Question & Answer
QUESTION

Prove the given theorem for a matrix A:
$A\left( adj.A \right)=\left( adj.A \right).A=\left| A \right|{{I}_{n}}$

ANSWER Verified Verified
Hint: Here, we can use the formula for finding the inverse of a matrix which is given as ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$. By applying some operations to the formula of the inverse of the matrix, we can prove the given theorem.

Complete step-by-step answer:

The adjoint of a matrix A is defined as the transpose of the cofactor matrix C of A which can be given as:
$adj\left( A \right)={{C}^{T}}$

Suppose, we have a $n\times n$ matrix A. The (i, j) minor of A, denoted by ${{M}_{ij}}$ , is the determinant of the $\left( n-1 \right)\times \left( n-1 \right)$ matrix that results from deleting row i and column j of A. The cofactor matrix of A is the $n\times n$ matrix C whose (i, j) entry is the (i, j) cofactor of A, which is the (i, j) – minor times a sign factor:
$C={{\left\{ {{\left( -1 \right)}^{i+j}}{{M}_{ij}} \right\}}_{1\le i,j\le n}}$

Determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformation described the matrix. It is denoted by det(A) or|A|.

Any matrix A of order $n\times n$ is called invertible if there exists an $n\times n$ square matrix B such that $AB=BA={{I}_{n}}$, where ${{I}_{n}}$ denotes the $n\times n$ identity matrix.
We know that the inverse of a matrix A is given as:
 ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}...........\left( 1 \right)$
On multiplying equation (1) by A on both sides, we get:
$A.{{A}^{-1}}=\dfrac{A.adj\left( A \right)}{\left| A \right|}$
Or, we can also write it as:
$\begin{align}
  & {{A}^{-1}}.A=\dfrac{adj\left( A \right).A}{\left| A \right|} \\
 & \Rightarrow I=\dfrac{adj\left( A \right).A}{\left| A \right|} \\
 & \Rightarrow I=\dfrac{A.adj\left( A \right)}{\left| A \right|} \\
\end{align}$

From above, we can say that:
$\left| A \right|.I=A.adj\left( A \right)$ and $\left| A \right|.I=adj\left( A \right).A$
Therefore, $A\left( adjA \right)=\left( adjA \right).A=\left| A \right|.{{I}_{n}}$
Hence, this is the required proof of the theorem.

Note: Students should note that whenever a matrix is multiplied with its inverse, we get an identity matrix. Students should also remember the formula for finding the inverse of a matrix in order to prove the given theorem.