Which one of the following is incorrect ? (a) If A is a square matrix of order n, and λ is a scalar, then Adj

Question 95 of 118

Q. Which one of the following is incorrect ?

(a) If A is a square matrix of order n, and

\lambda

is a scalar, then Adj

(\lambda A)=\lambda^n

(Adj A).

(b) Adjoint of a symmetric matrix is also a symmetric matrix.

(d) Adjoint of a diagonal matrix is also a diagonal matrix.

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Concept understanding — Adjoint and Inverse of a Matrix

For a square matrix $A=[a_{ij}]$ of order $n$ , the cofactor of $a_{ij}$ is the signed minor $A_{ij}=(-1)^{i+j}M_{ij}$ (the minor $M_{ij}$ is the determinant left after deleting row $i$ and column $j$ ). Replace every entry of $A$ by its cofactor to get the cofactor matrix; its transpose is the adjoint, $\operatorname{adj}A$ .

Theorem (the central identity). For every square matrix $A$ of order $n$ ,

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

This follows from Laplace expansion: a row's entries dotted with their own cofactors reproduce $|A|$ , while a row's entries dotted with a different row's cofactors always give $0$ -- so the product matrix is $|A|$ on the diagonal and $0$ off it.

Definition of the inverse. A square matrix $B$ with $AB=BA=I_n$ is called the inverse of $A$ , written $A^{-1}$ . The inverse, when it exists, is unique. $A^{-1}$ exists if and only if $A$ is non-singular ( $|A|\ne0$ ): dividing the central identity by $|A|$ (possible exactly when $|A|\ne0$ ) gives the working formula

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

A singular matrix ( $|A|=0$ ) has no inverse.

Worked illustration (order 2). For $A=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$ , the cofactors are $A_{11}=d,\ A_{12}=-c,\ A_{21}=-b,\ A_{22}=a$ , so $\operatorname{adj}A=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}$ (swap the diagonal entries, negate the off-diagonal ones) and $A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}$ whenever $ad-bc\ne0$ .

Standing laws of inverses (for non-singular $A,B$ of the same order, $\lambda\ne0$ a scalar):

$|A^{-1}|=\dfrac{1}{|A|}$ .
$(A^T)^{-1}=(A^{-1})^T$ .
$(\lambda A)^{-1}=\dfrac1\lambda A^{-1}$ .
Left/right cancellation: $AB=AC\Rightarrow B=C$ ; $BA=CA\Rightarrow B=C$ (pre/post-multiply by $A^{-1}$ ) -- this fails when $A$ is singular.
Reversal law: $(AB)^{-1}=B^{-1}A^{-1}$ (note the order flips, exactly as for transposes).
Double inverse: $(A^{-1})^{-1}=A$ .

Six adjoint identities (non-singular $A$ , order $n$ ): (i) $\operatorname{adj}(A^{-1})=(\operatorname{adj}A)^{-1}=\dfrac{A}{|A|}$ ; (ii) $|\operatorname{adj}A|=|A|^{n-1}$ ; (iii) $\operatorname{adj}(\operatorname{adj}A)=|A|^{n-2}A$ ; (iv) $\operatorname{adj}(\lambda A)=\lambda^{n-1}\operatorname{adj}A$ ; (v) $|\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2}$ ; (vi) $(\operatorname{adj}A)^T=\operatorname{adj}(A^T)$ ; and for two non-singular matrices of the same order, $\operatorname{adj}(AB)=(\operatorname{adj}B)(\operatorname{adj}A)$ (order reverses, exactly like the inverse and the transpose). …

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The adjoint scaling law is $\text{Adj}(\lambda A)=\lambda^{n-1}\text{Adj}(A)$ , not $\lambda^{n}\text{Adj}(A)$ , so statement (a) is the incorrect one.

Each entry of $\text{Adj}(A)$ is a cofactor of $A$ , obtained from an $(n-1)\times(n-1)$ minor.
If every entry of $A$ is scaled by $\lambda$ to form $\lambda A$ , each $(n-1)\times(n-1)$ minor (a determinant of $n-1$ rows/columns, each scaled by $\lambda$ ) scales by $\lambda^{n-1}$ .
Hence the correct identity is $\text{Adj}(\lambda A)=\lambda^{n-1}\text{Adj}(A)$ , so the claim in (a) that it equals $\lambda^{n}\text{Adj}(A)$ is false.
Statement (b) is a standard true property: the adjoint (transpose of the cofactor matrix) of a symmetric matrix is symmetric. …

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Q.Which one of the following is incorrect ?

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