Adjoint and Inverse of a Matrix

The adjoint of a matrix is needed to calculate the inverse of a matrix.

Adjoint of a Matrix

The adjoint of a matrix $A$ is the transpose of the cofactor matrix of $A$. The adjoint of a square matrix $A$ is denoted by adj $A$. Let $A=[a_{ij}]$ be a square matrix of order $n$.

Steps involved in finding the adjoint of a matrix are:

• Find the minor matrix $M$ of all the elements of matrix $A$.
• Find the cofactor matrix $C$ of all the minor elements of matrix $M$
• Find the adj $A$ by taking the transpose of the cofactor matrix $C$.

Adjoint of a $3\ X\ 3$ Matrix

$A={\begin{bmatrix}2&-1&3\\0&5&2\\1&-1&-2\end{bmatrix}}$ Step 1: Find the minor matrix $M$ of all the elements of matrix $A$.

Step 1: Find the minor matrix $M$ of all the elements of matrix $A$.

Row 1:

Minor of $2={\begin{vmatrix}5&2\\-1&-2\end{vmatrix}}=(-10-(-2))=-10+2=-8$

Minor of $-1={\begin{vmatrix}0&2\\1&-2\end{vmatrix}}=(0-2)=-2$

Minor of $3={\begin{vmatrix}0&5\\1&-1\end{vmatrix}}=(0-5))=-5$

Row 2:

Minor of $0={\begin{vmatrix}-1&3\\-1&-2\end{vmatrix}}=(2-(-3))=2+3=5$

Minor of $5={\begin{vmatrix}2&3\\1&-2\end{vmatrix}}=(-4-3)=-7$

Minor of $2={\begin{vmatrix}2&-1\\1&-1\end{vmatrix}}=(-2-(-1))=-1$

Row 3:

Minor of $1={\begin{vmatrix}-1&3\\5&2\end{vmatrix}}=(-2-15)=-17$

Minor of $-1={\begin{vmatrix}2&3\\0&2\end{vmatrix}}=(4-0)=4$

Minor of $-2={\begin{vmatrix}2&-1\\0&5\end{vmatrix}}=(10-0)=10$

Minor of Matrix $A$ is $M={\begin{bmatrix}-8&-2&-5\\5&-7&-1\\-17&4&10\end{bmatrix}}$

Step 2: Find the cofactor matrix $C$ of all the minor elements of matrix $M$

To find the cofactors of $3\ X\ 3$ matrix, the corresponding minors should be multiplied by the signs below according to their position.

$C={\begin{bmatrix}+&-&+\\-&+&-\\+&-&+\end{bmatrix}}$

Minor of Matrix $A$ is $M={\begin{bmatrix}-8&-2&-5\\5&-7&-1\\-17&4&10\end{bmatrix}}$

Cofactor of Matrix A is $C={\begin{bmatrix}-8&2&-5\\-5&-7&1\\-17&-4&10\end{bmatrix}}$

Step 3: Find the adj $A$ by taking the transpose of the cofactor matrix $C$

Adjoint of Matrix A is adj $A$ = Transpose of the Cofactor Matrix $C$ $={\begin{bmatrix}-8&-5&-17\\2&-7&-4\\-5&1&10\end{bmatrix}}$

Inverse of a Matrix

The inverse of a matrix $A$, which is represented as $A^{-1}$, is found using the adjoint of a matrix.

A^-1 = (1/|A|) × adj(A). Here, $A^{-1}=\left[{\frac {1}{\begin{vmatrix}A\end{vmatrix}}}\right]\times adj(A)$

Here

• ${\begin{vmatrix}A\end{vmatrix}}$= the determinant of $A$

• $adj(A)$= adjoint of $A$

Inverse of a $3\ X\ 3$ Matrix

determinant of $A=$ ${\begin{vmatrix}A\end{vmatrix}}={\begin{vmatrix}2&-1&3\\0&5&2\\1&-1&-2\end{vmatrix}}=2(-10-(-2))-(-1)(0-2)+3(0-5)=2(-8)-2-15=-33$

Adjoint of Matrix $A=$ $adj(A)$ $={\begin{bmatrix}-8&-5&-17\\2&-7&-4\\-5&1&10\end{bmatrix}}$

Inverse of matrix $A=$$A^{-1}=\left[{\frac {1}{\begin{vmatrix}A\end{vmatrix}}}\right]\times adj(A)$

$A^{-1}=\left[{\frac {1}{-33}}\right]\times {\begin{bmatrix}-8&-5&-17\\2&-7&-4\\-5&1&10\end{bmatrix}}={\begin{bmatrix}{\frac {8}{33}}&{\frac {5}{33}}&{\frac {17}{33}}\\-{\frac {2}{33}}&{\frac {7}{33}}&{\frac {4}{33}}\\{\frac {5}{33}}&-{\frac {1}{33}}&-{\frac {10}{33}}\end{bmatrix}}$