Chapter 7, Problem 110CR

To determine

To solve: the system of linear equations using an inverse matrix (if possible).

Answer to Problem 110CR

  $\begin{array}{c}x=10.\\ y=4.\\ z=3.\end{array}$

Explanation of Solution

Given information: Given linear equations are;

  $\begin{array}{c}-x+4y-2z=0.\\ 2x-9y+5z=-1.\\ -x+5y-4z=-2.\end{array}$

Formula used:

If A is the given Matrix then,

  $A^{-1}=\frac{Adj(A)}{\left|A\right|}.$

Matrix A times Matrix B = Matrix C.

So, Matrix B = (Inverse of Matrix A ) times Matrix C .

Calculation:

Let, Matrix A = $\left[\begin{array}{ccc}-1& 4& -2\\ 2& -9& 5\\ -1& 5& -4\end{array}\right]$ .

Matrix B = $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ .

Matrix C = $\left[\begin{array}{c}0\\ -1\\ -2\end{array}\right]$ .

So, $\begin{array}{l}B=A^{-1}\times C\\ \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=A^{-1}\times \left[\begin{array}{c}0\\ -1\\ -2\end{array}\right]------------(1).\end{array}$

To find inverse of Matrix A.

Determinant of A is -1.

  $\left|A\right|=\left[\begin{array}{ccc}-1& 4& -2\\ 2& -9& 5\\ -1& 5& -4\end{array}\right]=-1.$

Cofactor matrix of A ;

  $\left[\begin{array}{ccc}11& 3& 1\\ 6& 2& 1\\ 2& 1& 1\end{array}\right]$ .

Adjoint Matrix is the transpose of the cofactor matrix is shown below.

  $Adj(A)=\left[\begin{array}{ccc}11& 6& 2\\ 3& 2& 1\\ 1& 1& 1\end{array}\right]$

Hence the inverse of Matrix A;

  $\begin{array}{l}{A}^{-1}=\frac{1}{-1}\left[\begin{array}{ccc}11& 6& 2\\ 3& 2& 1\\ 1& 1& 1\end{array}\right].\\ A^{-1}=\left[\begin{array}{ccc}-11& -6& -2\\ -3& -2& -1\\ -1& -1& -1\end{array}\right].\end{array}$

From equation (1) $\begin{array}{c}\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}-11& -6& -2\\ -3& -2& -1\\ -1& -1& -1\end{array}\right]\times \left[\begin{array}{c}0\\ -1\\ -2\end{array}\right]=\left[\begin{array}{c}10\\ 4\\ 3\end{array}\right].\\ x=10.\\ y=4.\\ z=3.\end{array}$