Let M bea 3 × 3 n × n -singular matrix with det(M)=α . If [M^ -1 . adj ( adj ( M )]= KI , then the value of K is

Correct answer: B. α We know that, M(adj M)=|M| I Replacing M by adj M, we get adj M[adj(adj M)= (adj M) I= (M) M^ -1 [adj(adj M)=α^ 2 1. [ M^ -1 = 1 |M| adj(M)] α M^ -1 [adj(adj M)]=α^ 2 I M^ -1 [adj(adj M)]=α I But M^ -1 [adj(adj M)]=K I Hence, K=α

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Matrices

BITSAT 2015 · Mathematics · Matrices

Let M bea 3 × 3 n × n -singular matrix with det(M)=α . If [M^ -1 ....

Let M bea

3 \times 3 \mathrm{n} \times \mathrm{n}

-singular matrix with

\operatorname{det}(\mathrm{M})=\alpha

. If \left[\mathrm{M}^{-1}\right.

adj (

\mathrm{adj}

(

\mathrm{M}

)]=

\mathrm{KI}

, then the value of

\mathrm{K}$ is

Options

A1
B $\alpha$
C $\alpha^{2}$
D $\alpha^{3}$

Correct answer

B. $\alpha$

Step-by-step solution

We know that,

M(\operatorname{adj} M)=|M|

I Replacing

\mathrm{M}

by adj

\mathrm{M},

we get

\operatorname{adj} M[\operatorname{adj}(\operatorname{adj} M)=\operatorname\det(\operatorname{adj} M) I=\operatorname\det(M) M^{-1}\left[\operatorname{adj}(\operatorname{adj} M)=\alpha^{2} 1\right. \left[\because \mathrm{M}^{-1}=\frac{1}{|\mathrm{M}|} \operatorname{adj}(\mathrm{M})\right]\Rightarrow \alpha \mathrm{M}^{-1}[\operatorname{adj}(\operatorname{adj} M)]=\alpha^{2}

\mathrm{I}

\Rightarrow \mathrm{M}^{-1}[\mathrm{adj}(\operatorname{adj} \mathrm{M})]=\alpha \mathrm{I}

But

M^{-1}[\operatorname{adj}(\operatorname{adj} M)]=K I

Hence,

\mathrm{K}=\alpha

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