Question

If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]

Hint:

For \( A^{-1} \), use \( A^{-1} = \frac{1}{\det A} \text{Adj}(A) \) and verify by computing \( A A^{-1} = I \).

Answer 1

To solve the problem, we are given a matrix \( A \) and we need to compute its inverse \( A^{-1} \). Then, using the inverse, solve the system of linear equations.

1. Writing the Matrix Form of the System:
We rewrite the system of equations as a matrix equation \( AX = B \), where:

\( A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \)

2. Finding the Inverse \( A^{-1} \):
We calculate \( A^{-1} \) using the adjoint method or row reduction. The inverse is given as:

\( A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A) \)
Computing the determinant of \( A \):

\[ |A| = 1 \cdot \left((-1)(1) - (-2)(-1)\right) - 2 \cdot \left((-2)(1) - (-2)(0)\right) + 0 \cdot (\text{any}) \\ = 1(-1 - 2) - 2(-2) = -3 + 4 = 1 \]
So the determinant is 1.
Now computing the adjugate (cofactor matrix transpose), we get:
\[ A^{-1} = \begin{bmatrix} -3 & -2 & -4 \\ -2 & 1 & 2 \\ -1 & -2 & -3 \end{bmatrix} \]

3. Solving for \( X = A^{-1}B \):
Now multiply \( A^{-1} \) with \( B \):

\[ X = A^{-1}B = \begin{bmatrix} -3 & -2 & -4 \\ -2 & 1 & 2 \\ -1 & -2 & -3 \end{bmatrix} \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} \]
Performing the multiplication:
\[ x = -3(10) + (-2)(8) + (-4)(7) = -30 - 16 - 28 = -74 \\ y = -2(10) + 1(8) + 2(7) = -20 + 8 + 14 = 2 \\ z = -1(10) + (-2)(8) + (-3)(7) = -10 - 16 - 21 = -47 \]

Final Answer:
The solution to the system is: \( x = -74, \, y = 2, \, z = -47 \)