Question
Let $A = \left| \begin{matrix} 1 & 3 & - 1 \\ 2 & 1 & \alpha \\ 0 & 1 & - 1 \end{matrix} \right|$ be a singular matrix. Let $f(x) = \int_{0}^{x}\mspace{2mu}\left( t^{2} + 2t + 3 \right)dt,x \in \lbrack 1,\alpha\rbrack$. If $M$ and $m$ are respectively the maximum and the minimum values of $f$ in $\lbrack 1,\alpha\rbrack$ then $3(M - m)$ is equal to :
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Solution
Sol. [Key Concept: Singularity $det\ A = 0$ gives $\alpha = 3$; then $f(x) = \int_{0}^{x}(t^{2} + 2t + 3)\ dt$ is increasing on $\lbrack 1,3\rbrack$, so $M = f(3),\ m = f(1)$.]
$${|A| = 0 }{\Rightarrow 1.( - 1 - \alpha) - 3( - 2 - 0) - 1(2 - 0) = 0 }{\Rightarrow \alpha = 3 }{f(x) = \int_{0}^{x}\mspace{2mu}\left( t^{2} + 2t + 3 \right)dt;x \in \lbrack 1,3\rbrack }{f(x) = \frac{x^{3}}{3} + x^{2} + 3x }$$$f^{'}(x) = x^{2} + 2x + 3\ $ where $D < 0$
$\Rightarrow f(x)$ is strictly increasing
$${M = f(3) = 9 + 9 + 9 = 27 }{m = f(1) = \frac{1}{3} + 1 + 3 = \frac{13}{3} }{3(M - m) = 3\left( 27 - \frac{13}{3} \right) = 68 }$$
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