<p>If \[\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix},\] then the inverse of \(\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}\) is:</p>
<p>\(\begin{bmatrix} 1 & 0 \\ 12 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & -13 \\ 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & -12 \\ 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 0 \\ 13 & 1 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Recognize that matrices of the form [1, k; 0, 1] multiply such that their top-right entries add: the product gives [1, 1+2+3+...+(n-1); 0, 1]. Use this to find n, then apply the standard inversion formula for 2×2 matrices.
<p><strong>Step 1:</strong> Recognize the pattern. Matrices of the form [1, k; 0, 1] satisfy:</p><p>[1, a; 0, 1] · [1, b; 0, 1] = [1, a+b; 0, 1]</p><p><strong>Step 2:</strong> Apply this pattern to the product:</p><p>[1, 1; 0, 1] · [1, 2; 0, 1] · [1, 3; 0, 1] · ... · [1, n-1; 0, 1] = [1, 1+2+3+...+(n-1); 0, 1]</p><p><strong>Step 3:</strong> The top-right entry equals n(n-1)/2. Given that this equals 78:</p><p>n(n-1)/2 = 78</p><p>n(n-1) = 156</p><p>n² - n - 156 = 0</p><p>(n-13)(n+12) = 0</p><p>Since n > 0, we have n = 13</p><p><strong>Step 4:</strong> Find the inverse of [1, 13; 0, 1]. For a matrix [1, k; 0, 1], the inverse is [1, -k; 0, 1]:</p><p>The inverse of [1, 13; 0, 1] is [1, -13; 0, 1]</p><p>∴ Answer: C</p>
Correct Answer: C