Identity and Inverse Matrices

The n×n identity matrix I has ones down the main diagonal (upper-left corner to lower-right corner) and zeroes elsewhere. The 3×3 identity matrix, for example, is

I = $\displaystyle \left[\vphantom{
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
}\right.$$\displaystyle \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
}\right]$

The inverse of an n×n matrix A is an n×n matrix B satisfying AB = I. To find the inverse of an invertible matrix A, enter A with ``-1'' as a superscript and apply Evaluate. (As an alternative, leave the insertion point anywhere inside the matrix A, and from the Matrices submenu, choose Inverse.)

$\blacktriangleright$ Evaluate

$\left(\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right.$$\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}$$\left.\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right)^{{-1}}_{}$  =  $\left(\vphantom{
\begin{array}{rr}
-%
\dfrac{7}{13}\vspace{6pt} & \dfrac{6}{13} \\
\dfrac{8}{13} & -\dfrac{5}{13}
\end{array}
}\right.$$\begin{array}{rr}
-%
\dfrac{7}{13}\vspace{6pt} & \dfrac{6}{13} \\
\dfrac{8}{13} & -\dfrac{5}{13}
\end{array}$$\left.\vphantom{
\begin{array}{rr}
-%
\dfrac{7}{13}\vspace{6pt} & \dfrac{6}{13} \\
\dfrac{8}{13} & -\dfrac{5}{13}
\end{array}
}\right)$

To check that this matrix satisfies the defining property, evaluate the product.

$\left(\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right.$$\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}$$\left.\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right)$ $\left(\vphantom{
\begin{array}{rr}
-\dfrac{7}{13}\vspace{6pt} & \dfrac{6}{13} \\
\dfrac{8}{13} & -\dfrac{5}{13}
\end{array}
}\right.$$\begin{array}{rr}
-\dfrac{7}{13}\vspace{6pt} & \dfrac{6}{13} \\
\dfrac{8}{13} & -\dfrac{5}{13}
\end{array}$$\left.\vphantom{
\begin{array}{rr}
-\dfrac{7}{13}\vspace{6pt} & \dfrac{6}{13} \\
\dfrac{8}{13} & -\dfrac{5}{13}
\end{array}
}\right)$ =  $\left(\vphantom{
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
}\right.$$\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
}\right)$

The operation Evaluate Numerically gives you a numerical approximation of the inverse. The accuracy of this numerical approximation depends on properties of the matrix, as well as on the choices you have made in the Settings menu.

$\blacktriangleright$ Evaluate Numerically

$\left(\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right.$$\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}$$\left.\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right)^{{-1}}_{}$ =  $\left(\vphantom{
\begin{array}{rr}
-.53846 & .46154 \\
.61538 & -.38462
\end{array}
}\right.$$\begin{array}{rr}
-.53846 & .46154 \\
.61538 & -.38462
\end{array}$$\left.\vphantom{
\begin{array}{rr}
-.53846 & .46154 \\
.61538 & -.38462
\end{array}
}\right)$

Checking the product of a matrix with its inverse gives you an idea of the degree of accuracy of the approximation.

$\blacktriangleright$ Evaluate

$\left(\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right.$$\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}$$\left.\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right)$$\left(\vphantom{
\begin{array}{rr}
-.53846 & .46154 \\
.61538 & -.38462
\end{array}
}\right.$$\begin{array}{rr}
-.53846 & .46154 \\
.61538 & -.38462
\end{array}$$\left.\vphantom{
\begin{array}{rr}
-.53846 & .46154 \\
.61538 & -.38462
\end{array}
}\right)$ =  $\left(\vphantom{
\begin{array}{rr}
.99998 & -.00002 \\
-.00002 & .99998
\end{array}
}\right.$$\begin{array}{rr}
.99998 & -.00002 \\
-.00002 & .99998
\end{array}$$\left.\vphantom{
\begin{array}{rr}
.99998 & -.00002 \\
-.00002 & .99998
\end{array}
}\right)$

Since $\left(\vphantom{ A^n}\right.$An$\left.\vphantom{ A^n}\right)^{{-1}}_{}$ = $\left(\vphantom{ A^{-1}}\right.$A-1$\left.\vphantom{ A^{-1}}\right)^{n}_{}$, you can compute negative powers of invertible matrices.

