Let $M = [\begin{matrix} a & b \\ c & d \end{matrix}], A = [\begin{matrix} 1 \\ - 1 \end{matrix}], B = [\begin{matrix} - 1 \\ 2 \end{matrix}], C = [\begin{matrix} 1 \\ 0 \end{matrix}]$ and satisfying the conditions $M A = B a n d M^{2} A = C .$ If $x_{1} a n d x_{2} (x_{1} > x_{2})$ are two values of $x$ for which $\det (M - x I) = 0$ where $I$ is the identity matrix of order $2 \times 2$ then the value of $5 x_{1} + 2 x_{2}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

– 3

– 12

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$M A = B \Rightarrow a - b = - 1..... (1) c - d = 2..... (2) M^{2} A = C \Rightarrow M B = C \Rightarrow - a + 2 b = 1..... (3) - c + 2 d = 0..... (4)$
From (1) (2) & (3), we get $a = - 1, b = 0, c = 4, d = 2$
$∴ M = [\begin{matrix} - 1 & 0 \\ 4 & 2 \end{matrix}]$
$\det (M - x I) = 0 \begin{array}{l} \Rightarrow | \begin{matrix} - 1 - x & 0 \\ 4 & 2 - x \end{matrix} | = 0 \\ \Rightarrow x^{2} - x - 2 = 0 \end{array} \Rightarrow x_{1} = 2, x_{2} = - 1 5 x_{1} + 2 x_{2} = 8$