Let A be a 3 $\times$ 3 matrix such that
adj A = $$\left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 0 & 2 \cr
1 & { - 2} & { - 1} \cr
} } \right]$$ and B = adj(adj A).
If |A| = $\lambda$ and |(B-1)T| = $\mu$ , then the ordered pair,
(|$\lambda$|, $\mu$) is equal to :
Solution
$$adj\,A = \left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 0 & 2 \cr
1 & { - 2} & { - 1} \cr
} } \right]$$<br><br>$B = adj\,(adj\,A)$<br><br>$= |A{|^{n - 2}}A$<br><br>$= |A{|^{3 - 2}}.A$ [As here n = 3]<br><br>$= |A|.A$ .....(1)<br><br>Now, $$|adj\,A| = \left[ {\matrix{
2 & { - 1} & 1 \cr
{ - 1} & 0 & 2 \cr
1 & { - 2} & { - 1} \cr
} } \right]$$<br><br>$= + 2(0 + 4) + 1(1 - 2) + 1(2 - 0)$<br><br>$= 8 - 1 + 2$<br><br>$= 9$<br><br>Also we know, |adj A| = |A|<sup>n$-$1</sup><br><br>$\therefore$ Here |adj A| = |A|<sup>3 $-$ 1</sup> = |A|<sup>2</sup><br><br>$\therefore$ |A|<sup>2</sup> = 9<br><br>$\Rightarrow$ |A| = $\pm$3<br><br>Given, |A| = $\lambda$<br><br>$\therefore$ $\lambda$ = $\pm$3<br><br>$\Rightarrow$ |$\lambda$| = 3<br><br>From (1)<br><br>B = ($\pm$3)A<br><br>Given, |(B<sup>$-$1</sup>)<sup>T</sup>| = $\mu$<br><br>$\Rightarrow$ |(B<sup>T</sup>)<sup>$-$1</sup>| = $\mu$<br><br>$\Rightarrow$ ${1 \over {|{B^T}|}} = \mu$<br><br>$\Rightarrow$ ${1 \over {|B|}} = \mu$ [As |B<sup>T</sup>| = |B|]
<br><br>$\Rightarrow$ ${1 \over {| \pm 3A|}} = \mu$
<br><br>$\Rightarrow$ ${1 \over {{{\left( { \pm 3} \right)}^3}\left| A \right|}} = \mu$ [As |KA| = K<sup>n</sup> |A|]<br><br>$\Rightarrow {1 \over {( \pm 27)|A|}} = \mu$<br><br>$\Rightarrow \mu = {1 \over {( \pm 27)( \pm 3)}} = \pm {1 \over {81}}$<br><br>$\therefore$ $(|\lambda |,\,\mu ) = \left( {3,\, \pm {1 \over {81}}} \right)$<br><br>Here correct option will be $\left( {3,\,{1 \over {81}}} \right)$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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