Question

a. Write each linear system as a matrix equation in the form AX = B. b. Solve the system using the inverse that is given for the coefficient matrix. ﻿{x−y+z=82y−z=−72x+3y=1The inverse of [1−1102−1230] is [33−1−2−21−4−52].\begin{cases}
x - y + z = 8 \
2y - z = -7 \
2x + 3y = 1
\end{cases}
\
	ext{The inverse of }
\begin{bmatrix}
1 & -1 & 1 \
0 & 2 & -1 \
2 & 3 & 0
\end{bmatrix}
	ext{ is }
\begin{bmatrix}
3 & 3 & -1 \
-2 & -2 & 1 \
-4 & -5 & 2
\end{bmatrix}	ext{.}⎩⎨⎧​x−y+z=82y−z=−72x+3y=1​The inverse of ​102​−123​1−10​​ is ​3−2−4​3−2−5​−112​​.

Accepted Answer

Step 1: Write the system of equations in matrix form $$AX = B$$, where $$A is $$the coefficient matrix, $$X is $$the column matrix of variables, and $$B is $$the constants matrix. For the system: 
$$\begin{cases} x - y + z = 8 \ 2y - z = -7 \ 2x + 3y = 1 \end{cases}$$
The coefficient matrix $$A is$$:
$$A = \begin{bmatrix} 1 & -1 & 1 \ 0 & 2 & -1 \ 2 & 3 & 0 \end{bmatrix}$$
The variable matrix $$X is$$:
$$X = \begin{bmatrix} x \ y \ z \end{bmatrix}$$
The constants matrix $$B is$$:
$$B = \begin{bmatrix} 8 \ -7 \ 1 \end{bmatrix}$$ Step 2: Confirm the inverse matrix $$A^{-1} is $$given as:
$$A^{-1} = \begin{bmatrix} 3 & 3 & -1 \ -2 & -2 & 1 \ -4 & -5 & 2 \end{bmatrix}$$ Step 3: Use the matrix equation $$X = A$$^{-1}$$B to $$find the solution vector $$X. $$This means multiplying the inverse matrix $$A^{-1} by $$the constants matrix $$B. $$Step 4: Perform the matrix multiplication:
$$X = \begin{bmatrix} 3 & 3 & -1 \ -2 & -2 & 1 \ -4 & -5 & 2 \end{bmatrix} 	imes \begin{bmatrix} 8 \ -7 \ 1 \end{bmatrix}$$
Multiply each row of $$A^{-1} by $$the column matrix $$B to $$get each variable $$x$$, $$y$$, and $$z. $$Step 5: Write the resulting matrix $$X as $$the solution to the system, where the first element is $$x$$, the second is $$y$$, and the third is $$z.$$

Verified video answer for a similar problem:

Key Concepts

Matrix Representation of Linear Systems

Matrix Inverse and Its Role in Solving Systems

Matrix Multiplication for Solution Computation