use the inverse of the coefficient matrix to solve the following system of equations. -x - y - z = 3 4y + 5z…

Explanation:

Step1: Write the system in matrix form

The system of equations (-x - y - z=3), (4y + 5z=17), (4x+3y + 3z=-15) can be written as (AX = B), where (A=\begin{bmatrix}-1&-1&-1\0&4&5\4&3&3\end{bmatrix}), (X=\begin{bmatrix}x\y\z\end{bmatrix}) and (B=\begin{bmatrix}3\17\-15\end{bmatrix}).

Step2: Find the inverse of matrix (A)

First, find the determinant of (A), (\text{det}(A)=-1\times\begin{vmatrix}4&5\3&3\end{vmatrix}+1\times\begin{vmatrix}0&5\4&3\end{vmatrix}- 1\times\begin{vmatrix}0&4\4&3\end{vmatrix}) [ \begin{align*} \text{det}(A)&=-1\times(12 - 15)+1\times(0 - 20)-1\times(0 - 16)\ &=-1\times(-3)+1\times(-20)-1\times(-16)\ &=3-20 + 16\ &=-1 \end{align*} ] Then, find the co - factor matrix of (A): (C=\begin{bmatrix}\begin{vmatrix}4&5\3&3\end{vmatrix}&-\begin{vmatrix}0&5\4&3\end{vmatrix}&\begin{vmatrix}0&4\4&3\end{vmatrix}\-\begin{vmatrix}-1&-1\3&3\end{vmatrix}&\begin{vmatrix}-1&-1\4&3\end{vmatrix}&-\begin{vmatrix}-1&-1\4&3\end{vmatrix}\\begin{vmatrix}-1&-1\4&5\end{vmatrix}&-\begin{vmatrix}-1&-1\0&5\end{vmatrix}&\begin{vmatrix}-1&-1\0&4\end{vmatrix}\end{bmatrix}=\begin{bmatrix}-3&20&-16\0&1&1\-1&5&-4\end{bmatrix}) The adjoint of (A), (\text{adj}(A)=C^{T}=\begin{bmatrix}-3&0&-1\20&1&5\-16&1&-4\end{bmatrix}) The inverse of (A), (A^{-1}=\frac{1}{\text{det}(A)}\text{adj}(A)=\begin{bmatrix}3&0&1\-20&-1&-5\16&-1&4\end{bmatrix})

Step3: Solve for (X)

Since (X = A^{-1}B), we have (X=\begin{bmatrix}3&0&1\-20&-1&-5\16&-1&4\end{bmatrix}\begin{bmatrix}3\17\-15\end{bmatrix}) [ \begin{align*} X&=\begin{bmatrix}3\times3+0\times17 + 1\times(-15)\-20\times3+(-1)\times17+(-5)\times(-15)\16\times3+(-1)\times17 + 4\times(-15)\end{bmatrix}\ &=\begin{bmatrix}9-15\-60 - 17+75\48-17-60\end{bmatrix}\ &=\begin{bmatrix}-6\-2\-29\end{bmatrix} \end{align*} ]

Answer:

(x=-6,y = - 2,z=-29)