[ 0 a b 0 ]^4=1 , then

Correct answer: (D) a b=1. Here, [ 0 a b 0 ]^2= [ 0 a b 0 ] [ 0 a b 0 ]^2 = [ 0+a b 0+0 0+0 a b+0 ]= [ a b 0 0 a b ] Similarly, [ 0 a b 0 ]^4= [ a b 0 0 a b ] [ a b 0 0 a b ] = [ a^2 b^2+0 0+0 0+0 0+a^2 b^2 ]= [ a^2 b^2 0 0 a^2 b^2 ] Now, [ 0 a b 0 ]^4=I ⇒ [ a^2 b^2 0 0 a^2 b^2 ]= [ 1 0 0 1 ] ⇒ a^2 b^2=1 ⇒ a b=1

Q72 / 120 · VITEEE 2008 · Mathematics

$\left[\begin{array}{ll}0 & \mathrm{a} \\ \mathrm{b} & 0\end{array}\right]^{4}=1$, then

A
$\mathrm{a}=1=2 \mathrm{~b}$
B
$\mathrm{a}=\mathrm{b}$
C
$a=b^{2}$
Correct answer
D
$a b=1$

Solution

$\begin{array}{l} \text { Here, }\left[\begin{array}{ll} 0 & a \\ b & 0 \end{array}\right]^{2}=\left[\begin{array}{ll} 0 & a \\ b & 0 \end{array}\right]\left[\begin{array}{ll} 0 & a \\ b & 0 \end{array}\right]^{2} \\ =\left[\begin{array}{ll} 0+a b & 0+0 \\ 0+0 & a b+0 \end{array}\right]=\left[\begin{array}{ll} a b & 0 \\ 0 & a b \end{array}\right] \end{array}$ $\begin{array}{l} \text { Similarly, }\left[\begin{array}{ll} 0 & a \\ b & 0 \end{array}\right]^{4}=\left[\begin{array}{ll} a b & 0 \\ 0 & a b \end{array}\right]\left[\begin{array}{ll} a b & 0 \\ 0 & a b \end{array}\right] \\ =\left[\begin{array}{ll} a^{2} b^{2}+0 & 0+0 \\ 0+0 & 0+a^{2} b^{2} \end{array}\right]=\left[\begin{array}{ll} a^{2} b^{2} & 0 \\ 0 & a^{2} b^{2} \end{array}\right] \end{array}$ Now, $\begin{array}{l} {\left[\begin{array}{ll} 0 & a \\ b & 0 \end{array}\right]^{4}=I \Rightarrow\left[\begin{array}{ll} a^{2} b^{2} & 0 \\ 0 & a^{2} b^{2} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ \Rightarrow a^{2} b^{2}=1 \Rightarrow a b=1 \end{array}$

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

$\left[\begin{array}{ll}0 & \mathrm{a} \\ \mathrm{b} & 0\end{array}\right]^{4}=1$, then

Solution

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