use the inverse of the coefficient matrix to solve the following system of equations. 2x - y - 2z = 6 2y +…

Explanation:

Step1: Write the coefficient matrix A and the constant - vector b

The coefficient matrix $A=\begin{bmatrix}2&-1&-2\0&2&5\-5&3&6\end{bmatrix}$, and the constant - vector $b=\begin{bmatrix}6\-13\-18\end{bmatrix}$.

Step2: Find the inverse of matrix A

First, find the determinant of $A$: [ \begin{align*} \det(A)&=2\times\begin{vmatrix}2&5\3&6\end{vmatrix}-(-1)\times\begin{vmatrix}0&5\-5&6\end{vmatrix}-2\times\begin{vmatrix}0&2\-5&3\end{vmatrix}\ &=2\times(2\times6 - 3\times5)+1\times(0\times6+5\times5)-2\times(0\times3 + 5\times2)\ &=2\times(12 - 15)+25-2\times10\ &=2\times(-3)+25 - 20\ &=-6 + 25-20\ &=-1 \end{align*} ] Next, find the adjoint of $A$. The co - factor matrix of $A$: $C=\begin{bmatrix}\begin{vmatrix}2&5\3&6\end{vmatrix}&-\begin{vmatrix}0&5\-5&6\end{vmatrix}&\begin{vmatrix}0&2\-5&3\end{vmatrix}\-\begin{vmatrix}-1&-2\3&6\end{vmatrix}&\begin{vmatrix}2&-2\-5&6\end{vmatrix}&-\begin{vmatrix}2&-1\-5&3\end{vmatrix}\\begin{vmatrix}-1&-2\2&5\end{vmatrix}&-\begin{vmatrix}2&-2\0&5\end{vmatrix}&\begin{vmatrix}2&-1\0&2\end{vmatrix}\end{bmatrix}=\begin{bmatrix}-3&25&10\0&2&1\-1&-10&4\end{bmatrix}$ The adjoint of $A$, $\text{adj}(A)=C^T=\begin{bmatrix}-3&0&-1\25&2&-10\10&1&4\end{bmatrix}$ The inverse of $A$, $A^{-1}=\frac{1}{\det(A)}\text{adj}(A)=\begin{bmatrix}3&0&1\-25&-2&10\-10&-1&-4\end{bmatrix}$

Step3: Solve the system $x = A^{-1}b$

[ \begin{align*} x&=\begin{bmatrix}3&0&1\-25&-2&10\-10&-1&-4\end{bmatrix}\begin{bmatrix}6\-13\-18\end{bmatrix}\ &=\begin{bmatrix}3\times6+0\times(-13)+1\times(-18)\-25\times6-2\times(-13)+10\times(-18)\-10\times6-1\times(-13)-4\times(-18)\end{bmatrix}\ &=\begin{bmatrix}18 - 18\-150 + 26-180\-60 + 13 + 72\end{bmatrix}\ &=\begin{bmatrix}0\-304\25\end{bmatrix} \end{align*} ]

Answer:

$x = 0,y=-304,z = 25$