1. **State the problem:** We need to show that the matrix $$A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$$ is idempotent. This means we want to verify that $$A^2 = A$$. 2. **Recall the definition:** A matrix $$A$$ is idempotent if $$A^2 = A$$, where $$A^2$$ is the matrix product of $$A$$ with itself. 3. **Calculate $$A^2$$:** Multiply $$A$$ by $$A$$: $$ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \times \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} $$ Calculate each element: - First row, first column: $$2\times2 + (-2)\times(-1) + (-4)\times1 = 4 + 2 - 4 = 2$$ - First row, second column: $$2\times(-2) + (-2)\times3 + (-4)\times(-2) = -4 - 6 + 8 = -2$$ - First row, third column: $$2\times(-4) + (-2)\times4 + (-4)\times(-3) = -8 - 8 + 12 = -4$$ - Second row, first column: $$-1\times2 + 3\times(-1) + 4\times1 = -2 - 3 + 4 = -1$$ - Second row, second column: $$-1\times(-2) + 3\times3 + 4\times(-2) = 2 + 9 - 8 = 3$$ - Second row, third column: $$-1\times(-4) + 3\times4 + 4\times(-3) = 4 + 12 - 12 = 4$$ - Third row, first column: $$1\times2 + (-2)\times(-1) + (-3)\times1 = 2 + 2 - 3 = 1$$ - Third row, second column: $$1\times(-2) + (-2)\times3 + (-3)\times(-2) = -2 - 6 + 6 = -2$$ - Third row, third column: $$1\times(-4) + (-2)\times4 + (-3)\times(-3) = -4 - 8 + 9 = -3$$ 4. **Form the resulting matrix:** $$ A^2 = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} $$ 5. **Compare $$A^2$$ with $$A$$:** They are equal. 6. **Conclusion:** Since $$A^2 = A$$, matrix $$A$$ is idempotent.