Question

Use the product of matrices  \(\begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix}\) \(\begin{bmatrix} 0 & 1 & 2 \\ -7 & -7 & -7 \\ 7 & 5 & -4 \end{bmatrix}\)  to solve the following system of equations: \[ x + 2y - 3z = 6 \] \[ 3x + 2y - 2z = 3 \] \[ 2x - y + z = 2 \]

Hint:

For solving systems of linear equations using matrices, use \( X = A^{-1}C \) when \( A \) is invertible, and compute \( A^{-1} \) from the given product matrices.

Answer 1

Step 1: Compute the matrix product. Given matrices \( A \) and \( B \), compute \( AB \):  \(AB = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix}\) \(\begin{bmatrix} 0 & 1 & 2\\  -7 & -7 & -7 \\ 7 & 5 & -4 \end{bmatrix}\) = \(\begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{bmatrix}\) = 7I 
 Since \( AB = 7I \), it follows that \( A^{-1} = \frac{1}{7} B \). 

Step 2: Compute \( A^{-1} \).  \(A^{-1} = \frac{1}{7} \begin{bmatrix} 0 & 1 & 2 \\ -7 & -7 & -7 \\ 7 & 5 & -4 \end{bmatrix}\)

Step 3: Convert system into matrix form. The given system can be written as: \[ AX = C \] where \(A = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix}\), \(\quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\), \(\quad C = \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix}\) 

 Thus,  \(X = A^{-1} C \)

Step 4: Compute \( X \).  \(X = \frac{1}{7} \begin{bmatrix} 0 & 1 & 2 \\ -7 & -7 & -7 \\ 7 & 5 & -4 \end{bmatrix} \begin{bmatrix} 6 \\ 3 \\ 2 \end{bmatrix}\)

Multiplying the matrices:  \(X = \frac{1}{7} \begin{bmatrix} (0 \times 6) + (1 \times 3) + (2 \times 2) \\ (-7 \times 6) + (-7 \times 3) + (-7 \times 2) \\ (7 \times 6) + (5 \times 3) + (-4 \times 2) \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 3 + 4 \\ -42 -21 -14 \\ 42 + 15 -8 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 7 \\ -77 \\ -49 \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ -5 \end{bmatrix}\)

Conclusion: Thus, the solution to the given system is: 

\[x = 1, \quad y = -5, \quad z = -5\]