A$ with $|\text{adj } A| = 64$.

   Tip

   For a $2 \times 2$ matrix, $|\text{adj } A| = |A|$ (since $n-1 = 1$). For a $3 \times 3$, it’s $|A|^2$. For a $4 \times 4$, it’s $|A|^3$, and so on. The exponent is always one less than the order.

   ✓Final answer

   The value of $|A|$ is $8$ or $-8$, so the correct option is (D).

3. CBSE 2026Set 65/3/11 markMCQ
   Q.If $B(\text{adj } B) = \begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \end{bmatrix}$, then the value of $\det(B^{-1})$ is: (A) $\frac{1}{3}$ (B) $\frac{1}{9}$ (C) $3$ (D) $9$

   ›Reveal solutionSolution

   By recognizing the given matrix product $B(\text{adj } B)$ as $(\det B)I$, we find $\det B = \frac{1}{3}$. Then, using the property $\det(B^{-1}) = \frac{1}{\det B}$, we calculate $\det(B^{-1}) = 3$.

   The problem asks for the determinant of the inverse of matrix $B$, given a relationship involving $B$ and its adjoint. To solve this, we need to recall two fundamental properties of matrices and their determinants.

   The first key idea is the relationship between a square matrix $A$, its adjoint $\text{adj } A$, and its determinant $\det A$. This relationship is a cornerstone of matrix theory and is often used to define the inverse of a matrix. It states that the product of a matrix and its adjoint is equal to the determinant of the matrix multiplied by the identity matrix.

   The second key idea is how the determinant of an inverse matrix relates to the determinant of the original matrix. If a matrix $A$ is invertible, then the determinant of its inverse, $A^{-1}$, is simply the reciprocal of the determinant of $A$.

   Let's apply these concepts step-by-step.

   1. Identify the fundamental matrix property.

      We are given the equation $B(\text{adj } B) = \begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \end{bmatrix}$.

      The crucial property connecting a square matrix $A$ with its adjoint is:

      $A(\text{adj } A) = (\det A)I$

      where $I$ is the identity matrix of the same order as $A$.

      From the given $3 \times 3$ matrix on the right-hand side, we can infer that $B$ is a $3 \times 3$ matrix. Thus, $I$ is the $3 \times 3$ identity matrix:

      $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

   2. Determine $\det(B)$ from the given equation.

      Let's rewrite the given right-hand side in terms of the identity matrix:

      $\begin{bmatrix} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \frac{1}{3}I$.

      Now, substitute this back into the original equation:

      $B(\text{adj } B) = \frac{1}{3}I$.

      Comparing this with the fundamental property $B(\text{adj } B) = (\det B)I$, we can directly equate the scalar multiples of $I$:

      $(\det B)I = \frac{1}{3}I$.

      This implies that $\det B = \frac{1}{3}$.

   3. Calculate $\det(B^{-1})$ using the value of $\det(B)$.

      We need to find $\det(B^{-1})$. For any invertible matrix $A$, the determinant of its inverse is given by:

      $\det(A^{-1}) = \frac{1}{\det(A)}$

      This property holds true as long as $\det(A) \neq 0$, which is a condition for $A$ to be invertible. Since $\det(B) = \frac{1}{3} \neq 0$, matrix $B$ is indeed invertible.

      Substitute the value of $\det(B)$ we found:

      $\det(B^{-1}) = \frac{1}{\det(B)} = \frac{1}{\frac{1}{3}}$.

      Simplifying the expression:

      $\det(B^{-1}) = 3$.

      Comparing this result with the given options, option (C) matches our answer.

   ✓Final answer

   The value of $\det(B^{-1})$ is $\boxed{3}$.

4. CBSE 2026Set V11 markMCQ
   Q.For the matrix $A = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$ the value of $|adj\ A|$

   (a) 25

   (b) 5

   (c) 0

   (d) 1

   ›Reveal solutionSolution

   $|\operatorname{adj}A|=|A|^{n-1}=|A|$ for a $2\times2$ matrix, and $|A|=25$; answer (a).

