<p>Given
\[\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix}\cdots\begin{bmatrix}1 & n-1\\0 & 1\end{bmatrix} = \begin{bmatrix}1 & 78\\0 & 1\end{bmatrix}\]
Then find the inverse of matrix \(A = \begin{bmatrix}1 & 13\\0 & 1\end{bmatrix}\). What is \(A^{-1}\)?</p>
<p>\(\begin{bmatrix}1 & 13\\0 & 1\end{bmatrix}\)</p>
<p>\(\begin{bmatrix}1 & -13\\0 & 1\end{bmatrix}\)</p>
<p>\(\begin{bmatrix}1 & 0\\13 & 1\end{bmatrix}\)</p>
<p>\(\begin{bmatrix}1 & 0\\-13 & 1\end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Matrices of the form [1, k; 0, 1] multiply by adding their upper-right entries: the product equals [1, 1+2+3+...+(n-1); 0, 1]. Use this pattern to find n, then apply the standard formula for 2×2 matrix inversion.
<p><strong>Step 1: Identify the pattern.</strong> Matrices of form [1, k; 0, 1] multiply as:</p><p>[1, k₁; 0, 1] · [1, k₂; 0, 1] = [1, k₁+k₂; 0, 1]</p><p><strong>Step 2: Find the sum.</strong> The product equals [1, 1+2+3+...+(n-1); 0, 1] = [1, n(n-1)/2; 0, 1] = [1, 78; 0, 1]</p><p>Therefore: n(n-1)/2 = 78 → n(n-1) = 156 → n = 13</p><p><strong>Step 3: Verify matrix A.</strong> A = [1, 13; 0, 1] appears in the product (when k=13)</p><p><strong>Step 4: Compute A⁻¹.</strong> For a 2×2 matrix [1, k; 0, 1]:</p><p>det(A) = 1(1) - 13(0) = 1</p><p>A⁻¹ = (1/det(A)) · [1, -k; 0, 1] = [1, -13; 0, 1]</p><p>∴ Answer: <strong>A⁻¹ = [1, -13; 0, 1]</strong> or equivalently <strong>[1, -13; 0, 1]</strong></p>
Correct Answer: B