$\blacktriangleright$ Evaluate

$\left(\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right.$$\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}$$\left.\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right)^{{3}}_{}$ = $\left(\vphantom{
\begin{array}{cc}
941 & 942 \\
1256 & 1255
\end{array}
}\right.$$\begin{array}{cc}
941 & 942 \\
1256 & 1255
\end{array}$$\left.\vphantom{
\begin{array}{cc}
941 & 942 \\
1256 & 1255
\end{array}
}\right)$

        

$\left(\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right.$$\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}$$\left.\vphantom{
\begin{array}{cc}
5 & 6 \\
8 & 7
\end{array}
}\right)^{{-3}}_{}$ = $\left(\vphantom{
\begin{array}{cc}
-\frac{1255}{2197}\vspace{6pt} & \frac{942}{2197} \\
\frac{1256}{2197} & -\frac{941}{2197}
\end{array}
}\right.$$\begin{array}{cc}
-\frac{1255}{2197}\vspace{6pt} & \frac{942}{2197} \\
\frac{1256}{2197} & -\frac{941}{2197}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
-\frac{1255}{2197}\vspace{6pt} & \frac{942}{2197} \\
\frac{1256}{2197} & -\frac{941}{2197}
\end{array}
}\right)$

$\left(\vphantom{
\begin{array}{cc}
941 & 942 \\
1256 & 1255
\end{array}
}\right.$$\begin{array}{cc}
941 & 942 \\
1256 & 1255
\end{array}$$\left.\vphantom{
\begin{array}{cc}
941 & 942 \\
1256 & 1255
\end{array}
}\right)$$\left(\vphantom{
\begin{array}{cc}
-\frac{1255}{2197}\vspace{6pt} & \frac{942}{2197} \\
\frac{1256}{2197} & -\frac{941}{2197}
\end{array}
}\right.$$\begin{array}{cc}
-\frac{1255}{2197}\vspace{6pt} & \frac{942}{2197} \\
\frac{1256}{2197} & -\frac{941}{2197}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
-\frac{1255}{2197}\vspace{6pt} & \frac{942}{2197} \\
\frac{1256}{2197} & -\frac{941}{2197}
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
}\right.$$\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
}\right)$

The preceding product demonstrates that A-3 = $\left(\vphantom{ A^3}\right.$A3$\left.\vphantom{ A^3}\right)^{{-1}}_{}$.

The m×n matrix with every entry equal to zero is the identity for addition; that is, for any m×n matrix A,

A + 0 = 0 + A = A

and the additive inverse of a matrix A is the matrix $\left(\vphantom{
-1}\right.$ -1$\left.\vphantom{
-1}\right)$A.

$\blacktriangleright$ Evaluate

$\left[\vphantom{
\begin{array}{cc}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2} \\
a_{3,1} & a_{3,2}
\end{array}
}\right.$$\begin{array}{cc}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2} \\
a_{3,1} & a_{3,2}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2} \\
a_{3,1} & a_{3,2}
\end{array}
}\right]$ + $\left[\vphantom{
\begin{array}{cc}
-a_{1,1} & -a_{1,2} \\
-a_{2,1} & -a_{2,2} \\
-a_{3,1} & -a_{3,2}
\end{array}
}\right.$$\begin{array}{cc}
-a_{1,1} & -a_{1,2} \\
-a_{2,1} & -a_{2,2} \\
-a_{3,1} & -a_{3,2}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
-a_{1,1} & -a_{1,2} \\
-a_{2,1} & -a_{2,2} \\
-a_{3,1} & -a_{3,2}
\end{array}
}\right]$ =  $\left[\vphantom{
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}
}\right.$$\begin{array}{cc}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}$$\left.\vphantom{
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}
}\right]$