If left[ {begin{array}{*{20}{c}}{2 + x}&3&41&{ - 1}&2x&1&{ - 5}end{array}} r

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3 and 4 .Determinants and Matrices

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If $\left[ {\begin{array}{*{20}{c}}{2 + x}&3&4\\1&{ - 1}&2\\x&1&{ - 5}\end{array}} \right]$is a singular matrix, then $x$ is

A

$\frac{{13}}{{25}}$

B

$ - \frac{{25}}{{13}}$

C

$\frac{5}{{13}}$

D

$\frac{{25}}{{13}}$

Solution

(b) Given, $\left| {\,\begin{array}{*{20}{c}}{2 + x}&3&4\\1&{ – 1}&2\\x&1&{ – 5}\end{array}\,} \right|\, = 0$

==> $(2 + x)(5 – 2) – 3( – 5 – 2x) + 4(1 + x) = 0$

==> $6 + 3x + 15 + 6x + 4 + 4x = 0$

==> $13x = – 25 \Rightarrow x = – \frac{{25}}{{13}}$.

Standard 12

Mathematics

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