If, for a given square matrix \(A\), a matrix \(B\) exists such that \(A\,B=B\,A=I\), then \(B\) is called the inverse of \(A\). The notation for this matrix is \( A^{-1}\). In this case we say that \(A\) is invertible, or regular.
If a matrix is not invertible, then we speak of a singular matrix.
- If \(A\, B = I\), then \(B\) the inverse of \(A\).
- If \(B\, A = I\), then \(B\) the inverse of \(A\).
We speak of the inverse and not of an inverse because the inverse of an invertible matrix is unique. For, suppose that both \(B\) and \(C\) are square matrices that satisfy the conditions for an inverse of \(A\): \[A\,B = B\,A= I\quad\text{and}\quad A\,C = C\,A = I\] Then \[ B = B\, I = B \,( A\, C) = (B\, A)\, C = I\, C=C\] so \(B = C\).
Later we will give a simple proof of the fact that for square matrices \( A\,B=I\) we also have \(B\,A=I\). Because matrix multiplication is not commutative, this property is not obvious.
The proof is based on the interpretation of an #(m\times n)#-matrix #A# as a map assigning to a column vector #x# of length #n# the column vector #A x# of length #m#. If #m=n# and #A\,B=I_n# for a square matrix #B#, then #B# is injective and so (as we shall see later) surjective, so that, for each column vector #x#, there is a column vector #y# with #x = B y#. Consequently, \[B\,A x = B\,A\,B y = By = x\] Because the matrix #B\, A# is completely determined by its images on all the column vectors of length #n#, this gives #B\, A = I_n#.
Like the number zero, the (square) zero matrix \(O\) has no inverse: for every \(B\) of the same size as #O# we have \(O\,B=B\,O=O\). But unlike the case of real numbers, there are non-invertible matrices distinct from the zero matrix. Two examples of square singular matrices are \[\matrix{1 & 1\\ 1 & 1}\text{ and }\matrix{1 & 2\\ 2 & 4}\]
\(\matrix{-5 & -2 \\ 3 & 1 \\ }\) is the inverse of \(\matrix{1 & 2 \\ -3 & -5 \\ }\).
After all, \[\matrix{1 & 2 \\ -3 & -5 \\ } \,\matrix{-5 & -2 \\ 3 & 1 \\ } = \matrix{1 & 0\\ 0 & 1}\] Verify this by yourself.
If #B\,A = I#, then #B# is called a left inverse of #A# and if #A\,B = I#, then #B# is called a right inverse of #A#. Thus, the last two statements of the theorem contend that a left inverse and a right inverse of #A# are also inverses of #A#.
Here are some calculation rules for the inverse of a matrix.
- Let \(A\) and \(B\) be \((n\times n)\)-matrices. If \(A\) and \(B\) are invertible, then their product, \(A\, B\), is also invertible, with inverse \[(A\, B)^{-1}=B^{-1} A^{-1}\]
- If \(A\) is an invertible matrix, then its transposed matrix, \(A^{\top}\), is also invertible, with inverse \[ \left(A^{\top}\right)^{-1}=\left(A^{-1}\right)^{\top}\]
The first rule follows from \[(A\,B)\,(B^{-1}A^{-1})=A\,(B\,B^{-1})\,A^{-1}=A\,(I\,A^{-1})=A\,A^{-1}=I \]
and similarly \((B^{-1}A^{-1})(A\,B)=I\).
The second rule can also be verified directly: \[ \begin{aligned} A^{\top}\left(A^{-1}\right)^{\top} &=\left(A^{-1}A\right)^{\top}=I^{\top}=I\\ \\ \left(A^{-1}\right)^{\top} A^{\top} &=\left(A\,A^{-1}\right)^{\top}=I^{\top}=I \end{aligned}\]
For square matrices of dimension #2# we have an explicit expression for the inverse:
If \(A=\matrix{a & b \\ c & d}\), then \(A\) is invertible if and only if \(a\,d-b\,c\neq 0\). In this case \[A^{-1}=\frac{1}{a\,d-b\,c}\matrix{d & -b \\ -c & a}\]
The expression \(a\,d-b\,c\) is called the determinant of the matrix \(A=\matrix{a & b \\ c & d}\) and is denoted as \(\text{det}(A)\).
In order to determine the inverse of the general \((2\times 2)\)-matrix \( A=\matrix{a & b \\ c & d}\) we need to find scalars \(p, q, r, s\) such that \[ \matrix{a & b \\ c & d} \matrix{p & q \\ r & s}= \matrix{1 & 0 \\ 0 & 1}\] In other words: \[ \matrix{a\,p+b\,r & a\,q +b\,s \\ c\,p+d\,r & c\,q +d\,s }= \matrix{1 & 0 \\ 0 & 1}\] So we are dealing with two systems of equations, each with two unknowns, namely \[\lineqs{a\,p+b\,r\!\!\! &= 1 \\ c\,p+d\,r\!\!\! &= 0}\qquad \lineqs{a\,q+b\,s \!\!\! &= 1 \\ c\,q+d\,s \!\!\! &= 0}\] The associated augmented matrices are \[\left(\begin{array}{rr|r} a & b & 1\\ c & d & 0\end{array}\right)\quad\text{and}\quad \left(\begin{array}{rr|r} a & b & 0\\ c & d & 1\end{array}\right)\] Using elementary row operations we can reduce the augmented matrices to echelon form: \[\left(\begin{array}{rr|r} a & b & 1\\ 0 & a\,d-b\,c & -c\end{array}\right)\quad\text{and}\quad \left(\begin{array}{rr|r} a & b & 0\\ 0 & a\,d-b\,c & a\end{array}\right)\] Note that we apply the same row operations left and right. There is only one solution if and only if \(a\,d-b\,c\neq 0\), in which case the reduced echelon form provides the solution \[\matrix{p & q \\ r & s}= \frac{1}{a\,d-b\,c}\matrix{d & -b \\ -c & a}\]
The theorem says that the matrix #A# is invertible if and only if \(\text{det}(A)\neq 0\). Later the determinant will also be defined as a function for square matrices of greater finite dimensions as well. In theorem Inveritibility in terms of determinant, we will see that the equivalence of invertibility of #A# and \(\text{det}(A)\neq 0\) will remain valid.
