Determinants — Class 12 Maths Solution

(i) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&4\\5&1\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2x}&4\\6&x\end{array}} \right|$

$\Rightarrow$ $2 - 20 = 2{x^2} - 24$ $\Rightarrow$ $- 18 = 2{x^2} - 24$

$\Rightarrow$ $2{x^2} = 24 - 18$ $\Rightarrow$ $2{x^2} = 6$
$\Rightarrow$ ${x^2} = 3$ $\Rightarrow$ $x = \pm \sqrt 3$

(ii) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\4&5\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&3\\{2x}&5\end{array}} \right|$

$\Rightarrow$ $2 \times 5 - 4 \times 3 = 5 \times x - 2 \times 3$ $\Rightarrow$ $10 - 12 = 5x - 6x$

$\Rightarrow$ $- 2 = - x \Rightarrow x = 2$ .

8. If $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&2\\{18}&x\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&2\\{18}&6\end{array}} \right|$, then x is equal to

(A) 6

(B) $\pm 6$

(C) -6

(D) 0

(B) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&2\\{18}&x\end{array}} \right| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}6&2\\{18}&6\end{array}} \right|$

$\Rightarrow$ ${x^2} - 36 = 36 - 36 \Rightarrow {x^2} = 36 \Rightarrow x = \pm 6$

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