Download App

Determinants — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDeterminants5 Marks
Question
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
✓

Answer

$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$\text{L.H.S}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}$
$=\text{xyz}\begin{vmatrix}1&1&1\\\text{x}&\text{y}&\text{z}\\\text{x}^3&\text{y}^3&\text{z}^3\end{vmatrix}$
$=\text{xyz}\begin{vmatrix}0&1&0\\\text{x}-\text{y}&\text{y}&\text{z}-\text{y}\\\text{x}^3-\text{y}^3&\text{y}^3&\text{z}^3-\text{y}^3\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})\begin{vmatrix}0&1&0\\1&\text{y}&1\\\text{x}^2+\text{y}^2+\text{xy}&\text{y}^3&\text{z}^2+\text{y}^2+\text{zy}\end{vmatrix}$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[\text{z}^2+\text{y}^2+\text{zy}-\text{x}^2-\text{y}^2-\text{xy}]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[(\text{z}-\text{x})(\text{z}+\text{x})+\text{y}(\text{z}-\text{x})]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})(\text{z}-\text{x})[\text{z}+\text{x}+\text{y}]$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$

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 "Solve the Following Question.(5 Marks)" from Determinants→Determinants — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Show that the points $2\hat{\text{i}},-\hat{\text{i}}-4\hat{\text{j}}\text{ and }-\hat{\text{i}}+4\hat{\text{j}}$ form an isosceles triangle.
→
Find the feasible solution of the following inequations graphically.x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
→
Solve the following equation for x:
$\cot^{-1}\text{x}-\cot^{-1}(\text{x}+2)=\frac{\pi}{12},\text{x}>0$
→
At what point of the curve $y=x^2$ does the tangent make an angle of $45^{\circ}$ with the $x$-axis?
→
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ show that $\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $\text{A}^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}.$
→
A particle is moving in a straigth line such that its distance at any time t is given by $\text{S}=\frac{\text{t}^{4}}{4}-2\text{t}^{3}+4\text{t}^{2}-7$. Find when its velocity is maximum and acceleration minimum.
→
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
$\text{f}(\text{x})=4\text{x}-\frac{\text{x}^{2}}{2}\ \text{in}\ [2,4,5]$
→
Find the area of the region included between:

$y^2=4 a x$ and $y=x$

→
Show that $\begin{vmatrix}\text{x}+1&\text{x}+2&\text{x}+\text{a}\\\text{x}+2&\text{x}+3&\text{x}+\text{b}\\\text{x}+3&\text{x}+4&\text{x}+\text{c}\end{vmatrix}=0$ where a, b, c are in A.P.
→
Evaluate the following integrals:
$\int(\text{x}-2)\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$
→