   $A=\begin{pmatrix}5&0\\0&5\end{pmatrix}\Rightarrow |A|=5\cdot5-0=25.$

   For an $n\times n$ matrix, $|\operatorname{adj}A|=|A|^{\,n-1}$. Here $n=2$, so

   $|\operatorname{adj}A|=|A|^{1}=25.$

   ✓Final answer

   (a) 25

5. CBSE 2026Set A1 markMCQ
   Q.If $A = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$ then $\operatorname{adj} A =$

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

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

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

   (d) none of these

   ›Reveal solutionSolution

   For a $2\times2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\operatorname{adj} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

   Given $A = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}$, apply the rule (swap $a,d$; negate $b,c$):

   $\operatorname{adj} A = \begin{bmatrix} 2 & -(-5) \\ -(-1) & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}.$

   ✓Final answer

   (a) $\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$.

6. CBSE 2025Set X11 markMCQ
   Q.Let $A$ be a nonsingular matrix of order $3 \times 3$, then $|adj\, A|$ is equal to

   (a) $|A|$

   (b) $3|A|$

   (c) $|A|^3$

   (d) $|A|^2$

   ›Reveal solutionSolution

   Determinant of the adjoint of a $3\times3$ matrix — correct option is (d).

   A key property of the adjoint is $A\,(\text{adj}\,A)=|A|\,I_n$. Taking determinants gives $|A|\,|\text{adj}\,A|=|A|^n$, so for a nonsingular $A$, $|\text{adj}\,A|=|A|^{n-1}$. With order $n=3$, this equals $|A|^{3-1}=|A|^2$.

   ✓Final answer

   (d) $|A|^2$

7. CBSE 2025Set E1 markMCQ
   Q.Adjoint matrix of matrix $\begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix} =$

   (a) $\begin{bmatrix} 4 & -5 \\ -3 & 2 \end{bmatrix}$

   (b) $\begin{bmatrix} 4 & -3 \\ -5 & 2 \end{bmatrix}$

   (c) $\begin{bmatrix} 4 & 5 \\ 3 & 2 \end{bmatrix}$

   (d) $\begin{bmatrix} 4 & 3 \\ 5 & 2 \end{bmatrix}$

   ›Reveal solutionSolution

   For a $2\times2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{adj} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

   For $A=\begin{bmatrix} 2 & 3 \\ 5 & 4 \end{bmatrix}$, the adjoint is the transpose of the cofactor matrix. For a $2\times2$ this reduces to interchanging the leading-diagonal entries and changing the sign of the off-diagonal entries:

   $\text{adj}(A) = \begin{bmatrix} 4 & -3 \\ -5 & 2 \end{bmatrix}.$

   ✓Final answer

   (B) $\begin{bmatrix} 4 & -3 \\ -5 & 2 \end{bmatrix}$.

8. CBSE 2025Set A1 mark
   Q.If $A = \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$, then write the value of $|adj(A)|$.

   ›Reveal solutionSolution

   For an $n\times n$ matrix, $|adj(A)| = |A|^{n-1}$; here $n=2$ so $|adj(A)|=|A|$.

   First compute $|A|$ for $A=\begin{bmatrix}1&2\\3&4\end{bmatrix}$:

   $|A| = 1(4)-2(3) = 4-6 = -2$

   For a square matrix of order $n$, the standard identity is $|adj(A)| = |A|^{n-1}$. Here $n=2$, so:

   $|adj(A)| = |A|^{2-1} = |A| = -2$

   (Check directly: $adj(A) = \begin{bmatrix}4&-2\\-3&1\end{bmatrix}$, and $|adj(A)| = 4(1)-(-2)(-3) = 4-6=-2$.) ✓

   ✓Final answer

   $|adj(A)| = -2$.

9. CBSE 2025Set ANNUAL1 markMCQ
   Q.Let A be a nonsingular square matrix of order $3\times 3$. Then $|\text{adj }A|$ is equal to -

   (a) $|A|^2$

   (b) $|A|^3$

   (c) $|A|$

   (d) $2|A|$

   ›Reveal solutionSolution

   For an $n\times n$ nonsingular matrix, $|\text{adj}\,A|=|A|^{n-1}$.

   This follows from the identity $A\cdot(\text{adj}\,A)=|A|\,I_n$, which on taking determinants gives $|A|\cdot|\text{adj}\,A|=|A|^n$, so $|\text{adj}\,A|=|A|^{n-1}$ (valid since $A$ is nonsingular, $|A|\ne0$).

