JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

Let

{A}

and

{B}

be two matrices such that

A=\left[a_{i j}\right]_{2 \times 3}=\left[\begin{array}{ccc}x & 1 & 2 \\ 3 x & -1 & \frac{1}{4 x^2+1}\end{array}\right]

and

B=\left[b_{i j}\right]_{3 \times 2}=\left[\begin{array}{cc}\frac{1}{x^2} & \frac{1}{x} \\ 2 x & 2 \\ 3 & x\end{array}\right]

where

x > 0

. If matrix

C

is defined as

C=\left[c_{i j}\right]_{2 \times 2}

, where

c_{i j}=\sum\limits_{{r=1}}^3 a_{i r} b_{r j}, \forall 1 \leq i, j \leq 2

, then the minimum value of

\Delta(x)=\sum\limits_{{1 \leq i \leq j \leq 2}} c_{i j}

View answer and explanation

Correct option: A

\frac{57}{4}

Option A (

\frac{57}{4}

) is correct because after calculating the matrix C and summing its elements, the minimum value of the resulting expression is

\frac{57}{4}

More Questions from MATRICES

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\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|