Question

Given: \[ A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix} \]
Find \( A^{-1} \) and hence solve the following system of equations:
\[ \begin{cases} 3x + 2y + z = 2000 \\ 4x + y + 3z = 2500 \\ x + y + z = 900 \end{cases} \]

Answer 1

Step 1: Write the system in matrix form:
\[ AX = B, \quad \text{so} \quad X = A^{-1}B \] where \[ A = \begin{bmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 2000 \\ 2500 \\ 900 \end{bmatrix} \]
Step 2: Find the determinant of \( A \):
\[ |A| = 3(1 \cdot 1 - 3 \cdot 1) - 2(4 \cdot 1 - 3 \cdot 1) + 1(4 \cdot 1 - 1 \cdot 1) = -5 \] Step 3: Find the cofactor matrix \( C \) and its transpose:
\[ C = \begin{bmatrix} -2 & -1 & 3 \\ -1 & 2 & -1 \\ 5 & -5 & -5 \end{bmatrix} \quad \Rightarrow \quad C^T = \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix} \] So, \[ A^{-1} = \frac{1}{|A|} C^T = -\frac{1}{5} \begin{bmatrix} -2 & -1 & 5 \\ -1 & 2 & -5 \\ 3 & -1 & -5 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 2 & 1 & -5 \\ 1 & -2 & 5 \\ -3 & 1 & 5 \end{bmatrix} \] Step 4: Multiply \( A^{-1}B \):
\[ X = A^{-1}B = \frac{1}{5} \begin{bmatrix} 2 & 1 & -5 \\ 1 & -2 & 5 \\ -3 & 1 & 5 \end{bmatrix} \begin{bmatrix} 2000 \\ 2500 \\ 900 \end{bmatrix} \] Compute each entry:

• \( x = \frac{1}{5} [2(2000) + 1(2500) - 5(900)] = \frac{1}{5}(4000 + 2500 - 4500) = \frac{2000}{5} = 400 \)
• \( y = \frac{1}{5} [1(2000) - 2(2500) + 5(900)] = \frac{1}{5}(2000 - 5000 + 4500) = \frac{1500}{5} = 300 \)
• \( z = \frac{1}{5} [-3(2000) + 1(2500) + 5(900)] = \frac{1}{5}(-6000 + 2500 + 4500) = \frac{1000}{5} = 200 \)

Final Answer:
\[ x = 400, \quad y = 300, \quad z = 200 \]