Download App

Determinants — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDeterminants5 Marks
Question
If $\triangle=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix},$ $\triangle_1=\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix},$ then prove that $\triangle+\triangle_1=0$
✓

Answer

$\triangle+\triangle_1=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&\text{yz}&\text{x}\\1&\text{zx}&\text{y}\\1&\text{xy}&\text{z} \end{vmatrix}$
$[$Interchanging rows and coloumns in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\1&\text{y}&\text{zx}\\1&\text{z}&\text{xy} \end{vmatrix}$
$[$Applying $\text{C}_2\leftrightarrow\text{C}_3$ in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{y}-\text{x}&\text{y}^2-\text{x}^2\\0&\text{z}-\text{x}&\text{z}^2-\text{x}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\0&\text{y}-\text{x}&\text{zx}-\text{yz}\\0&\text{z}-\text{x}&\text{xy}-\text{yz}\end{vmatrix}
[$ Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1]$
$=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{y}+\text{x}\\0&1&\text{z}+\text{x}\end{vmatrix}-(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{yz}\\0&1&-\text{z}\\0&1&-\text{y}\end{vmatrix}
[$Taking $(y - x)$ common from $R_2$ and $(z - x)$ common from $R_3]$
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}+\text{x}-\text{y}-\text{x})-(\text{y}-\text{x})(\text{z}-\text{x})(-\text{y}+\text{z})
[$Expanding along first column$]$
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}-\text{y})(1-1)$
$=0$
$\therefore\ \triangle+\triangle_1=0$

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

Evaluate the following intergrals:
$\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
→
In a shop X, 30 tins of pure ghee and 40 tins of adulterated ghee which look alike, are kept for sale while in shop Y, similar 50 tins of pure ghee and 60 tins of adulterated ghee are there. One tin of ghee is purchased from one of the randomly selected shops and is found to be adulterated. Find the probability that it is purchased from shop Y. What measures should be taken to stop adulteration?
→
Differentiate the following functions with respect to x:
$10^{\log\sin\text{x}}$
→
For each of the differential equations given in find a particular solution satisfying the given condition:
$\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}=\sin\text{x};\text{y}=0\ \text{when x}=\frac{\pi}{3}$
→
Write the values of 'a' for which the following distribution of probabilities becomes a probability distributioin:
$\text{X}=\text{x}_\text{i}:$ $-2$ $-1$ $0$ $1$
$\text{P}(\text{X}=\text{x}_\text{i}):$ $\frac{1-\text{a}}{4}$ $\frac{1+2\text{a}}{4}$ $\frac{1-2\text{a}}{4}$ $\frac{1+\text{a}}{4}$
→
Find the shortest distance between the following pairs of parallel lines whose equations are:
$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}\big)+\lambda\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(4\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
→
Prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}$
→
Solve the following equation for x:
$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+1}\Big)=\frac{\pi}{4}$
→
If $\int\limits_{0}^{\text{k}}\frac{1}{2+8\text{x}^2}\text{ dx}=\frac{\pi}{16},$ find the value of k.
→
If $\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$ show that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$
→