Download App

MATRICES — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMATRICES5 Marks
Question
Find the matrix A satisfying the matrix equation:$\begin{bmatrix}2&1\\3&2\end{bmatrix}\text{A}\begin{bmatrix}-3&2\\5&-3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$
✓

Answer

We have, $\begin{bmatrix}2&1\\3&2\end{bmatrix}\text{A}\begin{bmatrix}-3&2\\5&-3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ Or PAQ = I, where $\text{P}=\begin{bmatrix}2&1\\3&2\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}-3&2\\5&-3\end{bmatrix}$ $\therefore\ \text{P}^{-1}\text{PAQ}=\text{P}^{-1}\text{I}$ $\Rightarrow\ \text{IAQ}=\text{P}^{-1}$ $\Rightarrow\ \text{AQ}=\text{P}^{-1}$ $\Rightarrow\ \text{AQQ}^{-1}=\text{P}^{-1}\text{Q}^{-1}$ $\Rightarrow\ \text{AI}=\text{P}^{-1}\text{Q}^{-1}$ $\Rightarrow\ \text{A}=\text{P}^{-1}\text{Q}^{-1}$ Now adj. $\text{P}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}$ and $|\text{P}|=1$ $\therefore\ \text{P}^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}$ Also adj. $\text{Q}=\begin{bmatrix}-3&-2\\-5&-3\end{bmatrix}$ and $|\text{Q}|=-1$ $\therefore\ \text{Q}^{-1}=\begin{bmatrix}3&2\\5&3\end{bmatrix}$$\Rightarrow\ \text{A}=\text{P}^{-1}\text{Q}^{-1}$
$=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}\begin{bmatrix}3&2\\5&3\end{bmatrix}$
$=\begin{bmatrix}6-5&4-3\\-9+10&-6+6\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}$

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 MATRICES→MATRICES — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Evaluate the following integrals:
$\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
→
Prove that $\begin{vmatrix}\text{bc}-\text{a}^2&\text{ca}-\text{b}^2&\text{ab}-\text{c}^2\\\text{ca}-\text{b}^2&\text{ab}-\text{c}^2&\text{bc}-\text{a}^2\\\text{ab}-\text{c}^2&\text{bc}-\text{a}^2&\text{ca}-\text{b}^2\end{vmatrix}$ is divisible by $(a + b + c)$ and find the quotient.
→
Find the general solution of $(x+y) \frac{d y}{d x}=1$
→
Differentiate the following functions with respect to x:
$\text{e}^{\sin\text{x}}+(\tan\text{x})^\text{x}$
→
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} 2 & -1 & 3 \\ 1 & 2 &4 \\ 3 & 1 & 1 \end{bmatrix}$
→
If f(x) is a continuous function defind on [-a, a], then prove that:
$\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$ 
→
If $\vec{\text{a}}$ be the position vector whose tip is (5, -3), find the coordinates of a point B such that $\overrightarrow{\text{AB}}=\vec{\text{a}}$, the coordinates of A being (4, -1).
→
If $\text{A}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix},$ find the value of $\lambda$ so that $\text{A}^2=\lambda\text{A}-2\text{I}.$ Hence, find $A^{-1}.$
→
Evaluate the following integrals:
$\int\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
→
An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 6 heads.
→