The inverse of a matrix

Matrices: Matrices

The inverse of a matrix

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 $B$ is the inverse of $A$ .
If $B\, A = I$ , then $B$ is 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 $B = C$ .

There is further evidence that for square matrices $A\,B=I$ we also have $B\,A=I$ . Because matrix multiplication is not commutative, this property is not obvious.
So, a weaker condition on $B$ suffices to conclude that $B$ is the inverse of a matrix $A$ :

If $A\, B = I$ , then $B$ the inverse of $A$ .
If $B\, A = I$ , then $B$ the inverse of $A$ .

Examples

$\matrix{10 & 3 \\ -3 & -1 \\ }$ is the inverse of $\matrix{1 & 3 \\ -3 & -10 \\ }$ .

After all, $\matrix{1 & 3 \\ -3 & -10 \\ } \,\matrix{10 & 3 \\ -3 & -1 \\ } = \matrix{1 & 0\\ 0 & 1}$ Verify this!

New example

Singular matrices 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 $\matrix{1 & 1\\ 1 & 1}\text{ and }\matrix{1 & 2\\ 2 & 4}$

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: $\begin{aligned} A^{\top}\left(A^{-1}\right)^{\top} &=\left(A^{-1}A\right)^{\top}=I^{\top}=I\\ \\ \left(A^{-1}\right)^{\top} A^{\top} &=\left(A\,A^{-1}\right)^{\top}=I^{\top}=I \end{aligned}$

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 $\text{det}(A)$ . The theorem says so that the matrix is invertible if and only if $\text{det}(A)\neq 0$ .

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 $\matrix{p & q \\ r & s}= \frac{1}{a\,d-b\,c}\matrix{d & -b \\ -c & a}\tiny.$

The theorem says so that the matrix is invertible if and only if $\det(A)\neq 0$ . This is the correct definition of the determinant true for square matrices of different dimensions.

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 row 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 row 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.

If you have looked at the proof of the previous theorem, you might have an idea how to get started. Suppose that we want to compute the inverse $X$ of matrix $A$ . Then we must solve $A\,X=I$ . Consider the unknown matrix $X$ as a matrix consisting of $n$ column vectors $x_1, \ldots, x_n$ . The product matrix $A\,X$ then has columns $A\,x_1, \ldots, A\,x_n$ , and these need to be equal to the columns of the unit matrix, that is to say, be equal to the unit vectors $e_1, \ldots, e_n$ .

We find in this way $n$ systems of equations $A\,x_1=e_1, \ldots, A\,x_n=e_n$ . These are all systems of linear equations that have $A$ as a coefficient matrix. We can solve these equations all at once (simultaneously) by bringing the matrix $(A|I)=\left(\begin{array}{cccc|cccc} a_{11} & a_{12} & \cdots & a_{1n} & 1 & 0 &\cdots & 0\\ a_{21} & a_{22} & \cdots & a_{2n} & 0 & 1 &\ddots & 0\\ \vdots & \vdots & & \vdots & \vdots & \ddots & \ddots & 0\\ a_{n1} & a_{n2} & \cdots & a_{nn} & 0 & \cdots &\cdots & 1\end{array}\right)$ into reduced row echelon form. If the rank of the system is less than $n$ is, then $A$ has no inverse. If the rank of the system is $n$ , then the reduced system has the form $\left(\begin{array}{cccc|cccc} 1 & 0 &\cdots & 0 & b_{11} & b_{12} & \cdots & b_{1n} \\ 0 & 1 &\ddots & 0 & b_{21} & b_{22} & \cdots & b_{2n} \\ \vdots & \ddots & \ddots & 0 & \vdots & \vdots & & \vdots \\ 0 & \cdots &\cdots & 1 & b_{n1} & b_{n2} & \cdots & a_{nn} \end{array}\right)$ We find in this way (unique) solutions of the system $A\,x_i = e_i$ , namely the columns of the matrix $B = (b_{ij})_{1\le i,j\le n}$ . The matrix $B$ thus satisfies $A\,B = I$ .

Is the following matrix invertible?
$\matrix{ 1 &-2 & -1 \\ -1 &18 &13 \\ 1 &-6 &-20}$

The matrix is invertible and its inverse is given by $\matrix{\frac{141}{128}&\frac{17}{128}&\frac{1}{32}\\\frac{7}{256}&\frac{19}{256}&\frac{3}{64}\\\frac{3}{64}&-\frac{1}{64}&-\frac{1}{16}}$

We extend the matrix with an identity matrix. Row reduction then gives
$\begin{aligned}\left(\begin{array}{ccc|ccc} 1&-2&-1&1 & 0 & 0 \\ -1&18&13&0 &1 &0\\ 1&-6&-20&0 &0 &1\\ \end{array} \right)&\sim \left( \begin{array}{ccc|ccc} 1 &-2&-1&1 & 0 & 0\\ 0 &16&12&1& 1 & 0 \\ 0 &-4&-19&-1& 0 & 1 \\ \end{array} \right) &{\blue{\begin{array}{ccc} \mbox{}\\ R_2 +R_1\\ R_3 -R_1 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &-2&-1 &1 & 0 & 0 \\ 0 &1 &\frac{3}{4}&\frac{1}{16}&\frac{1}{16} &0 \\ 0 &-4&-19&-1& 0 & 1 \\ \end{array} \right) &{\blue{\begin{array}{ccc} \mbox{}\\ \frac{1}{16}R_2\\ \mbox{} \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{1}{2}&\frac{9}{8}&\frac{1}{8}&0 \\ 0 &1 &\frac{3}{4}&\frac{1}{16}&\frac{1}{16}&0 \\ 0 &0 &-16&-\frac{3}{4}&\frac{1}{4}&1\\ \end{array} \right) &{\blue{\begin{array}{ccc} {}R_1 +2 R_2\\ \mbox{}\\ R_3 +4 R_1 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{1}{2}&\frac{9}{8}&\frac{1}{8}&0\\ 0 &1 &\frac{3}{4}&\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &\frac{3}{64}&-\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\blue{\begin{array}{ccc} {}\\ {}\\ -\frac{1}{16}R_3 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &0 &\frac{141}{128}&\frac{17}{128}&\frac{1}{32}\\ 0 &1 &0 &\frac{7}{256}&\frac{19}{256}&\frac{3}{64}\\ 0 &0 &1 &\frac{3}{64}&-\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\blue{\begin{array}{ccc} {}R_1 -\frac{1}{2} R_3\\ R_2 -\frac{3}{4} R_3\\ {} \end{array}}} \end{aligned}$ The reduced row 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.

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