Download App

3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
If $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$ $(i = \sqrt { - 1} ),$then ${A^{ - 1}}$=
  • A
    $\left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$
  • ✓
    $\left[ {\begin{array}{*{20}{c}}{ - i}&0\\0&{ - 2i}\end{array}} \right]$
  • C
    $\left[ {\begin{array}{*{20}{c}}i&0\\0&{2i}\end{array}} \right]$
  • D
    $\left[ {\begin{array}{*{20}{c}}0&i\\{2i}&0\end{array}} \right]$
✓

Answer

Correct option: B.
$\left[ {\begin{array}{*{20}{c}}{ - i}&0\\0&{ - 2i}\end{array}} \right]$
b
(b) For $A = \left[ {\begin{array}{*{20}{c}}i&0\\0&{i/2}\end{array}} \right]$,

$adj\,(A) = \left[ {\begin{array}{*{20}{c}}{i/2}&0\\0&i\end{array}} \right]$ and $|A| = - \frac{1}{2}$.

$\therefore $ ${A^{ - 1}} = \frac{1}{\Delta }(adj\,A) = \frac{1}{{ - 1/2}}\,\left[ {\begin{array}{*{20}{c}}{i/2}&0\\0&i\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - i}&0\\0&{ - 2i}\end{array}} \right]$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 3 and 4 . determinant and metrices→3 and 4 . determinant and metrices — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The graph of the function $f(x)=\frac{\cos x}{\cos 2 x}$ in the domain $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ is
→
If $\int_{}^{} {\frac{{dx}}{{1 + \sin x}} = \tan \left( {\frac{x}{2} + a} \right) + b} $, then
→
Let $\mathrm{f}: R \rightarrow R$ and $\mathrm{g}: R \rightarrow R$ be defined as $f(x)=\left\{\begin{array}{lll}\log _e x & , & x>0 \\ e^{-x} & , & x \leq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{lll} x & , & x \geq 0 \\ e^{x} & , & x <  0\end{array}\right.$ Then $gof:R \to R$ is . . . . 
→
If X follows a binomial distribution with parameter $\text{n}=100$ and $\text{p}=\frac{1}{3},$ then P(X = r) is maximum when r = 
→
$\mathop \smallint \limits_0^{1.5} x\left[ {{x^2}} \right]dx = $
→
The range of the function, $\text{f(x)}=(1+\sec^{-1}\text{x})(1+\cos^{-1}\text{x})$ is:
  1. $(-\infty,\infty)$
  2. $(-\infty,0]\cup[4.\infty)$
  3. $\big\{0,(1+\pi^2)\big\}$
  4. $[1.(1+\pi)^2]$
→
Solution of the differential equation, $y\,dx - x\,dy + x{y^2}dx = 0$ can be
→
The shortest distance between the lines $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to $............$
→
Let $( a , b ) \subset(0,2 \pi)$ be the largest interval for which $\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta) > 0, \theta \in(0,2 \pi)$ holds. If $\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}$ $\left(x^2-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to:
→
Let $k$ be a non-zero real number 

If $f(x) = {\rm{ }}\left\{ {\begin{array}{*{20}{c}}
{\frac{{\left( {{e^x} - 1} \right)^2}}{{\sin {\mkern 1mu} \left( {\frac{x}{k}} \right){\mkern 1mu} \log {\mkern 1mu} \left( {1 + \frac{x}{4}} \right)}}{\mkern 1mu} ,{\mkern 1mu} x \ne 0}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 12{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ,x{\mkern 1mu} {\mkern 1mu}  = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }
\end{array}} \right.$ 

is a continuous function then the value of $k$ is

→