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Section 3.10 Inverses

The matrix inverse of a matrix \(A\text{,}\) denoted \(A^{-1}\text{,}\) is the matrix such that when multiplied by the matrix A the result is the identity matrix. (The identity matrix is the matrix with ones down the diagonal and zeroes everywhere else.)

For \(2\times2\) matrices, if

\begin{equation} A=\begin{pmatrix} a\amp b\\ c\amp d \end{pmatrix}\tag{3.10.1} \end{equation}


\begin{equation} A^{-1}={1\over\det(A)}\begin{pmatrix} d\amp -b\\ -c\amp a \end{pmatrix}\text{.}\tag{3.10.2} \end{equation}

For \(3\times3\) matrices \(B\text{,}\) \(B^{-1}\) is the transpose of the matrix made of all cofactors of \(B\text{,}\) divided by the determinant of \(B\text{.}\) This is easier said in symbols, so if

\begin{equation} B=\begin{pmatrix} a\amp b\amp c\\ d\amp e\amp f\\ g\amp h\amp i \end{pmatrix}\tag{3.10.3} \end{equation}


\begin{equation} B^{-1}={1\over\det(B)} \begin{pmatrix} \begin{vmatrix} e\amp f\\ h\amp i \end{vmatrix} \amp -\begin{vmatrix} b\amp c\\ h\amp i \end{vmatrix} \amp \begin{vmatrix} b\amp c\\ e\amp f \end{vmatrix}\\ \\ -\begin{vmatrix} d\amp f\\ g\amp i \end{vmatrix} \amp \begin{vmatrix} a\amp c\\ g\amp i \end{vmatrix}\amp -\begin{vmatrix} a\amp c\\ d\amp f \end{vmatrix}\\ \\ \begin{vmatrix} d\amp e\\ g\amp h \end{vmatrix}\amp -\begin{vmatrix} a\amp b\\ g\amp h \end{vmatrix} \amp \begin{vmatrix} a\amp b\\ d\amp e \end{vmatrix} \end{pmatrix}\tag{3.10.4} \end{equation}

In both cases, the inverse only exists if the determinant is nonzero.

Suppose that

\begin{equation} Q=\begin{pmatrix} 1\amp 2\\ 3\amp 4 \end{pmatrix}\text{.}\tag{3.10.5} \end{equation}

What is \(Q^{-1}\text{?}\) Verify that \(QQ^{-1}\) is the identity matrix.


Using (3.10.2), we have \(\det(Q)=(1)(4)-(2)(3)=-2\text{,}\) so

\begin{equation} Q^{-1} = -\frac12 \begin{pmatrix}4\amp -2\\-3\amp 1\end{pmatrix} = \begin{pmatrix}-2\amp 1\\\frac32\amp -\frac12\end{pmatrix}\tag{3.10.6} \end{equation}

so that

\begin{align} QQ^{-1} \amp= \begin{pmatrix}1\amp 2\\3\amp 4\end{pmatrix} \begin{pmatrix}-2\amp 1\\\frac32\amp -\frac12\end{pmatrix}\notag\\ \amp= \begin{pmatrix} (1)(-2)+(2)(\frac32)\amp(1)(1)+(2)(-\frac12) \\ (3)(-2)+(4)(\frac32)\amp(3)(1)+(4)(-\frac12) \\ \end{pmatrix} = \begin{pmatrix}1\amp0\\0\amp1\end{pmatrix}\tag{3.10.7} \end{align}