AcademyLinear Systems
Academy
Inverse Computation
Level 1 - Linear Algebra topic page in Linear Systems.
Principle
Gauss-Jordan elimination can compute the inverse of a square matrix. The method row-reduces the augmented matrix \([A\mid I]\). If the left side becomes \(I\), then the right side becomes \(A^{-1}\).
Notation
\(A\)
a square matrix
\(I_n\)
the n by n identity matrix
\(A^{-1}\)
the inverse of A
\([A\mid I_n]\)
the augmented matrix used to compute an inverse
\(\sim\)
row-equivalent
The Core Method
To compute an inverse:
- Form \([A\mid I_n]\).
- Use Gauss-Jordan elimination on the whole augmented matrix.
- If the left block reduces to \(I_n\), read the inverse from the right block.
Inverse computation
\[[A\mid I_n]\sim[I_n\mid A^{-1}]\]
If the left block cannot be reduced to \(I_n\), then \(A\) is singular and \(A^{-1}\) does not exist.
Worked Cases
Question
Use row reduction to invert \(A=\begin{pmatrix}1&2\\3&7\end{pmatrix}\).
Answer
Start with \([A|I]=\begin{pmatrix}1&2&|&1&0\\3&7&|&0&1\end{pmatrix}\). Apply \(R_2\leftarrow R_2-3R_1\): \(\begin{pmatrix}1&2&|&1&0\\0&1&|&-3&1\end{pmatrix}\). Clear above the second pivot with \(R_1\leftarrow R_1-2R_2\): \(\begin{pmatrix}1&0&|&7&-2\\0&1&|&-3&1\end{pmatrix}\). Therefore \(A^{-1}=\begin{pmatrix}7&-2\\-3&1\end{pmatrix}\).
Examples
Question
Verify that \(\begin{pmatrix}7&-2\\-3&1\end{pmatrix}\) is the inverse of \(\begin{pmatrix}1&2\\3&7\end{pmatrix}\).
Answer
Multiply: \(\begin{pmatrix}1&2\\3&7\end{pmatrix}\begin{pmatrix}7&-2\\-3&1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). The reverse product also gives \(I_2\). Therefore the computed matrix is the inverse.