<p>Let <span>\(A = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}\)</span> and <span>\(\text{adj } A = \begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}\)</span>. Find the element \(A_{13}\) of \(A^{-1}\).</p>
Step-by-Step Solution
Key Concept: For an upper triangular matrix, the determinant is the product of diagonal elements. The inverse of a matrix is found using the adjugate formula.
<p><strong>Step 1:</strong> Calculate \(|A|\). Since \(A\) is an upper triangular matrix, \(|A| = 1 \cdot 1 \cdot 1 = 1\).</p><p><strong>Step 2:</strong> Use the formula \(A^{-1} = \frac{1}{|A|} \text{adj } A\).</p><p><strong>Step 3:</strong> \(A^{-1} = \frac{1}{1} \begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 & 7 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}\)</p><p><strong>Step 4:</strong> The element \(A_{13}\) of \(A^{-1}\) is \(7\).</p>
Correct Answer: 7