Download App

Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants5 Marks
Question
Prove that:
$\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}=4\text{abc}$
✓

Answer

$\text{L.H.S}=\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2&\text{c}^2\\\text{a}^2&\text{b}^2+\text{c}^2&\text{a}^2\\\text{b}^2&\text{b}^2&\text{c}^2+\text{a}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&-2\text{b}^2&-2\text{a}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}(-\text{a}^2)\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\\text{b}^2+\text{c}^2-\text{a}^2&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2(\text{c}^2+\text{a}^2-\text{b}^2)\}+\text{b}^2\{0+2\text{a}^2(\text{b}^2+\text{c}^2-\text{a}^2)\}\big]$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2\text{c}^2-2\text{b}^2\text{a}^2+2\text{b}^4\}+\text{b}^2\{2\text{a}^2\text{b}^2+2\text{a}^2\text{c}^2-2\text{a}^4\}\big]$
$=\frac{1}{\text{abc}}\big[2\text{a}^2\text{b}^2\text{c}^2+2\text{a}^4\text{b}^2-2\text{a}^4\text{b}^4+2\text{a}^2\text{b}^4+2\text{a}^2\text{b}^2\text{c}^2-2\text{a}^4\text{b}^2\big]$
$=\frac{1}{\text{abc}}4\text{a}^2\text{b}^2\text{c}^2$
$=4\text{abc}$
$=\text{R.H.S}$

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

If A and B are two events such that $2\text{P(A)}=\text{P(B)}=\frac{5}{13}$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{5}$ find $\text{P}(\text{A}\cap\text{B}).$
→
Find the vector and Cartesian equations of the line through the point (1, 2, – 4) and perpendicular to the two lines.
$\overrightarrow{\text{r}} = (8\hat{\text{i}} - 19\hat{\text{j}} + 10\hat{\text{k}})+\lambda(3\hat{\text{i}} - 16\hat{\text{j}} + 7\hat{\text{k})}$ and $\overrightarrow{\text{r}} = (15\hat{\text{i}} - 29\hat{\text{j}} + 5\hat{\text{k}})+\mu(3\hat{\text{i}} - 8\hat{\text{j}} + 5\hat{\text{k})}.$
→
If $\vec{\alpha}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\beta}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}},$ then express $\vec{\beta}$ in the form of $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2,$ where $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.
→
Differentiate the following functions from first principles:
$\text{e}^{\sqrt{2\text{x}}}$
→
Prove that :$\tan ^{-1}\left\{\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}\right\}=\frac{\pi}{4}+\frac{x}{2}, 0 < x <\frac{\pi}{2}$
→
Find the shortest distance between the lines
$\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}$
→
If  $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}},\vec{\text{b}}=-3\hat{\text{i}}+4\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}},$ compute $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\vec{\text{c}}$ and $\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{c}}\big)$ and verify that these are not equal.
→
Solve the matrix equations:
$\begin{bmatrix}2\text{x}&3\end{bmatrix}\begin{bmatrix}1&2\\-3&0\end{bmatrix}\begin{bmatrix}\text{x}\\8\end{bmatrix}=0$
→
At what point of the curve $y = x^2$ does the tangent make an angle of $45^\circ$ with the $x-$axis?
→
If $\text{y}=\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2},$ prove that $\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
→