1. **Problem statement:** Given matrix $$ M = \begin{bmatrix} 3 & -6 & 2 \\ 6 & 2 & -3 \\ 2 & 3 & 6 \end{bmatrix} $$ (i) Show that $$MM^T = \lambda I$$ where $$I$$ is the identity matrix. (ii) Hence determine $$M^{-1}$$. 2. **Calculate $$MM^T$$:** $$ M^T = \begin{bmatrix} 3 & 6 & 2 \\ -6 & 2 & 3 \\ 2 & -3 & 6 \end{bmatrix} $$ Now $$ MM^T = \begin{bmatrix} 3 & -6 & 2 \\ 6 & 2 & -3 \\ 2 & 3 & 6 \end{bmatrix} \begin{bmatrix} 3 & 6 & 2 \\ -6 & 2 & 3 \\ 2 & -3 & 6 \end{bmatrix} $$ Compute each element: - Top-left element: $$3\cdot 3 + (-6)\cdot (-6) + 2\cdot 2 = 9 + 36 + 4 = 49$$ - Top-middle element: $$3\cdot 6 + (-6)\cdot 2 + 2\cdot (-3) = 18 - 12 - 6 = 0$$ - Top-right element: $$3\cdot 2 + (-6)\cdot 3 + 2\cdot 6 = 6 - 18 + 12 = 0$$ - Middle-left element: $$6\cdot 3 + 2\cdot (-6) + (-3)\cdot 2 = 18 - 12 - 6 = 0$$ - Middle element: $$6\cdot 6 + 2\cdot 2 + (-3)\cdot (-3) = 36 + 4 + 9 = 49$$ - Middle-right element: $$6\cdot 2 + 2\cdot 3 + (-3)\cdot 6 = 12 + 6 - 18 = 0$$ - Bottom-left element: $$2\cdot 3 + 3\cdot (-6) + 6\cdot 2 = 6 - 18 + 12 = 0$$ - Bottom-middle element: $$2\cdot 6 + 3\cdot 2 + 6\cdot (-3) = 12 + 6 - 18 = 0$$ - Bottom-right element: $$2\cdot 2 + 3\cdot 3 + 6\cdot 6 = 4 + 9 + 36 = 49$$ Therefore, $$ MM^T = \begin{bmatrix} 49 & 0 & 0 \\ 0 & 49 & 0 \\ 0 & 0 & 49 \end{bmatrix} = 49 I $$ which shows that $$ MM^T = \lambda I $$ with $$ \lambda = 49 $$. 3. **Find $$M^{-1}$$:** Since $$ MM^T = 49 I $$, the matrix $$M$$ is invertible. Multiply both sides by $$ M^{-1} $$ on the left: $$ M^{-1}MM^T = M^{-1} 49 I $$ This simplifies to $$ M^T = 49 M^{-1} $$ Therefore, $$ M^{-1} = \frac{1}{49} M^T $$ Explicitly, $$ M^{-1} = \frac{1}{49} \begin{bmatrix} 3 & 6 & 2 \\ -6 & 2 & 3 \\ 2 & -3 & 6 \end{bmatrix} $$ **Final answers:** (i) $$ MM^T = 49 I $$ (ii) $$ M^{-1} = \frac{1}{49} M^T = \frac{1}{49} \begin{bmatrix} 3 & 6 & 2 \\ -6 & 2 & 3 \\ 2 & -3 & 6 \end{bmatrix} $$