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

(aA)⁻¹ = (1)/(a) A⁻¹, where a is any real number and A is a square matrix.

NCERT Exemplar · Q49

(aA)^{-1} = \dfrac{1}{a} A^{-1}

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Concept understanding — Adjoint Matrix Property

The Adjoint Matrix and Its Central Property

You have a square matrix $A$ and want to find $A^{-1}$ . There is a clean route through the adjoint (or adjugate) of $A$ — a matrix built from the cofactors of $A$ that has a beautiful relationship with $A$ itself.

The intuition

For a $2\times2$ matrix you already know the inverse:

A=\begin{pmatrix}a&b\\c&d\end{pmatrix},\quad A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.

That second matrix, $\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$ , is exactly $\operatorname{adj}(A)$ . The adjoint is the object you multiply by $1/\det(A)$ to recover the inverse.

Note

In Indian exam usage, "adjoint" always means the adjugate — the transpose of the cofactor matrix — not the Hermitian conjugate.

Building the adjoint

For each entry $a_{ij}$ of an $n\times n$ matrix, the cofactor is

C_{ij}=(-1)^{i+j}M_{ij},

where $M_{ij}$ is the minor (the determinant left after deleting row $i$ and column $j$ ). Collect the cofactors into the cofactor matrix, then transpose:

\operatorname{adj}(A)=[C_{ij}]^{T},

so the $(i,j)$ entry of $\operatorname{adj}(A)$ is $C_{ji}$ .

The central property

A\cdot\operatorname{adj}(A)=\operatorname{adj}(A)\cdot A=\det(A)\,I_n.

Why? The $(i,i)$ entry of $A\,\operatorname{adj}(A)$ is $a_{i1}C_{i1}+\dots+a_{in}C_{in}$ — precisely the expansion of $\det(A)$ along row $i$ . An off-diagonal $(i,j)$ entry is the expansion of a determinant with two equal rows, which is $0$ .

Consequences

If $\det(A)\neq0$ : $\;A^{-1}=\dfrac{1}{\det(A)}\operatorname{adj}(A)$ .
If $\det(A)=0$ : $\;A\cdot\operatorname{adj}(A)=O$ , so the adjoint annihilates $A$ .
Order fact: $\;\det\big(\operatorname{adj}(A)\big)=\det(A)^{\,n-1}$ . …

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False — the identity $(aA)^{-1}=\tfrac{1}{a}A^{-1}$ is valid only when $a\neq0$ (and $A$ is invertible); the word "any" wrongly includes $a=0$ , where $aA$ has no inverse.

What the statement claims

It says $(aA)^{-1}=\tfrac{1}{a}A^{-1}$ for any real number $a$ . To decide true or false we must check whether it holds for every allowed $a$ .

Where it is true

When $a\neq0$ and $A$ is invertible, the formula is a genuine identity. Verify by multiplying:

(aA)\left(\tfrac{1}{a}A^{-1}\right) = \left(a\cdot\tfrac{1}{a}\right)(AA^{-1}) = 1\cdot I = I,

and similarly on the other side, so $\tfrac{1}{a}A^{-1}$ really is the inverse of $aA$ .

Where it breaks …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KEAM 2022Set eng-2022-P2-B14 marksMCQ
$A$
›Reveal solutionSolution
$\mathrm{adj}(A)=\begin{bmatrix}10&4\\-2&-1\end{bmatrix}$ .

Concept and Intuition

Since $A^{-1}=\frac{\mathrm{adj}(A)}{|A|}$ , we have $\mathrm{adj}(A)=|A|\,A^{-1}$ ; find $|A|$ from $|A^{-1}|=1/|A|$ .

Step-by-Step Solution

$A^{-1}=\frac12\begin{bmatrix}-10&-4\\2&1\end{bmatrix}$ .

$|A^{-1}|=\left(\frac12\right)^2\big((-10)(1)-(-4)(2)\big)=\frac14(-10+8)=-\frac12$ .

$|A|=\frac{1}{|A^{-1}|}=-2$ . …
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KEAM 2024Set eng-2024-06084 marksMCQ
$A=\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}$
›Reveal solutionSolution

Use $A\,(\mathrm{adj}\,A)=(\det A)I$ with $\det A=1$ .

For $A=\begin{bmatrix}3 & 7\\ 2 & 5\end{bmatrix}$ , $\det A=15-14=1$ .

The fundamental identity is $A\,(\mathrm{adj}\,A)=(\det A)\,I=1\cdot I=I$ .

Therefore …
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KEAM 2024Set eng-2024-06074 marksMCQ
$A$
›Reveal solutionSolution

Use $|kA|=k^n|A|$ and $|\mathrm{adj}\,A|=|A|^{n-1}$ for $n=3$ . …

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KEAM 2025Set eng-2025-04234 marksMCQ
$A$
›Reveal solutionSolution
With |B| = 135, |adj B| = 135^2, so |adj B|/|3A| = 135.

Concept and Intuition

For an n x n matrix, |adj B| = |B|^(n-1); here n = 3 so it is |B|^2. Also |3A| = 3^3 |A| = |B|. The ratio collapses to |B|.

