Download App

DETERMINANTS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDETERMINANTS5 Marks
Question
$\text{Let A} = \begin{bmatrix}3&7\\2&5\end{bmatrix}\text{and B} = \begin{bmatrix}6&8\\7&9\end{bmatrix}.$Verify that $(AB)^{-1} = B^{-1}A^{-1}.$
✓

Answer

$\text{Given}:\ \text{Matrix A}=\begin{bmatrix}3&7\\2&5\end{bmatrix}$ 
$\therefore\ \text{|A|}=\begin{vmatrix}3&7\\2&5\end{vmatrix}=15-14=1\neq0$
$\therefore\ \text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj. A}=\frac{1}{1}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\begin{bmatrix}5&-7\\-2&3\end{bmatrix}$
$\text{Matrix B}=\begin{bmatrix}6&8\\7&9\end{bmatrix}$ $\therefore\ \text{|B|}=\begin{vmatrix}6&8\\7&9\end{vmatrix}=54-56=-2\neq0$
$\therefore\ \text{B}^{-1}=\frac{1}{\text{|B|}}\text{adj. B}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}$
$\text{Now}\ \text{AB}=\begin{bmatrix}3&7\\2&5\end{bmatrix}\begin{bmatrix}6&8\\7&9\end{bmatrix}=\begin{bmatrix}18+49&24+63\\12+35&16+45\end{bmatrix}=\begin{bmatrix}67&87\\47&61\end{bmatrix}$
$\therefore\ \text{|AB|}=\begin{vmatrix}67&87\\47&61\end{vmatrix}=67(61)-87(47)=4087-4089=-2\neq0$
$\text{Now}\ \text{L.H.S.}=\left(\text{AB}\right)^{-1}=\frac{1}{\text{|AB|}}\text{adj. (AB)}=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(1)$
$\text{R.H.S.}=\text{B}^{-1}\text{A}^{-1}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}45+16&-63-24\\-35-12&49+18\end{bmatrix}$
$=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(2)$
From eq. $(1)$ and $(2),$ we get $\text{L.H.S. = R.H.S.}$
$\Rightarrow (AB)^{-1} = B^{-1}A^{-1}$

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 "5 Marks Questions" from DETERMINANTS→DETERMINANTS — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find a point on the parabola $y = (x - 4)^2,$ where the tangent is parallel to the chord joining $(4, 0)$ and $(5, 1).$
→
Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to 2/3 of the diameter of the sphere.
→
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\sin\text{x}}+\big(\sin\text{x}\big)^\text{x}$
→
Show that the points whose position vectors are as given below are collinear:
$2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$
→
Find the shortest distance between the lines $I_1$ and $l_2$
$\vec{r}=(-\hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})$ and $\vec{r}=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\mu(\hat{\imath}-2 \hat{\jmath}+\hat{k})$where $\lambda$ and $\mu$ are parameters.
→
Solve the following differential equation:
$\text{xy}\log\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\Big\{\text{y}^2-\text{x}^2\log\Big(\frac{\text{y}}{\text{x}}\Big)\Big\}\text{dy}=0$
→
Evaluate the following integrals:
$\int(\text{x}+2)\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
→
A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl's displacement from her initial point of departure.
→
Evaluate the following integrals:
$\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}+\sqrt[9]{9-\text{x}}}\text{ dx}$
→
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\tan\text{x})^{\cot\text{x}}+(\cot\text{x})^{\tan\text{x}}$
→