Question

If \[ A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}, \] show that \( A^2 = A \).

Hint:

To verify \( A^2 = A \), multiply the matrix by itself and check if the result equals the original matrix.

Answer 1

Compute \( A^2 = A \times A \): \[ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}. \] Calculate each element: \[ \text{Row 1, Column 1} = 2 \times 2 + (-2) \times (-1) + (-4) \times 1 = 4 + 2 - 4 = 2, \] \[ \text{Row 1, Column 2} = 2 \times (-2) + (-2) \times 3 + (-4) \times (-2) = -4 - 6 + 8 = -2, \] \[ \text{Row 1, Column 3} = 2 \times (-4) + (-2) \times 4 + (-4) \times (-3) = -8 - 8 + 12 = -4, \] \[ \text{Row 2, Column 1} = (-1) \times 2 + 3 \times (-1) + 4 \times 1 = -2 - 3 + 4 = -1, \] \[ \text{Row 2, Column 2} = (-1) \times (-2) + 3 \times 3 + 4 \times (-2) = 2 + 9 - 8 = 3, \] \[ \text{Row 2, Column 3} = (-1) \times (-4) + 3 \times 4 + 4 \times (-3) = 4 + 12 - 12 = 4, \] \[ \text{Row 3, Column 1} = 1 \times 2 + (-2) \times (-1) + (-3) \times 1 = 2 + 2 - 3 = 1, \] \[ \text{Row 3, Column 2} = 1 \times (-2) + (-2) \times 3 + (-3) \times (-2) = -2 - 6 + 6 = -2, \] \[ \text{Row 3, Column 3} = 1 \times (-4) + (-2) \times 4 + (-3) \times (-3) = -4 - 8 + 9 = -3. \] Therefore, \[ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = A. \]