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Inverse Matrix Formula

Inverse of a matrix is an important operation in the case of a square matrix. It is applicable only for a square matrix. To calculate the inverse, one has to find out the determinant and adjoint of that given matrix. Adjoint is given by the transpose of cofactor of the particular matrix. The formula to find out the inverse of a matrix is given as,
Inverse Matrix Formula

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Inverse Matrix Problems

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Let us discuss how to find out inverse of a matrix.

Solved Examples

Question 1: Find the inverse of $\begin{bmatrix}
5 &6 \\
 -1& 2
\end{bmatrix}$?

Solution:
 
Let A = $\begin{bmatrix}
5 &6 \\
 -1& 2
\end{bmatrix}$ be the given matrix.

$A^{-1}$ = $\frac{adj(A)}{\left | A \right |}$

To find out the adj(A), first we have to find out cofactor(A).

So, cofactor(A) = $\begin{bmatrix}
2&-6 \\
  1& 5
\end{bmatrix}$

adj(A) = $[cofactor(A)]^{T}$

adj(A) = $[cofactor(A)]^{T}$ = $\begin{bmatrix}
2&-6 \\
  1& 5
\end{bmatrix}^{T}$

adj(A) = $\begin{bmatrix}
2&1 \\
  -6& 5
\end{bmatrix}$

Then, $\left | A \right|$ = (5$\times$2)-(-1$\times$6) = 10 + 6 = 16

$A^{-1}$ = $\frac{adj(A)}{\left | A \right |}$ = $\frac{\begin{bmatrix}
2&1 \\
  -6& 5
\end{bmatrix}}{16}$


 

Question 2: Find out the inverse of  $\begin{bmatrix}
1 &-1  &2 \\
 4&0  &6 \\
 0&1  &-1
\end{bmatrix}$?

Solution:
 
Let A = $\begin{bmatrix}
1 &-1  &2 \\
 4&0  &6 \\
 0&1  &-1
\end{bmatrix}$ be the given matrix.

$A^{-1}$ = $\frac{adj(A)}{\left | A \right |}$

To find out the adj(A), first we have to find out cofactor(A).
a11 = -6, a12 = 4, a13 = 4
a21 = 1, a22 = -1, a23 = -1
a13 = -6, a32 = 2, a33 = 4

So, cofactor(A) = $\begin{bmatrix}
  -6&4  &4 \\
  1&-1  &-1 \\
  -6&2  &4
\end{bmatrix}$

adj(A) = $[cofactor(A)]^{T}$

adj(A) = $[cofactor(A)]^{T}$ = $\begin{bmatrix}
  -6&4  &4 \\
  1&-1  &-1 \\
  -6&2  &4
\end{bmatrix}^{T}$

adj(A) = $\begin{bmatrix}
  -6&1  &-6 \\
  4&-1  &2 \\
  4&-1  &4
\end{bmatrix}$

Then, $\left | A \right |$ = 1(0-6)+1(-4-0)+2(4-0) = -6-4+8 = -2

$A^{-1}$ = $\frac{adj(A)}{\left | A \right |}$ = $\frac{\begin{bmatrix}
  -6&1  &-6 \\
  4&-1  &2 \\
  4&-1  &4
\end{bmatrix}}{-2}$


 

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