Matrices & Determinants
Matrices
nta_pyq_2025_jan
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 the 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: Notice $a_{ij}=(\sqrt{2})^{i+j}=2^{(i+j)/2}.$ The third row of $A^{2}$ is the third row of $A$ multiplied by $A$, so each entry is a sum of three products with nice powers of $\sqrt{2}.$
Row $3$ of $A$: $(4,\,4\sqrt{2},\,8)$. Columns sums of $A$:
col $1$: $2+2\sqrt{2}+4=6+2\sqrt{2}$;
col $2$: $2\sqrt{2}+4+4\sqrt{2}=4+6\sqrt{2}$;
col $3$: $4+4\sqrt{2}+8=12+4\sqrt{2}.$
$(A^{2})_{31}=4(2)+4\sqrt{2}(2\sqrt{2})+8(4)=8+16+32=56.$
$(A^{2})_{32}=4(2\sqrt{2})+4\sqrt{2}(4)+8(4\sqrt{2})=8\sqrt{2}+16\sqrt{2}+32\sqrt{2}=56\sqrt{2}.$
$(A^{2})_{33}=4(4)+4\sqrt{2}(4\sqrt{2})+8(8)=16+32+64=112.$
Sum of row $3$: $56+56\sqrt{2}+112=168+56\sqrt{2}.$
$\alpha=168,\ \beta=56,\ \alpha+\beta=224.$
Correct Answer: 2