The inverse of [ * 20 c 2& - 3 - 4 &2 ]is - Vidyadip

MCQ

The inverse of $\left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&2\end{array}} \right]$is

✓
$\frac{{ - 1}}{8}\,\left[ {\begin{array}{*{20}{c}}2&3\\4&2\end{array}} \right]$
B
$\frac{{ - 1}}{8}\,\left[ {\begin{array}{*{20}{c}}3&2\\2&4\end{array}} \right]$
C
$\frac{1}{8}\,\left[ {\begin{array}{*{20}{c}}2&3\\4&2\end{array}} \right]$
D
$\frac{1}{8}\,\left[ {\begin{array}{*{20}{c}}3&2\\2&4\end{array}} \right]$

✓

Answer

Correct option: A.

$\frac{{ - 1}}{8}\,\left[ {\begin{array}{*{20}{c}}2&3\\4&2\end{array}} \right]$

a
(a) Let $A = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&2\end{array}} \right]\,,{\rm{ }}\therefore \,|A|\, = \,\left| {\,\begin{array}{*{20}{c}}2&{ - 3}\\{ - 4}&2\end{array}\,} \right| = 4 - 12 = - 8$

The matrix of cofactors of the elements of $A $ viz.

$\left[ {\begin{array}{*{20}{c}}{{c_{11}}}&{{c_{12}}}\\{{c_{21}}}&{{c_{22}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&{ - ( - 4)}\\{ - ( - 3)}&2\end{array}} \right] = \left[ {\,\begin{array}{*{20}{c}}2&4\\3&2\end{array}\,} \right]$

$\therefore $ $adjA = $ transpose of the matrix of cofactors of elements of $A = \left[ {\begin{array}{*{20}{c}}2&3\\4&2\end{array}} \right]$ ${A^{ - 1}} = \frac{1}{\Delta }(adj\,A) = \frac{1}{{ - 8}}\,\left[ {\begin{array}{*{20}{c}}
2&3 \\
4&2
\end{array}} \right]$

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

Consider a matrix $A =\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta\end{array}\right]$, where $\alpha, \beta, \gamma$ are three distinct natural numbers. If $\frac{\operatorname{det}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=2^{32} \times 3^{16}$, then the number of such $3 -$ tuples $(\alpha, \beta, \gamma)$ is $.....$

The area of the region enclosed by $y \leq 4 x^{2}, x^{2} \leq 9 y$ and $y \leq 4$, is equal tọ

${d \over {dx}}\left( {{{{{\cot }^2}x - 1} \over {{{\cot }^2}x + 1}}} \right) = $

Let $f : R \rightarrow R$ be a differentiable function such that $f \left(\frac{\pi}{4}\right)=\sqrt{2}, f \left(\frac{\pi}{2}\right)=0$ and $f ^{\prime}\left(\frac{\pi}{2}\right)=1$ and let $g(x)=\int\limits_{x}^{\pi / 4}\left(f^{\prime}(t) \sec t+\tan t \operatorname{sectf}(t)\right) d t$ for $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then $\lim\limits _{ x \rightarrow\left(\frac{\pi}{2}\right)^{-}} g ( x )$ is equal to

$\int_{}^{} {\frac{{{x^2}{{\tan }^{ - 1}}{x^3}}}{{1 + {x^6}}}\;dx} $ is equal to

In a square matrix $A$ of order $3, a_{i i}'s$ are the sum of the roots of the equation $x^2 - (a + b)x + ab= 0$; $a_{i , i + 1}'s$ are the product of the roots, $a_{i , i - 1}'s$ are all unity and the rest of the elements are all zero. The value of the det. $(A)$ is equal to

If $\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13}.$ then, $\text{P}(\overline{\text{A}}|\text{B})=$

$\frac{5}{9}$
$\frac{4}{9}$
$\frac{4}{13}$
$\frac{6}{13}$

If $x^py^q=(x+y)^{p+q}$,then $\frac {dy}{dx}=$

$|a \times i{|^2} + |a \times j{|^2} + |a \times k{|^2} = $

$\int1.\text{dx}=$

$\text{x}+\text{k}$
$1+\text{k}$
$\frac{\text{x}^2}{2}+\text{k}$
$\log\text{x}+\text{k}$