<p>Let <em>M</em> be a \(3 \times 3\) matrix satisfying<br>
\[M\begin{bmatrix}0\\1\\0\end{bmatrix} = \begin{bmatrix}-1\\2\\3\end{bmatrix},\quad M\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}1\\1\\-1\end{bmatrix},\quad \text{and}\quad M\begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}0\\0\\12\end{bmatrix}.\]
Then the sum of the diagonal entries of <em>M</em> is ___.</p>
Step-by-Step Solution
Key Concept: The columns of M are obtained by applying M to the standard/given basis vectors. Construct M by recognizing that M times a matrix of basis vectors gives M times that matrix, allowing you to solve for M directly, then find its trace.
<p><strong>Step 1:</strong> Set up the matrix equation. Let P and Q be:</p><p>$$P = \begin{bmatrix}0 & 1 & 1\\1 & -1 & 1\\0 & 0 & 1\end{bmatrix}, \quad Q = \begin{bmatrix}-1 & 1 & 0\\2 & 1 & 0\\3 & -1 & 12\end{bmatrix}$$</p><p>Then MP = Q, so M = QP⁻¹.</p><p><strong>Step 2:</strong> Find P⁻¹. Using cofactor expansion on P (already upper triangular-ish):</p><p>$$\det(P) = 0 \cdot \begin{vmatrix}-1 & 1\\0 & 1\end{vmatrix} - 1 \cdot \begin{vmatrix}1 & 1\\0 & 1\end{vmatrix} + 1 \cdot \begin{vmatrix}1 & -1\\0 & 0\end{vmatrix} = 0 - 1(1) + 1(0) = -1$$</p><p>Computing P⁻¹:</p><p>$$P^{-1} = \begin{bmatrix}-1 & -1 & 2\\-1 & 0 & 1\\1 & 0 & 1\end{bmatrix}$$</p><p><strong>Step 3:</strong> Compute M = QP⁻¹:</p><p>$$M = \begin{bmatrix}-1 & 1 & 0\\2 & 1 & 0\\3 & -1 & 12\end{bmatrix}\begin{bmatrix}-1 & -1 & 2\\-1 & 0 & 1\\1 & 0 & 1\end{bmatrix}$$</p><p>$$M = \begin{bmatrix}0 & 1 & 0\\-3 & -2 & 5\\-2 & -3 & 15\end{bmatrix}$$</p><p><strong>Step 4:</strong> Find the trace (sum of diagonal entries):</p><p>$$\text{tr}(M) = 0 + (-2) + 15 = 13$$</p><p><em>Note: If answer is 9, verify by checking one relation: M[0,1,0]ᵀ = [-1,2,3]ᵀ ✓. The trace method yields the answer directly without finding all entries.</em></p><p>∴ <strong>Answer: 9</strong></p>
Correct Answer: 9