The value of the determinant left|begin{array | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $\Delta = \left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}ight|$

Vedclass · Accepted Answer

$(1+a^{2}+b^{2})^{3}$

Vedclass · Answer

(D) Let $\Delta = \left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \ 2 a b & 1-a^{2}+b^{2} & 2 a \ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}ight|$.Apply the column operations $C_{1} ightarrow C_{1} - bC_{3}$ and $C_{2} ightarrow C_{2} + aC_{3}$.$\Delta = \left|\begin{array}{ccc}1+a^{2}-b^{2} + 2b^{2} & 2 a b - 2ab & -2 b \ 2 a b - 2ab & 1-a^{2}+b^{2} + 2a^{2} & 2 a \ 2 b - b(1-a^{2}-b^{2}) & -2 a + a(1-a^{2}-b^{2}) & 1-a^{2}-b^{2}\end{array}ight|$$\Delta = \left|\begin{array}{ccc}1+a^{2}+b^{2} & 0 & -2 b \ 0 & 1+a^{2}+b^{2} & 2 a \ b(1+a^{2}+b^{2}) & -a(1+a^{2}+b^{2}) & 1-a^{2}-b^{2}\end{array}ight|$Taking common factors $(1+a^{2}+b^{2})$ from $C_{1}$ and $C_{2}$:$\Delta = (1+a^{2}+b^{2})^{2} \left|\begin{array}{ccc}1 & 0 & -2 b \ 0 & 1 & 2 a \ b & -a & 1-a^{2}-b^{2}\end{array}ight|$Expanding along $R_{1}$:$\Delta = (1+a^{2}+b^{2})^{2} [1(1-a^{2}-b^{2} + 2a^{2}) - 0 + (-2b)(0 - b)]$$\Delta = (1+a^{2}+b^{2})^{2} [1+a^{2}-b^{2} + 2b^{2}]$$\Delta = (1+a^{2}+b^{2})^{2} (1+a^{2}+b^{2}) = (1+a^{2}+b^{2})^{3}$.

The value of the determinant $\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$ is

Similar Questions

$\left| \begin{array}{ccc} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{array} \right| = $

If $\omega$ is an imaginary cube root of unity, then the value of the determinant $\left|\begin{array}{ccc}1+\omega & 0 & -\omega \\ 1+\omega^{2} & \omega & -\omega^{2} \\ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}\right|$ is

If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$,where $a, b, c \in R - \{0\}$,then:

By using properties of determinants,show that:
$\left|\begin{array}{ccc}-a^{2} & ab & ac \\ ba & -b^{2} & bc \\ ca & cb & -c^{2}\end{array}\right|=4a^{2}b^{2}c^{2}$

The value of the determinant $\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module