Adjoint and Inverse of a Matrix - Practice Questions & MCQ

Adjoint of a Matrix

Adjoint of a matrix A is the transpose of cofactor matrix of the matrix A . Cofactor matrix of matrix A is a matrix which has same order as that of A and has element $C_{i j}$ in place of $a_{i j}$. E.g.,

$
\text { let } A=\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]
$

and let cofactor of every element is

$
\left[\begin{array}{lll}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{array}\right]
$

then Adjoint of A is

$
A^{\prime}=\left[\begin{array}{lll}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{array}\right]^{\prime}=\left[\begin{array}{lll}
C_{11} & C_{21} & C_{31} \\
C_{12} & C_{22} & C_{32} \\
C_{13} & C_{23} & C_{33}
\end{array}\right]
$
 

Properties of adjoint of a matrix

1. If A is a square matrix of order n, then
$(\operatorname{Adj} A) A=A(\operatorname{Adj} A)=|A| \mathbb{I}_n$, or product of a matrix and its adjoint is commutative.
Proof:
Let, $\mathrm{A}=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$, then adj $\mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]$
where, $A_{i j}$ is co - factor of $a_{i j}$
Since the sum of the product of elements of a row (or a column) with corresponding cofactors is equal to $|\mathrm{A}|$ and otherwise zero, we have

$
\mathrm{A}(\operatorname{adj} \mathrm{~A})=\left[\begin{array}{ccc}
|A| & 0 & 0 \\
0 & |A| & 0 \\
0 & 0 & |A|
\end{array}\right]=|\mathrm{A}|\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=|\mathrm{A}| \mathrm{I}
$
If $A$ is a singular matrix of order $n$, then

$
(\operatorname{Adj} A) A=A(\operatorname{Adj} A)=O \text { (null matrix }) \quad(\text { As }|A|=0)
$

2. If $A$ be non-singular square matrix of order $n$, then $|\operatorname{Adj} A|=|A|^{n-1}$

Proof:

$
\mathrm{A}(\operatorname{Adj} \mathrm{~A})=|\mathrm{A}| \mathrm{I}_{\mathrm{n}}
$
Taking determinants on both sides

$
\begin{aligned}
& |\mathrm{A}(\operatorname{Adj} \mathrm{~A})|=\left||\mathrm{A}| \mathrm{I}_{\mathrm{n}}\right| \\
& |\mathrm{A}||(\operatorname{Adj} \mathrm{A})|=|\mathrm{A}|^{\mathrm{n}} \\
& |(\operatorname{adj} \mathrm{~A})|=|\mathrm{A}|^{\mathrm{n}-1}
\end{aligned}
$

3. If $A$ and $B$ are square matrices of order $n$, then, $\operatorname{adj}(A B)=(\operatorname{adj} B)(\operatorname{adj} A)$
4. If $A$ is a square matrix of order $n$, then, $(\operatorname{adj} A)^{\prime}=\operatorname{adj} A^{\prime}$

Properties of adjoint of matrix

4. If $A$ be a square non-singular matrix of order $n$, then $\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A$

Proof:

$
\begin{aligned}
& A(\operatorname{adj} A)=|A| \mathbb{I}_n \\
& \operatorname{replace} A \operatorname{by} \operatorname{adj} A, \text { then } \\
& (\operatorname{adj} A)(\operatorname{adj}(\operatorname{adj} A))=|\operatorname{adj} A| \mathbb{I}_n=|A|^{\mathrm{n}-1} \mathbb{I}_{\mathrm{n}} \\
& \text { Pre }-\operatorname{multiplying} \operatorname{both} \operatorname{sides} \text { by matrix } A \text {, then } \\
& A(\operatorname{adj} A)(\operatorname{adj}(\operatorname{adj} A))=A \mathbb{I}_n|A|^{\mathrm{n}-1}=A|A|^{\mathrm{n}-1} \\
& |A| \mathbb{I}_{\mathrm{n}}(\operatorname{adj}(\operatorname{adj} A))=A|A|^{\mathrm{n}-1} \\
& (\operatorname{adj}(\operatorname{adj} A))=A|A|^{\mathrm{n}-2}=|A|^{\mathrm{n}-2} \mathrm{~A}
\end{aligned}
$

5. If A is non-singular square matrix, then, $|\operatorname{adj}(\operatorname{adj} \mathrm{A})|=|\mathrm{A}|^{(\mathrm{n}-1)^2}$

Proof: from the previous property, we know that

$
\operatorname{adj}(\operatorname{adj} \mathrm{A})=|\mathrm{A}|^{(\mathrm{n}-2)} \mathrm{A}
$
Taking determinant on both sides,

$
\begin{aligned}
& |\operatorname{adj}(\operatorname{adj} \mathrm{A})|=\left||\mathrm{A}|^{(\mathrm{n}-2)} \mathrm{A}\right|=|\mathrm{A}|^{\mathrm{n}(\mathrm{n}-2)}|\mathrm{A}|=|\mathrm{A}|^{(\mathrm{n}-1)^2} \\
& \text { (using } \left.|k A|=k^n|A|\right)
\end{aligned}
$

6. If $A$ be a square matrix of order $n$ and $m$ be any natural number, then $\left(\operatorname{adj} A^m\right)=(\operatorname{adj} A)^m$
7. If $A$ is a square matrix of order $n$ and $k$ be a scalar, then, $\operatorname{adj}(k A)=k^{n-1} \cdot(\operatorname{adj} A)$