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JEE MATHEMATICS MCQ: MATRICES

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Analyzing +4 | -1

If

a_1, b_2, c_3 \neq 0

\Delta_1=\left|\begin{array}{ccc}a_1^2+b_1+c_1 & a_1 a_2+b_2+c_2 & a_1 a_3+b_3+c_3 \\ b_1 b_2+c_1 & b_2^2+c_2 & b_2 b_3+c_3 \\ c_3 c_1 & c_3 c_2 & c_3^2\end{array}\right|

and

\Delta_2=\left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|

, then

\frac{\Delta_1}{\Delta_2}

equal to

View answer and explanation

Correct option: A

a_1 b_2 c_3

Option A (

a_1 b_2 c_3

) is correct because after determinant operations, the ratio simplifies to this product.

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

are real numbers then the line

a x+b y+c=0

passes through the fixed point is The number of solutions of the system of equations

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

is If the equations

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

ax + 2y - b = 0

are consistent, then The value of the determinant

\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

is the cube root of unity, then

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

is equal to For any matrix

A

2 \times 2

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

|A|