Step-by-Step Solution

|B| = |3A| = 3^3 * |A| = 27 * 5 = 135.

|adj B| = |B|^(3-1) = 135^2.

|adj B| / |3A| = 135^2 / 135 = 135. …
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KEAM 2025Set eng-2025-04284 marksMCQ
$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -2 \end{bmatrix}$
›Reveal solutionSolution

For an $n\times n$ matrix, $|\mathrm{adj}(\mathrm{adj}\,A)|=|A|^{(n-1)^2}$ . Here $n=3$ , $|A|=-4$ , so $(-4)^4=256$ .

$A$ is upper triangular, so $|A|=1\cdot2\cdot(-2)=-4$ .

Using $|\mathrm{adj}\,A|=|A|^{n-1}$ twice, for $n=3$ : …
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KEAM 2025Set eng-2025-04274 marksMCQ
$A$
›Reveal solutionSolution

$|\text{adj}(\text{adj}\,A)|=|A|^{(n-1)^2}$ ; with $n=3,\ |A|=9$ this is $9^4=6561$ .

For an $n\times n$ matrix, $|\text{adj}\,A|=|A|^{n-1}$ . Applying twice, …
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KEAM 2024Set eng-2024-06094 marksMCQ
$B=\begin{pmatrix} 1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4 \end{pmatrix}$
›Reveal solutionSolution

Since $B=\text{adj}A$ for a $3\times3$ matrix, $|B|=|A|^2=16$ ; solving $2\alpha-6=16$ gives $\alpha=11$ .

For an $n\times n$ matrix, $|\text{adj}A|=|A|^{n-1}$ . With $n=3$ and $|A|=4$ , $|B|=4^2=16$ .

Compute $\det B$ for $B=\begin{pmatrix}1&\alpha&3\\1&3&3\\2&4&4\end{pmatrix}$ : …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$
›Reveal solutionSolution
The adjoint has trace 9 and b1·a2 = 3, so the ratio is 3.

Concept and Intuition

Compute the adjoint of the given matrix (transpose of the cofactor matrix); then read off the required diagonal sum and specific off-diagonal entries.

Step-by-Step Solution

For M = [[1,3,3],[1,4,3],[1,3,4]], det(M) = 1.

Cofactors give adj(M) = [[7,−3,−3],[−1,1,0],[−1,0,1]].

a1 = 7, b2 = 1, c3 = 1 ⇒ a1+b2+c3 = 9 (trace). …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$A = \begin{bmatrix} 5a & -b \\ 3 & 2 \end{bmatrix}$
›Reveal solutionSolution
The condition forces det(A) = 13, which is ≥ 0.

Concept and Intuition

A·adj(A) = det(A)·I always, so the condition A·adj A = AA^T means AA^T = det(A)·I. This requires the off-diagonal of AA^T to vanish and the diagonal to equal det(A).

Step-by-Step Solution

A = [[5a, −b],[3, 2]]. AA^T = [[25a²+b², 15a−2b],[15a−2b, 13]].

det(A)·I = [[10a+3b, 0],[0, 10a+3b]].

Off-diagonal: 15a − 2b = 0 ⇒ b = 7.5a.

Lower-right: 13 = det(A) = 10a + 3b ⇒ 32.5a = 13 ⇒ a = 0.4, b = 3. …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$4\times 4$
›Reveal solutionSolution
det(adj A) = det(A)³ = 5 has 3 complex cube-root solutions for det(A).

Concept and Intuition

For an n×n matrix, det(adj A) = (det A)^(n−1). With complex entries det(A) may be any complex number satisfying the resulting equation.

Step-by-Step Solution

n = 4, so det(adj A) = det(A)^(4−1) = det(A)³.

det(A)³ = 5. …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$A = \begin{vmatrix} 2 & 1 & -2 \\ 1 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix}$
›Reveal solutionSolution
$|B|=1024$ .

Concept and Intuition

For an $n\times n$ matrix, $|kM|=k^n|M|$ and $|\mathrm{adj}(A)|=|A|^{n-1}$ .

Step-by-Step Solution

Compute $|A|=2(1\cdot3-(-1)\cdot0)-1(1\cdot3-(-1)\cdot1)+(-2)(1\cdot0-1\cdot1)$ .

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

$B=|A|\,\mathrm{adj}(A)=4\,\mathrm{adj}(A)$ , so $|B|=4^3\,|\mathrm{adj}(A)|$ (since $n=3$ ).

$|\mathrm{adj}(A)|=|A|^{n-1}=4^2=16$ .

$|B|=64\times16=1024$ . …
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←Previous questionQ48 — (A³)⁻¹ = (A⁻¹)³, where A is a square matrix and |A| ≠ 0.→Next questionQ50 — |A⁻¹| ≠ |A|⁻¹, where A is a non-singular matrix.

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

Concept understanding — Adjoint Matrix Property

The Adjoint Matrix and Its Central Property

The intuition

Building the adjoint

The central property

Consequences

What the statement claims

Where it is true

Where it breaks …

Method: Testing a "For Any Value of $a$ " Matrix Identity Claim

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.(aA)−1=1aA−1(aA)^{-1} = \dfrac{1}{a} A^{-1}(aA)−1=a1​A−1, where aaa is any real number and AAA is a square matrix.

What the statement claims

Where it is true

Where it breaks …

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

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