Download App

Matrices — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMatrices3 Marks
Question
Find $x,$ if $\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
✓

Answer

Given $:\left[ {\begin{array}{*{20}{c}} x&{ - 5}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ 4 \\ 1 \end{array}} \right] = 0$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {x - 0 - 2}&{0 - 10 - 0}&{2x - 5 - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ 4 \\ 1 \end{array}} \right] = 0$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {x - 2}&{ - 10}&{2x - 8} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ 4 \\ 1 \end{array}} \right] = 0$
$\Rightarrow [(x - 2)x - 10(4) + (2x - 8)1] = 0$
$\Rightarrow [x^2 - 2x - 40 + 2x - 8] = 0$
$\Rightarrow {\left[ {{x^2} - 48} \right]_{1 \times 1}} = {\left[ 0 \right]_{1 \times 1}}$
Equating corresponding entries, we have
$x^2 - 48 = 0$
$ \Rightarrow x^2 = 48$
$ \Rightarrow x = \pm 4\sqrt 3$

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

Represent the following families of curves by forming the corresponding differential equation:
$\text{y}=4\text{ax}$
→
Find value of p when given lines $\frac{1-x}{3}=\frac{7 y-14}{3 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are perpendicular.
→
If f : A → B and g : B → C are onto functions, show that gof is a onto function.
→
For the following differntial equations verify that the accompanying function is a solution:
Differential equation Function
$\text{x}^3\frac{\text{d}{^2}\text{y}}{\text{dx}^2}=1$ $\text{y}=\text{ax}+\text{b}+\frac{1}{2\text{x}}$
→
If $\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}},$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$
→
Find the equation of rthe planes parallel to the plane x + 2y - 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).
→
Find the mean and standard deviation of the following probability distributions:
$\text{x}_\text{i}$
$0$
$1$
$2$
$3$
$4$
$5$
$\text{p}_\text{i}$
$\frac{1}{6}$
$\frac{5}{18}$
$\frac{2}{9}$
$\frac{1}{6}$
$\frac{1}{9}$
$\frac{1}{18}$
→
By using the properties of definite integrals, evaluate the integral $\int\limits_{ - 5}^5 {\left| {x + 2} \right|dx} $
→
Evaluate the definite integral $\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$
→
Examine the following functions for continuity. $\text f(\text X)=\frac{\text X^{2} - 25}{\text {X} + 5}$
→