Let $A = left[ {begin{array}{*{20}{c}} p&{13} { - 13}&p end{array}} right] and B

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3 and 4 .Determinants and Matrices

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Let $A = \left[ {\begin{array}{{20}{c}}
p&{13}\\
{ - 13}&p
\end{array}} \right]$ and $B = \left[ {\begin{array}{{20}{c}}
{4q}&{85}\\
{ - 2}&1
\end{array}} \right]$ where $p,q \in N$. It is given that $\left| A \right| = \left| B \right|$ and $p,q \in[1,1000]$. Then total number of ordered pairs $(p,q)$ is

A

$31$

B

$35$

C

$41$

D

$23$

Solution

$p^{2}+169=4 q+170$

$p^{2}=4 q+1$

Let $\mathrm{p}=(2 \mathrm{k}+1) \Rightarrow \mathrm{q}=\mathrm{k}(\mathrm{k}+1)$

$1 \leq k(k+1) \leq 1000$

Number of possible values are $31 .$

Standard 12

Mathematics

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