Download App

Matrices — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMatrices3 Marks
Question
Find the matrix $X$ so that $X\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right]$
✓

Answer

Let $X = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]$
$\therefore \;\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} 2 \\ 5 \end{array}\;\;\begin{array}{*{20}{c}} 3 \\ 6 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7} \\ 2 \end{array}\;\;\begin{array}{*{20}{c}} { - 8} \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} { - 9} \\ 6 \end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}} {a + 4b} \\ {c + 4d} \end{array}\;\;\begin{array}{*{20}{c}} {2a + 5b} \\ {2c + 5d} \end{array}\;\;\begin{array}{*{20}{c}} {3a + 6b} \\ {3c + 6d} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7} \\ 2 \end{array}\;\;\begin{array}{*{20}{c}} { - 8} \\ 4 \end{array}\;\;\begin{array}{*{20}{c}} { - 9} \\ 6 \end{array}} \right]$
On solving a + 4b = -7 and  2a + 5b = -8 & c + 4d = 2 and 2c + 5d = 4
we get a = 1, b = -2, c = 2, d = 0
$X = \left[ {\begin{array}{*{20}{c}} 1&{ - 2} \\ 2&0 \end{array}} \right]$

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 "3 Marks Question" 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

Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),\;a > 0,\left( { - \frac{a}{{\sqrt 3 }} < x < \frac{a}{{\sqrt 3 }}} \right)$
→
$\int\frac{2\text{x}}{(2\text{x}+1)^2}\text{dx}$
→
If with reference to the right handed system of mutually perpendicular unit vectors $\hat{i}, \hat{j}$ and $\hat{k}, \vec{\alpha}=3 \hat{i}-\hat{j.}$
$\vec{\beta}=2 \hat{i}+\hat{j}-3 \hat{k.}$ then express $\vec{\beta}$ in the form $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, where $\vec{\beta}_1$ is $\|$ to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.
→
Find the mean and standard deviation of the following probability distributions:
$x_i$ $1$ $2$ $3$ $4$
$p_i$ $0.4$ $0.1$ $0.2$ $0.3$
→
Prove that $\cot\Big(\frac{\pi}{4}-2\cot^{-1}3\Big)=7$
→
Solve the following equation for x:
$\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)+\cot^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)=\frac{2\pi}{3},\text{x}>0$
→
Eight coins are thrown simultaneously. Find the chance of obtaining at least six heads.
→
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $\vec{\text{r}}.\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)=5\ \text{and}\ \vec{\text{r}}.\Big(3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=6.$
→
Prove that:
$\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{8}=\frac{\pi}{4}$
→
Let * be a binary operation on Q - {-1} defined by a * b = a + b + ab for all a, b ∈ Q - {-1}. Then,
Show that every element of Q − {−1} is invertible. Also, find the inverse of an arbitrary element.
→