Finding the inverse of a matrix can be seen as the solving of a system of linear equations. This means that we have a method for finding it:
In order to determine if the inverse of an \((n\times n)\)-matrix \(A\) exist and, if so, to calculate it, we write down the \((n\times 2n)\)-matrix \((A\,|\,I)\): \[(A\,|\,I)=\left(\begin{array}{cccc|cccc} a_{11} & a_{12} & \cdots & a_{1n} & 1 & 0 &\cdots & 0\\ a_{21} & a_{22} & \cdots & a_{2n} & 0 & 1 &\ddots & 0\\ \vdots & \vdots & & \vdots & \vdots & \ddots & \ddots & 0\\ a_{n1} & a_{n2} & \cdots & a_{nn} & 0 & \cdots &\cdots & 1\end{array}\right)\] We reduce this augmented matrix to reduced echelon form.
- If the reduced echelon form has the form \((I\,|\,B)\), meaning that the submatrix to the left of the vertical line is the identity matrix, then \(A\) has rank #n# and is invertible with inverse \(A^{-1}=B\).
- If the reduced echelon form is not of the form \((I\,|\,B)\), meaning that submatrix to the left of the vertical bar is not the identity matrix, then \(A\) has rank less than #n# and is not invertible.
In order to determine the inverse \(X\) of the matrix \(A\) we need to solve the matrix equation \(A\,X=I\). In the theory Matrix equations we have seen that this can be achieved by reducing the matrix \(\left(A\,|\,I\right)\) to reduced echelon form.
For the statements regarding the rank, observe that the rank of #A# is equal to #n# if and only if its reduced echelon form (that is, the matrix to the left of the vertical bar) is equal to the identity matrix #I#.
In particular, #A# is invertible if and only if the rank of #A# is equal to #n#.
Is the following matrix invertible?
\[
\matrix{
1 &4 & -3 \\
1 &20 &-3 \\
0 &0 &16}
\]
The matrix is invertible and its inverse is given by \[\matrix{\frac{5}{4}&-\frac{1}{4}&\frac{3}{16}\\-\frac{1}{16}&\frac{1}{16}&0\\0&0&\frac{1}{16}}\]
We extend the matrix with an identity matrix. Row reduction then gives
\[\begin{aligned}\left(\begin{array}{ccc|ccc}
1&4&-3&1 & 0 & 0 \\
1&20&-3&0 &1 &0\\
0&0&16&0 &0 &1\\
\end{array}
\right)&\sim
\left(
\begin{array}{ccc|ccc}
1 &4&-3&1 & 0 & 0\\
0 &16&0&-1& 1 & 0 \\
0 &0&16&0& 0 & 1 \\
\end{array}
\right)
&{\color{blue}{\begin{array}{ccc}
\mbox{}\\
R_2 -R_1\\
\phantom{X}
\end{array}}}\\
&\sim
\left(
\begin{array}{ccc|ccc}
1 &4&-3 &1 & 0 & 0 \\
0 &1 &0&-\frac{1}{16}&\frac{1}{16} &0 \\
0 &0&16&0& 0 & 1 \\
\end{array}
\right)
&{\color{blue}{\begin{array}{ccc}
\mbox{}\\
\frac{1}{16}R_2\\
\mbox{}
\end{array}}}\\
&\sim
\left(
\begin{array}{ccc|ccc}
1 &0 &-3&\frac{5}{4}&-\frac{1}{4}&0 \\
0 &1 &0&-\frac{1}{16}&\frac{1}{16}&0 \\
0 &0 &16&0&0&1\\
\end{array}
\right)
&{\color{blue}{\begin{array}{ccc}
{}R_1 -4 R_2\\
\mbox{}\\
\phantom{X}
\end{array}}}\\
&\sim
\left(
\begin{array}{ccc|ccc}
1 &0 &-3&\frac{5}{4}&-\frac{1}{4}&0\\
0 &1 &0&-\frac{1}{16}&\frac{1}{16}&0\\
0 &0 &1 &0&0&\frac{1}{16} \\
\end{array}
\right)
&{\color{blue}{\begin{array}{ccc}
{}\\
{}\\
\frac{1}{16}R_3
\end{array}}}\\
&\sim
\left(
\begin{array}{ccc|ccc}
1 &0 &0 &\frac{5}{4}&-\frac{1}{4}&\frac{3}{16}\\
0 &1 &0 &-\frac{1}{16}&\frac{1}{16}&0\\
0 &0 &1 &0&0&\frac{1}{16} \\
\end{array}
\right)
&{\color{blue}{\begin{array}{ccc}
{}R_1 +3 R_3\\
\phantom{X}\\
{}
\end{array}}}
\end{aligned}
\] The reduced echelon form implies that the given matrix is invertible. The inverse matrix is the submatrix to the right of the identity matrix of the reduced echelon form.
The inverse of a matrix
