1 Procedure:

In order to invert a matrix, the following four steps can be used: ex2html_comment_mark>103 0.bean2

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Find the determinant

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Set up the matrix of cofactors

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Transpose the matrix of cofactors

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Divide by the determinant

For example, the matrix

$${\bf A} = \left( \begin{array}{rr} 1 & 2\ 1 & 5 \end{array} \right)$$

can be inverted by: ex2html_comment_mark>104 0.bean3

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$${\bf \vert A\vert} = (1\times 5) - (2\times 1) = 3$$

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$$\begin{eqnarray*} A_{ij} = \left[ \begin{array}{rr} (-1)^{2}\times 5 & (-1)^{3}... ...= \left( \begin{array}{rr} 5 & -1\ -2 & 1 \end{array} \right) \end{eqnarray*}$$

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$$A_{ij}^{\prime}= \left( \begin{array}{rr} 5 & -2\ -1 & 1 \end{array} \right)$$

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$${\bf A^{-1}}= \frac{1}{3}\left( \begin{array}{rr} 5 & -2\ ... ...\frac{2}{3} \ -\frac{1}{3} & \frac{1}{3} \end{array} \right)$$

To verify this, we can multiply AA$^{-1}$ to obtain the identity matrix:

$$\frac{1}{3}\left( \begin{array}{rr} 5 & -2\ -1 & 1 \end{ar... ... = \left(\begin {array}{rr} 1 & 0\ 0 & 1 \end{array} \right)$$

The result that ${\bf AA}^{-1} = {\bf I}$ may be used to solve the pair of simultaneous equations:

$$\begin{eqnarray*} x_{1} + 2x_{2} &=& 8\\ x_{1} + 5x_{2} &=& 17\\ \end{eqnarray*}$$

which may be written

$$\left( \begin{array}{rr} 1 & 2\ 1 & 5 \end{array} \right) ... ...right) = \left( \begin{array}{r} 8\ 17 \end{array} \right)$$

i.e.,

$$\begin{eqnarray*} {\bf Ax} &=& {\bf y} \\ \end{eqnarray*}$$

premultiplying both sides by the inverse of $\bf A$, we have

$$\begin{eqnarray*} {\bf A}^{-1}{\bf Ax} &=& {\bf A}^{-1}{\bf y} \\ {\bf x} &=&... ...ght) \\ &=& \left( \begin{array}{r} 2\ 3 \end{array} \right) \end{eqnarray*}$$

which may be verified by substitution. For a larger matrix it is more tedious to compute the inverse. Let us consider the matrix

$${\bf A}= \left( \begin{array}{rrr} 1 & 1 & 0\ 1 & 0 & 1\ 1 &-1 &0 \ \end{array} \right)$$

ex2html_comment_mark>105 0.bean3

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The determinant is

$$\begin{eqnarray*} {\bf \vert A\vert} &=& +1 \left\vert \begin{array}{rr} 0 & 1\... ...ray}{rr} 1 & 0\ 1 & -1 \end{array} \right\vert = +1 +1 +0 = 2 \end{eqnarray*}$$

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The matrix of cofactors is:

$$\begin{array}{llc} A_{ij} = \left[ \begin{array}{r}+ \ \... ...-1 \ 0 & 0 & 2\ 1 & -1 & -1 \end{array} \right) \end{array}$$

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The transpose is

$$A_{ij}^{\prime} = \left( \begin{array}{rrr} 1 & 0 & 1\\ 1 & 0 & -1\ -1 & 2 & -1\ \end{array} \right)$$

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Dividing by the determinant, we have

$${\bf A}^{-1} = \frac{1}{2} \left( \begin{array}{rrr} 1 & ... ...0 & .5\ .5 & 0 & -.5\ -.5 & 1 & -.5\\ \end{array} \right)$$

which may be verified by multiplication with A to obtain the identity matrix.