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UPSC [CSE 2019 P1] Question 1(d) : If $$A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$$ then show that $AB=6I_3$. Use this result to solve the following system of equations : $$\begin{array}{r} 2x+y+z=5\\ x-y=0\\ 2x+y-z=1\\ \end{array}$$ Solution : Given $A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$ $\implies AB=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix}\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$ $\implies AB=\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6 \end{bmatrix}=6\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}=6I...

UPSC [CSE 2019 P1] Question 1(a) : Let $f:\left[0,\dfrac{\pi}{2}\right]\to\textbf{R}$ be a continuous function such that $$f(x)=\frac{\cos ^2 x}{4x^2-\pi^2}, 0\le x<\frac{\pi}{2}$$ Find the value of $f\left(\dfrac{\pi}{2}\right).$  Solution : Given that $f:\left[0,\dfrac{\pi}{2}\right]\to\textbf{R}$ is a continuous function such that $f(x)=\dfrac{\cos ^2 x}{4x^2-\pi^2}, 0\le x<\dfrac{\pi}{2}$. Now, since $f$ is continuous on $\left[0,\dfrac{\pi}{2}\right]$, We have, $\displaystyle\lim_{x\to\pi/2^{-}}f(x)=f\left(\frac{\pi}{2}\right)$ $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{\cos ^2 x}{4x^2-\pi^2}\quad\quad\ldots\left(\frac{0}{0}\right)$ form $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{2\cos x(-\sin x)}{8x}$ $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\frac{-2\sin 2x}{8x}$ $\displaystyle\implies\quad f\left(\frac{\pi}{2}\right)=\lim_{x\to\pi/2^{-}}\fr...