Let A = begin{bmatrix} p & 13 -13 & p end{bm | vedclass

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(A) Given the matrices $A = \begin{bmatrix} p & 13 \ -13 & p \end{bmatrix}$ and $B = \begin{bmatrix} 4q & 85 \ -2 & 1 \end{bmatrix}$.Calculating the

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$31$

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(A) Given the matrices $A = \begin{bmatrix} p & 13 \ -13 & p \end{bmatrix}$ and $B = \begin{bmatrix} 4q & 85 \ -2 & 1 \end{bmatrix}$.Calculating the determinants,we have $|A| = p^2 - (13)(-13) = p^2 + 169$.Calculating $|B| = (4q)(1) - (85)(-2) = 4q + 170$.Given $|A| = |B|$,we have $p^2 + 169 = 4q + 170$,which simplifies to $p^2 = 4q + 1$.Since $p^2 = 4q + 1$,$p^2$ must be odd,implying $p$ is an odd integer. Let $p = 2k + 1$ for some integer $k \geq 0$.Substituting $p = 2k + 1$ into the equation: $(2k + 1)^2 = 4q + 1 \Rightarrow 4k^2 + 4k + 1 = 4q + 1 \Rightarrow 4k(k + 1) = 4q \Rightarrow q = k(k + 1)$.We are given $1 \leq q \leq 1000$ and $1 \leq p \leq 1000$.For $q = k(k + 1) \leq 1000$,we find $k^2 \approx 1000$,so $k \leq 31$.Since $p = 2k + 1$,if $k = 31$,$p = 2(31) + 1 = 63$,which is $\leq 1000$.Thus,$k$ can take values from $1$ to $31$ (since $q \geq 1$,$k \geq 1$).There are $31$ possible values for $k$,and each $k$ gives a unique pair $(p, q)$.Therefore,the total number of ordered pairs $(p, q)$ is $31$.

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

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