Let $A = \left[ {\begin{array}{*{20}{c}} {1\;\;2}\\ {3\;\;4} \end{array}} \right]$ and $\;B = \left[ {\begin{array}{*{20}{c}} {\;a\;\;0}\\ {0\;\;b} \end{array}} \right]$, $a,b\;N.$ then

There exist more than one but finite number of $B'$s such that $AB = BA$

There exist exactly one $B$ such $AB = BA$

There cannot exist any $B$ such that $AB = BA$

JEE MATHEMATICS MCQ: MATRICES

1 Easy

1 of 1

Analyzing +4 | -1

Let

A = \left[ {\begin{array}{cccccccccccccccccccc} {1\;\;2}\\ {3\;\;4} \end{array}} \right]

and

\;B = \left[ {\begin{array}{cccccccccccccccccccc} {\;a\;\;0}\\ {0\;\;b} \end{array}} \right]

a,b\;N.

then

View answer and explanation

Correct option: C

There exists infinitely many

B's

such that

AB = BA

Option C (There exists infinitely many

B's

such that

AB = BA

) is correct because if

a = b

, then

AB = BA

, and there are infinitely many possible values for

a

and

b

in the set of natural numbers.

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|