Matrices & Determinants
Matrix Multiplication
nta_pyq_2025_apr
Grade 12
Question:
Let $A = [a_{ij}]$ be a matrix of order $3 \times 3$, with $a_{ij} = (\sqrt{2})^{i+j}$. If the sum of all elements in the third row of $A^2$ is $\alpha + \beta\sqrt{2}$, $\alpha, \beta \in \mathbb{Z}$, then $\alpha + \beta$ is equal to:
Step-by-Step Solution
Key Concept: Write out $A$ explicitly with $(\sqrt{2})^{i+j}$ entries, compute $A^2 = A \cdot A$, focusing on the third row, and extract $\alpha$ and $\beta$.
$A^2 = 4 \begin{bmatrix}\cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots \\ (2+4+8) & (2\sqrt{2}+4\sqrt{2}+8\sqrt{2}) & (4+8+16)\end{bmatrix}$. Third row sum $= 4(14 + 14\sqrt{2} + 28) = 4(42 + 14\sqrt{2}) = 168 + 56\sqrt{2}$. $\alpha = 168, \beta = 56$, $\alpha + \beta = 224$.
Correct Answer: 224