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

Uttar Pradesh UpmspShort· 1mImportance★★★★★

88% · 129/146 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: $\;\det\big(\operatorname{adj}(A)\big)=\det(A)^{\,n-1}$ . …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

←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

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 …

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days