   Here $n=3$, so $|\text{adj}\,A|=|A|^{3-1}=|A|^2$.

   ✓Final answer

   The correct option is (a) $|A|^2$.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.If A is a square matrix of order 2×2 and |A| = 5, then |Adj.(A)| is:

    (a) 25

    (b) 125

    (c) 5

    (d) 10

    ›Reveal solutionSolution

    Use the identity $|\text{adj}(A)| = |A|^{n-1}$ for an $n\times n$ matrix.

    For a square matrix $A$ of order $n$, $|\text{adj}(A)| = |A|^{n-1}$.

    Here $n=2$ and $|A|=5$, so

    $|\text{adj}(A)| = 5^{2-1} = 5^1 = 5.$

    ✓Final answer

    $5$ — option (c).

11. CBSE 2024Set 65/3/11 markMCQ
    Q.Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ be a square matrix such that $\operatorname{adj} A = A$. Then $(a + b + c + d)$ is equal to: (A) $2a$ (B) $2b$ (C) $2c$ (D) $0$

    ›Reveal solutionSolution

    When the adjugate of a $2 \times 2$ matrix equals the matrix itself, the trace constraint forces $a + d = 1$, and the off-diagonal symmetry gives $b = c$; together these yield $a + b + c + d = 1 + 2b = 2a + 2b - 1$, but the determinant condition $ad - bc = 1$ combined with $\text{adj} A = A$ ultimately forces $a + b + c + d = 1$, which matches none of the options directly until we recognize the answer is (D) $0$ when the special case $a = d = \frac{1}{2}, b = c = 0$ is considered, or more generally the problem expects $d = 1 - a$ and $b = c = 0$.

    The adjugate (or adjoint) of a matrix encodes how the matrix transforms cofactors. For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the adjugate is constructed by swapping the diagonal entries, negating the off-diagonal ones, and transposing (though for $2 \times 2$ the transpose is automatic):

    $\operatorname{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$

    The condition $\operatorname{adj} A = A$ means the matrix is its own adjugate, a rare and highly constrained situation. This forces four simultaneous equations that interlock the entries.

    ---

    Setting up the equations

    1. Equate corresponding entries. From $\operatorname{adj} A = A$:

    $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$

    This gives:

    • $d = a$
    • $-b = b \implies 2b = 0 \implies b = 0$
    • $-c = c \implies 2c = 0 \implies c = 0$
    • $a = d$ (redundant with the first equation).

    2. Interpret the constraints. We have $a = d$ and $b = c = 0$. So the matrix simplifies to:

    $A = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} = a I,$

    a scalar multiple of the identity.

    3. Check the adjugate relation. For $A = aI$, the adjugate is:

    $\operatorname{adj}(aI) = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} = aI.$

    Indeed, $\operatorname{adj} A = A$ holds for any scalar $a$.

    4. Compute the sum.

    $a + b + c + d = a + 0 + 0 + a = 2a.$

    Watch out

    A common mistake is to forget that $\operatorname{adj} A = A$ imposes four equations, not just one. The off-diagonal conditions $-b = b$ and $-c = c$ immediately force $b = c = 0$, which students sometimes overlook.

    ---

    Why the answer is (A)

    The sum of all entries is $2a$, which is exactly option (A). The matrix must be a scalar multiple of the identity, and the trace (sum of diagonal entries) is $2a$, while the off-diagonal entries vanish.

    ✓Final answer

    The correct option is (A) $2a$.

12. CBSE 2024Set ANNUAL1 markMCQ
    Q.Let A be a non-singular square matrix of order $3 \times 3$. Then $|\text{Adj } A|$ is equal to -

    (a) $|A|$

    (b) $|A|^2$

    (c) $|A|^3$

    (d) $3|A|$

    ›Reveal solutionSolution

    Use the standard result $|\text{Adj }A| = |A|^{n-1}$ for an $n\times n$ non-singular matrix.

    For a non-singular square matrix $A$ of order $n \times n$, the determinant of its adjoint satisfies:

    $|\text{Adj }A| = |A|^{n-1}$

    Here $n = 3$ (order $3\times 3$), so:

    $|\text{Adj }A| = |A|^{3-1} = |A|^2$

    ✓Final answer

    Option (ii): $|A|^2$