Download App

DETERMINANTS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDETERMINANTS5 Marks
Question
Using properties of determinants, prove the following:
$\begin{vmatrix} \text{x} &\text{x}^{2} & \text{1 + px}^{3} \\ \text{y} & \text{y}^{2} & \text{1 + py}^{3} \\ \text{z} & \text{z}^{2} & \text{1 + pz}^{3} \end{vmatrix}=\text{(1 + pxyz) (x - y)(y - z)(z - x)}$.
✓

Answer

$\text{LHS} = \begin{vmatrix} \text{x} &\text{x}^{2} & \text{1 + px}^{3} \\ \text{y} &\text{y}^{2} & \text{1 + py}^{3}\\ \text{z} &\text{z}^{2} & \text{1 + pz}^{3} \end{vmatrix}=\begin{vmatrix} \text{x} & \text{x}^{2} & 1 \\ \text{y} & \text{y}^{2} & 1 \\ \text{z} & \text{z}^{2} & 1 \end{vmatrix}+\text{p}\begin{vmatrix} \text{x} & \text{x}^{2} & \text{x}^{3} \\ \text{y} & \text{y}^{2} & \text{y}^{3} \\ \text{z} & \text{z}^{2} & \text{z}^{3} \end{vmatrix}$
=$\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{1} & \text{y} & \text{y}^{2} \\ \text{1} & \text{z} & \text{z}^{2} \end{vmatrix}+\text{pxyz}\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{1} & \text{y} & \text{y}^{2} \\ \text{1} & \text{z} & \text{z}^{2} \end{vmatrix}$
= (1 + pxyz) $\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{1} & \text{y} & \text{y}^{2} \\ \text{1} & \text{z} & \text{z}^{2} \end{vmatrix}$
= (1 + pxyz) $\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{y - x} & \text{y}^{2}-\text{x}^{2} \\ \text{0} & \text{z - x} & \text{z}^{2}-\text{x}^{2} \end{vmatrix} \begin{matrix} \text{R}_{2}\rightarrow & \text{R}_{2}\text{ }-& \text{R}_{1} \\ \text{R}_{3}\rightarrow & \text{R}_{3}\text{ }-& \text{R}_{1} \\ \end{matrix}$
= (1 + pxyz) (x - y) (z - x) $\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{-1} &-\text{(x + y)} \\ \text{0} & \text{1} &\text{z + x} \end{vmatrix} $
= (1 + pxyz) (x - y) (z - x) $\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{0} &\text{z - y} \\ \text{0} & \text{1} &\text{z + x} \end{vmatrix}\text{R}_{2}\rightarrow\text{R}_{2}+\text{R}_{3} $
= (1 + pxyz) (x - y) (y - z) (z - x) $\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{0} &-\text{1} \\ \text{0} & \text{1} &\text{z + x} \end{vmatrix} $
= (1 + pxyz) (x - y) (y - z) (z - x).1 = RHS.

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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the equatoion of the passing through the points (1, -1, 2) and (2, -2, 2) and which is perpendicular to the plane 6x - 2y + 2z = 9.
→
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that:
  1. All the five cards are spades?
  2. Only 3 cards are spades?
  3. None is a spade?
→
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{{27}}$ of the volume of the sphere.
→
Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}+\lambda\big(3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}+\mu\big(-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)$
→
Using integeation, find ate area of the region bounded by the 2y = 5x + 7. x-axis and the lines x = 2 and x = 8.
→
Using differentials, find the approximate values of the following:
$(1.999)^5$
→
Let $\text{A} = \Re\times\Re$ and$\ast$ be the binary operation on A defined by $\text(a, b) \ast \text{(c, d)} = \text{(a + c, b + d)}. $ Prove that $\ast$ is commutative and associative. Find the identity element for $\ast$ on A. Also write the inverse element of the element (3, – 5) in A.
→
Solve the following differential equation
$\text{x}\frac{\text{dy}}{\text{dx}}+1=0;\text{y}(-1)=0$
→
Solve the following systems of linear equations by cramer's rule:
x - 4y - z = 11,
2x - 5y + 2z = 39,
-3x + 2y + z = 1
→
Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x - 3y + 2z - 5 = 0 and 2x - y + 3z - 1 = 0 and passing through (1, -2, 3).
→