Let A- a b c d such that A3-0- then a-d equals

The correct option is D 0 Given, A=[ a amp; b; c amp; d ] and A^3=0 #160; #160; A.A=[ a amp; b; c amp; d ][ ...

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Byju's Answer

Standard XII

Mathematics

Transpose of a Matrix

Let A= a ...

Question

Let $A = (\begin{matrix} a & b c & d \end{matrix})$ such that $A^{3} = 0$ , then $a + d$ equals

a d

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b c

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Solution

The correct option is D $0$
Given, $A = [\begin{matrix} a & b c & d \end{matrix}]$ and $A^{3} = 0$
$A . A = [\begin{matrix} a & b c & d \end{matrix}] [\begin{matrix} a & b c & d \end{matrix}] \Rightarrow [\begin{matrix} a^{2} + a c & a b + b d a c + c^{2} & c b + d^{2} \end{matrix}] A^{3} = A^{2} . A = [\begin{matrix} a^{3} + a^{2} c + a b c + b c d & a^{2} b + a d c + a b d + b d^{2} a^{2} c + c^{2} a + c^{2} d + c d^{2} & a b c + c^{2} + b c d + b d^{2} \end{matrix}] = [\begin{matrix} a^{2} (a + c) + b c (a + d) & a b (a + c) + b d (a + d) a c (a + c) + c d (a + d) & c b (a + c) + b d (a + d) \end{matrix}] = [\begin{matrix} (a + c) & (a + d) (a + c) & (c + d) \end{matrix}] [\begin{matrix} a^{2} + b c & a b + b d a c + c d & c b + b d \end{matrix}]$
Since, $A^{3} = 0 \Rightarrow (a + d) = 0$

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Let A=(abcd) such that A3=0, then a+d equals

The correct option is D 0Given, A=[abcd] and A3=0 A.A=[abcd][abcd]⇒[a2+acab+bdac+c2cb+d2]A3=A2.A=[a3+a2c+abc+bcda2b+adc+abd+bd2a2c+c2a+c2d+cd2abc+c2+bcd+bd2]=[a2(a+c)+bc(a+d)ab(a+c)+bd(a+d)ac(a+c)+cd(a+d)cb(a+c)+bd(a+d)]=[(a+c)(a+d)(a+c)(c+d)][a2+bcab+bdac+cdcb+bd]Since, A3=0⇒(a+d)=0

Let $A = (\begin{matrix} a & b c & d \end{matrix})$ such that $A^{3} = 0$ , then $a + d$ equals