Question
Find the value of p and q if:
$\left[\begin{array}{cc}2 p+1 & q^2-2 \\ 6 & 0\end{array}\right]=\left[\begin{array}{cc}p+3 & 3 q-4 \\ 5 q-q^2 & 0\end{array}\right]$.

Answer

$2p + 1 = p + 3;$
$2p - p = 3 - 1$
$p = 2 ...(1)$
$q^2 - 2 = 3q - 4$
$q^2 - 3q + 2 = 0$
$q^2 - 2q - q + 2 = 0$
$q(q - 2) - (q - 2) = 0$
$(q - 2) (q - 1) = 0 ...(2)$
$5q - q^2 = 6$
$q^2 - 5q + 6 = 0$
$(q - 3) (q - 2) = 0 ...(3)$
By equation $(2)$ and $(3)$
$q = 2$
$\Rightarrow p = 2, q = 2.$

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$=a\left(\frac{1}{t^2}+1\right)=\frac{a\left(t^2+1\right)}{t^2}$
$\text { Now } \frac{1}{ SP }+\frac{1}{ SQ }=\frac{1}{a\left(t^2+1\right)}+\frac{1 \times t^2}{a\left(t^2+1\right)}$
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