Let $A = \left[ {\begin{array}{*{20}{c}} {{x^2}}&6&8 \\ 3&{{y^2}}&9 \\ 4&5&{{z^2}} \end{array}} \right],\,\,B = \left[ {\begin{array}{*{20}{c}} {2x}&3&5 \\ 2&{2y}&6 \\ 1&4&{2z - 3} \end{array}} \right]$ . If ${\text{trace}}\,A = {\text{trace}}\,B$, then $2x + y + x$ is equal to

JEE MATHEMATICS MCQ: MATRICES

1 Medium

1 of 1

Application +4 | -1

Let

A = \left[ {\begin{array}{cccccccccccccccccccc} {{x^2}}&6&8 \\ 3&{{y^2}}&9 \\ 4&5&{{z^2}} \end{array}} \right],\,\,B = \left[ {\begin{array}{cccccccccccccccccccc} {2x}&3&5 \\ 2&{2y}&6 \\ 1&4&{2z - 3} \end{array}} \right]

. If

{\text{trace}}\,A = {\text{trace}}\,B

, then

2x + y + x

is equal to

View answer and explanation

Correct option: A

Option A (4) is correct because by equating the trace of A and B, we can solve for x, y, and z, and then calculate 2x + y + z.

Practice All Chapter MCQs →

\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|