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Chemistry LibreTexts

15.5: Matrix Inversion

  • Page ID
    106900
  • The inverse of a square matrix \(\mathbf{A}\), sometimes called a reciprocal matrix, is a matrix \(\mathbf{A}^{-1}\) such that \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}\), where \(\mathbf{I}\) is the identity matrix.

    It is easy to obtain \(\mathbf{A}^{-1}\) in the case of a \(2\times 2\) matrix:

    \[\mathbf{A}=\begin{pmatrix} a&b \\ c&d \end{pmatrix};\;\mathbf{A}^{-1}=\begin{pmatrix} e&f \\ g&h \end{pmatrix} \nonumber\]

    \[\begin{pmatrix} a&b \\ c&d \end{pmatrix}\begin{pmatrix} e&f \\ g&h \end{pmatrix}=\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} \nonumber\]

    \[\label{eq:matrices_inverse1} ae+bg=1\]

    \[\label{eq:matrices_inverse2} af+bh=0\]

    \[\label{eq:matrices_inverse3} ce+dg=0\]

    \[\label{eq:matrices_inverse4} cf+dh=1\]

    From Equations \ref{eq:matrices_inverse1} and \ref{eq:matrices_inverse3}: \(g=(1-ae)/b=-ce/d\rightarrow ae=cbe/d+1\rightarrow e\left(a-cb/d\right)=1\rightarrow e\left(ad-cb\right)=d\rightarrow e=d/(ad-cb)\). You can obtain expressions for \(f,g\) and \(h\) in a similar way to obtain:

    \[\mathbf{A}^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d&-b \\ -c&a \end{pmatrix} \nonumber\]

    Notice that the term \((ad-bc)\) is the determinant of \(\mathbf{A}\), and therefore \(\mathbf{A}^{-1}\) exists only if \(|\mathbf{A}|\neq 0\). In other words, the inverse of a singular matrix is not defined.

    If you think about a square matrix as an operator, the inverse “undoes” what the original matrix does. For example, the matrix \(\begin{pmatrix} -2&0 \\ 0&1 \end{pmatrix}\), when applied to a vector \((x,y)\), gives \((-2x,y)\):

    \[\begin{pmatrix} -2&0 \\ 0&1 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} -2x\\ y \end{pmatrix} \nonumber\]

    The inverse of \(\mathbf{A}\), when applied to \((-2x,y)\), gives back the original vector, \((x,y)\):

    \[\mathbf{A}^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d&-b \\ -c&a \end{pmatrix}\rightarrow \mathbf{A}^{-1}= -\frac{1}{2}\begin{pmatrix} 1&0 \\ 0&-2 \end{pmatrix} \nonumber\]

    \[-\frac{1}{2}\begin{pmatrix} 1&0 \\ 0&-2 \end{pmatrix} \begin{pmatrix} -2x\\ y \end{pmatrix}=\begin{pmatrix} x\\ y \end{pmatrix} \nonumber\]

    It is of course possible to calculate the inverse of matrices of higher dimensions, but in this course you will not be required to do so by hand.

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