Find Matrix Inverse Using Row Operations
Introduction
We present examples on how to find the inverse of a matrix using the three row operations listed below:
- Interchange two rows
- Add a multiple of one row to another
- Multiply a row by a non zero constant
An Inverse of a Matrix Using Row Reduction - Calculator - Calculator
Inverse of a Matrix
Let A be an n × n matrix. If matrix A-1 is the inverse of matrix A , then we have
A A-1 = In = A-1 A
where In is the n × n identity matrixConsider the matrix equation A A-1 = In where A-1 is the unknown. To find the inverse A-1 , we start with the augmented matrix [ A | In ] and then row reduce it. If matrix A is invertible, the row reduction will end with an augmented matrix in the form
[ In | A-1 ]
where the inverse A-1 is the n × n on the right side of the augmented matrix [ In | A-1 ].
Examples with Solutions
Example 1
Find the inverse of matrix
Solution to Example 1
Write the augmented matrix [ A | I2 ]
Let R1 and R2 be the first and the second rows of the above augmented matrix.
Write the above augmented matrix in reduced row echelon form .
The above augmented matrix has the form [ I2 | A-1 ] and therefore A-1 is given by
Example 2
Find the inverse of matrix
Solution to Example 2
Write the augmented matrix [ A | I3 ]
Let R1, R2 and R3 be the first, the second and the third rows respectively of the above augmented matrix.
Write the above augmented matrix in reduced row echelon form .
\( \)\( \)\( \)
The above augmented matrix has the form \( [ I_3 | A^{-1} ] \) and therefore \( A^{-1} \) is given by
\( A^{-1} = \begin{bmatrix}
-\frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\
2 & -1 & 1 \\
-\frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{bmatrix}
\)
Example 3
Find the inverse of matrix \( A = \begin{bmatrix}
0 & 1 & -1 & 1\\
2 & 2 & 0 & -2\\
1 & 1 & -2 & 0 \\
0 & 1 & 2 & 0
\end{bmatrix}
\).
Solution to Example 3
Write the augmented matrix \( [ A | I_4 ] \)
\( \begin{bmatrix}
0 & 1 & -1 & 1 & | & 1 & 0 & 0 & 0\\
2 & 2 & 0 & -2 & | & 0 & 1 & 0 & 0 \\
1 & 1 & -2 & 0 & | & 0 & 0 & 1 & 0\\
0 & 1 & 2 & 0 & | & 0 & 0 & 0 & 1 \\
\end{bmatrix}
\)
Write the above augmented matrix in reduced row echelon form .
Interchange \( R_1 \) and \( R_3 \)
Interchange \( R_2 \) and \( R_4 \)
The above augmented matrix has the form \( [ I_4 | A^{-1} ] \) and therefore \( A^{-1} \) is given by
\( A^{-1} = \begin{bmatrix}
-4 & -2 & 5 & 3 \\
2 & 1 & -2 & -1 \\
-1 & -1/2 & 1 & 1 \\
-2 & -3/2 & 3 & 2
\end{bmatrix}
\)
Example 4
Find the inverse of matrix \( A = \begin{bmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
1 & 4 & 6
\end{bmatrix}
\).
Solution to Example 4
Write the augmented matrix \( [ A | I_3 ] \)
\( \begin{bmatrix}
1 & 2 & 0 & | & 1 & 0 & 0\\
0 & 1 & 3 & | & 0 & 1 & 0\\
1 & 4 & 6 & | & 0 & 0 & 1
\end{bmatrix}
\)
Write the above augmented matrix in reduced row echelon form .
\(
\begin{matrix}
\\
\\
\color{red}{R_3 - R_1}\\
\end{matrix}
\begin{bmatrix}
1 & 2 & 0 & | & 1 & 0 & 0\\
0 & 1 & 3 & | & 0 & 1 & 0\\
0 & 2 & 6 & | & -1 & 0 & 1 \\
\end{bmatrix}
\)
\(
\begin{matrix}
\\
\\
\color{red}{R_3 - 2 R_2}\\
\end{matrix}
\begin{bmatrix}
1 & 2 & 0 & | & 1 & 0 & 0\\
0 & 1 & 3 & | & 0 & 1 & 0\\
0 & 0 & 0 & | & -1 & -2 & 1 \\
\end{bmatrix}
\)
The last row of the original matrix (on the left side) is all zeros and therefore the rows in the orgiginal matrix \( A \) are not linearly independent and hence the given matrix is NOT invertible.