Absolutely, let's walk through the solution step-by-step, but we'll start by clearly outlining how we approach such a problem.
### a. Write the linear system as a matrix equation in the form [tex]\( AX = B \)[/tex].
The given system of equations is:
[tex]\[
\begin{array}{rr}
x - y + z &= 0 \\
-2y + z &= -3 \\
-3x - 4y &= -8
\end{array}
\][/tex]
We can rewrite this system in matrix form as [tex]\( AX = B \)[/tex], where:
[tex]\[ A = \begin{bmatrix}
1 & -1 & 1 \\
0 & -2 & 1 \\
-3 & -4 & 0
\end{bmatrix} \][/tex]
is the coefficient matrix,
[tex]\[ X = \begin{bmatrix}
x \\
y \\
z
\end{bmatrix} \][/tex]
is the vector of variables, and
[tex]\[ B = \begin{bmatrix}
0 \\
-3 \\
-8
\end{bmatrix} \][/tex]
is the constants vector.
So the matrix equation [tex]\( AX = B \)[/tex] is:
[tex]\[
\begin{bmatrix}
1 & -1 & 1 \\
0 & -2 & 1 \\
-3 & -4 & 0
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix} =
\begin{bmatrix}
0 \\
-3 \\
-8
\end{bmatrix}.
\][/tex]
### b. Solve the system using the inverse that is given for the coefficient matrix.
We are given the inverse of the coefficient matrix [tex]\( A \)[/tex], which is:
[tex]\[
A^{-1} = \begin{bmatrix}
4 & -4 & 1 \\
-3 & 3 & -1 \\
-6 & 7 & -2
\end{bmatrix}
\][/tex]
To find the solution vector [tex]\( X \)[/tex], we multiply both sides of the equation [tex]\( AX = B \)[/tex] by [tex]\( A^{-1} \)[/tex]:
[tex]\[ A^{-1}AX = A^{-1}B \][/tex]
Since [tex]\( A^{-1}A \)[/tex] is the identity matrix [tex]\( I \)[/tex], this simplifies to:
[tex]\[ X = A^{-1}B \][/tex]
We will now perform the matrix multiplication [tex]\( A^{-1}B \)[/tex]:
[tex]\[
X = \begin{bmatrix}
4 & -4 & 1 \\
-3 & 3 & -1 \\
-6 & 7 & -2
\end{bmatrix}
\begin{bmatrix}
0 \\
-3 \\
-8
\end{bmatrix}
\][/tex]
Perform the multiplication for each element in [tex]\( X \)[/tex]:
1. For the first element of [tex]\( X \)[/tex]:
[tex]\[
4 \cdot 0 + (-4) \cdot (-3) + 1 \cdot (-8) = 0 + 12 - 8 = 4
\][/tex]
2. For the second element of [tex]\( X \)[/tex]:
[tex]\[
-3 \cdot 0 + 3 \cdot (-3) + (-1) \cdot (-8) = 0 - 9 + 8 = -1
\][/tex]
3. For the third element of [tex]\( X \)[/tex]:
[tex]\[
-6 \cdot 0 + 7 \cdot (-3) + (-2) \cdot (-8) = 0 - 21 + 16 = -5
\][/tex]
So, the solution vector [tex]\( X \)[/tex] is:
[tex]\[
X = \begin{bmatrix}
4 \\
-1 \\
-5
\end{bmatrix}.
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 4, \quad y = -1, \quad z = -5. \][/tex]
