Consider A= a_11 a_12 a_21 a_22 and B= 1 1 2 1 such that AB=BA, then the value of a_12/a_21+a_11/a_22 is

Correct answer: (C) 3/2. Given: AB=BA a_11 a_12 a_21 a_22 1 1 2 1 = 1 1 2 1 a_11 a_12 a_21 a_22 a_11+2a_12 a_11+a_12 a_21+2a_22 a_21+a_22 = a_11+a_21 a_12+a_22 2a_11+a_21 2a_12+a_22 a_11+2a_12=a_11+a_21 ⇒2a_12=a_21 a_11+a_12=a_12+a_22⇒a_11=a_22 a_21+2a_22=2a_11+a_21⇒a_11=a_22 a_21+a_22=2a_12+a_22⇒a_21=2a_12 a_12a_21=1/2, a_11a_22=1

Q5 / 75 · NTA Abhyas (JEE Main) 2020 · Mathematics · Matrices

Consider $A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]$ and $B = [\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix}]$ such that $A B = B A$ , then the value of $\frac{a_{12}}{a_{21}} + \frac{a_{11}}{a_{22}}$ is

A
$2$
B
$4$
Correct answer
C
$\frac{3}{2}$
D
$\frac{1}{\sqrt{2}}$

Solution

Given:

A B = B A

[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] [\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix}] = [\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix}] [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]

[\begin{matrix} a_{11} + 2 a_{12} & a_{11} + a_{12} \\ a_{21} + 2 a_{22} & a_{21} + a_{22} \end{matrix}] = [\begin{matrix} a_{11} + a_{21} & a_{12} + a_{22} \\ 2 a_{11} + a_{21} & 2 a_{12} + a_{22} \end{matrix}]

a_{11} + 2 a_{12} = a_{11} + a_{21}

\Rightarrow 2 a_{12} = a_{21}

a_{11} + a_{12} = a_{12} + a_{22} \Rightarrow a_{11} = a_{22}

a_{21} + 2 a_{22} = 2 a_{11} + a_{21} \Rightarrow a_{11} = a_{22}

a_{21} + a_{22} = 2 a_{12} + a_{22} \Rightarrow a_{21} = 2 a_{12}

\frac{a_{12}}{a_{21}} = \frac{1}{2}, \frac{a_{11}}{a_{22}} = 1

Hence, the correct answer is option: 'C'.

Consider A=a11a12a21a22 and B=1121 such that AB=BA, then the value of a12a21+a11a22 is

Solution

More Matrices Questions

Consider $A = [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]$ and $B = [\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix}]$ such that $A B = B A$ , then the value of $\frac{a_{12}}{a_{21}} + \frac{a_{11}}{a_{22}}$ is