Multiplicative Inverse of 3×3 Matrices
Unfortunately, we do not have a formula similar to the one for a [latex]2\text{}\times \text{}2[/latex] matrix to find the inverse of a [latex]3\text{}\times \text{}3[/latex] matrix. But, we can still find the inverse by using a systematic approach involving row operations. This method requires augmenting the given matrix with the identity matrix and performing a series of row operations to transform the original matrix into the identity matrix. The resulting augmented matrix will then have the inverse of the original matrix on its right side.
How To: Given a [latex]3\times 3[/latex] matrix, find the inverse
Write the original matrix augmented with the identity matrix on the right.
Use elementary row operations so that the identity appears on the left.
What is obtained on the right is the inverse of the original matrix.
Use matrix multiplication to show that [latex]A{A}^{-1}=I[/latex] and [latex]{A}^{-1}A=I[/latex].
Given the [latex]3\times 3[/latex] matrix [latex]A[/latex], find the inverse.
[latex]A=\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right][/latex]
Show Solution
Augment [latex]A[/latex] with the identity matrix, and then begin row operations until the identity matrix replaces [latex]A[/latex]. The matrix on the right will be the inverse of [latex]A[/latex].
[latex]\left[\begin{array}{ccc|ccc}\hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 3& \hfill 3& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 1\end{array}\right]\stackrel{\text{Interchange }{R}_{2}\text{ and }{R}_{1}}{\to }\left[\begin{array}{ccc|ccc}\hfill 3& \hfill 3& \hfill 1& \hfill 0& \hfill 1& \hfill 0\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 1\end{array}\right][/latex]
[latex]-{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 4& \hfill 1& \hfill 0& \hfill 0& \hfill 1\end{array}\right][/latex]
[latex]-{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\end{array}\right][/latex]
[latex]{R}_{3}\leftrightarrow {R}_{2}\to \left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\\ \hfill 2& \hfill 3& \hfill 1& \hfill 1& \hfill 0& \hfill 0\end{array}\right][/latex]
[latex]-2{R}_{1}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 3& \hfill 1& \hfill 3& \hfill -2& \hfill 0\end{array}\right][/latex]
[latex]-3{R}_{2}+{R}_{3}={R}_{3}\to\left[\begin{array}{ccc|ccc}\hfill 1& \hfill 0& \hfill 0& \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill 1& \hfill 0& \hfill -1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1& \hfill 6& \hfill -2& \hfill -3\end{array}\right][/latex]
Thus,
[latex]{A}^{-1}=B=\left[\begin{array}{ccc}-1& 1& 0\\ -1& 0& 1\\ 6& -2& -3\end{array}\right][/latex]
Analysis of the Solution
To prove that [latex]B={A}^{-1}[/latex], let’s multiply the two matrices together to see if the product equals the identity, if [latex]A{A}^{-1}=I[/latex] and [latex]{A}^{-1}A=I[/latex].
[latex]\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ A{A}^{-1} & =\left[\begin{array}{ccc}2& 3& 1\\ 3& 3& 1\\ 2& 4& 1\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}2\left(-1\right)+3\left(-1\right)+1\left(6\right)& 2\left(1\right)+3\left(0\right)+1\left(-2\right)& 2\left(0\right)+3\left(1\right)+1\left(-3\right)\\ 3\left(-1\right)+3\left(-1\right)+1\left(6\right)& 3\left(1\right)+3\left(0\right)+1\left(-2\right)& 3\left(0\right)+3\left(1\right)+1\left(-3\right)\\ 2\left(-1\right)+4\left(-1\right)+1\left(6\right)& 2\left(1\right)+4\left(0\right)+1\left(-2\right)& 2\left(0\right)+4\left(1\right)+1\left(-3\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\hfill \end{array}[/latex]
Given the [latex]3\times 3[/latex] matrix [latex]A[/latex], find the inverse.
[latex]A = \left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right][/latex]
Show Answer
Augment the matrix [latex]A[/latex] with the identity matrix:
[latex]\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}\right][/latex]
Perform row operations to try to transform the left side into the identity matrix:
Subtract 4 times row 1 from row 2:
[latex]R_2 \leftarrow R_2 - 4R_1[/latex]
[latex]\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}\right][/latex]
Subtract 7 times row 1 from row 3:
[latex]R_3 \leftarrow R_3 - 7R_1[/latex]
[latex]\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 0 & -6 & -12 & -7 & 0 & 1 \end{array}\right][/latex]
Subtract 2 times row 2 from row 3:
[latex]R_3 \leftarrow R_3 - 2R_2[/latex]
[latex]\left[\begin{array}{ccc|ccc} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & -3 & -6 & -4 & 1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 1 \end{array}\right][/latex]
The third row of the left side of the augmented matrix is all zeros. This indicates that the original matrix A is singular, meaning it does not have an inverse .