91Ó°ÊÓ

The inverse of the matrix is another matrix, which on multiplication with the given matrix gives the multiplicative identity. For a matrix A , its inverse is ${\mathit{A}}^{\mathbf{-}\mathbf{1}}$ and $\mathit{A}\mathbf{.}{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathit{A}}^{\mathbf{-}\mathbf{1}}\mathbf{.}\mathit{A}\mathbf{=}\mathbf{I}$, where , is the identity matrix.

The matrix whose determinant is non-zero and for which the inverse matrix can be calculated is called the invertible matrix.

The given matrices are_{$$\begin{gathered} \left\{\begin{array}{l}x-y+z=4 \\ 2x+y-z=-1 \\ 3x+2y+2z=5\\ \end{array}\right. \end{gathered}$$}

To solve the system of equations Mr=k , where is the matrix of coefficient, r is the unknown vector.

The matrix m and the vectors are M .

Find the co-factors of matrix M .$$\begin{gathered} =\left(1-11 \\ 21-1 \\ 322\right);k\left(4 \\ -1 \\ 5\right);r\left(x \\ y \\ z\right) \end{gathered}$$

$$\begin{gathered} C_{11}=\left(1-1 \\ 22\right) \\ {C}_{11}=4 \\ {C}_{12}=\left(2-1 \\ 32\right) \\ {C}_{12}=-7 \end{gathered}$$

Solve further.

$$\begin{gathered} C_{13}=\left(21 \\ 32\right) \\ {C}_{13}=1 \\ {C}_{21}=\left(-11 \\ 22\right) \\ {C}_{21}=4 \end{gathered}$$

Solve further.

$$\begin{gathered} C_{22}=\left(11 \\ 33\right) \\ {C}_{22}=-1 \\ {C}_{23}=-\left(1-1 \\ 32\right) \\ {C}_{23}=-5 \end{gathered}$$

Solve further.

$$\begin{gathered} C_{31}=\left(-11 \\ 11\right) \\ {C}_{31}=0 \\ {C}_{32}=\left(1-1 \\ 2-1\right) \\ {C}_{32}=3 \end{gathered}$$

Solve further.

$$\begin{gathered} C_{33}=\left(1-1 \\ 21\right) \\ {C}_{33}=3 \\ C=\left(4-71 \\ 4-13 \\ 033\right) \\ {C}^T=\left(440 \\ -7-13 \\ 1-53\right) \end{gathered}$$

Find the determinant of the matrix M .

$$\begin{gathered} det\left(M\right)= \\ \left|1-11 \\ 21-1 \\ 322\right| \\ =C_{11}+\left(-1\right)C_{12}+C_{13} \\ =1×4\left(-1\right)×\left(-7\right)+1×1 \\ =12 \end{gathered}$$

Hence, $$\begin{gathered} M^{-1}=\frac{1}{12}\left(440 \\ -7-13 \\ 1-53\right) \end{gathered}$$

Substitute the value of the parameters in the equation $M^{-1}k=r$.

$$\begin{gathered} \frac{1}{12}\left(440 \\ -7-13 \\ 1-53\right)\left(4 \\ -1 \\ 5\right)=\left(x \\ y \\ z\right) \\ \frac{1}{12}\left(12 \\ -12 \\ 24\right)=\left(x \\ y \\ z\right) \\ x=1 \\ y=-1 \\ z=2 \end{gathered}$$

$$\begin{gathered} \frac{1}{12}\left(440 \\ -7-13 \\ 1-53\right)\left(4 \\ -1 \\ 5\right)=\left(x \\ y \\ z\right) \\ \frac{1}{12}\left(12 \\ -12 \\ 24\right)=\left(x \\ y \\ z\right) \\ x=1 \\ y=-1 \\ z=2 \end{gathered}$$

Therefore,the given set of equations is written as,

$$\begin{gathered} M^{-1}=\left(440 \\ -7-13 \\ 0-53\right);x=1,y=-1,z=2 \end{gathered}$$.