• # question_answer For $\lambda \in R$, let $f(\lambda )=$ det $(A-\lambda I)$ where $A\left[ \begin{matrix} 1 & 2 \\ -1 & 3 \\ \end{matrix} \right]$and I is an identity matrix of order 2. The minimum value of $f(\lambda )$ is equal to A)  1                     B)  2 C)  4                                 D)  5

Correct Answer: A

Solution :

$|x|\,=\frac{-(m+6)\,\pm \,(m+2)}{2}\,(reject+sign)$ $|x|=-(m+4)$ For two distinct solution, m + 4 < 0 $\Rightarrow \,m<-4$ $\Rightarrow$ Number of integral values of m in (-10, 10] are {-9, -8, -7, -6, -5} i.e., 5 values Alternating: Since equation has two distinct solution and therefore product of the roots must be less than zero. $\Rightarrow \,2(m+4)<0\Rightarrow \,m<-4$. $f(\lambda )\,=\det \,(A-\lambda I)\,=\left| \begin{matrix} 1-\lambda & 2 \\ -1 & 3-\lambda \\ \end{matrix} \right|\,=(1-\lambda )\,(3-\lambda )\,+2$$={{\lambda }^{2}}-4\,\lambda +5={{(\lambda -2)}^{2}}+1$ Clearly, ${{f}_{\min }}\,(\lambda =2)=1$

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