Let A be a 3 [ ] 3 matrix such that adj A = [[ pmatrix 2 & - 1 & 1 - 1 & 0 & 2 1 & - 2 & -…

Question

Let A be a 3 $$ \times $$ 3 matrix such that

adj A = $$\left[ {\begin{pmatrix} 2 & { - 1} & 1  \
   { - 1} & 0 & 2  \
   1 & { - 2} & { - 1}  \ \end{pmatrix} } \right]$$ and B = adj(adj A).

If |A| = $$\lambda $$ and |(B-1)T| = $$\mu $$ , then the ordered pair,

(|$$\lambda $$|, $$\mu $$) is equal to :

Accepted Answer

$$adj\,A = \left[ {\begin{pmatrix} 2 & { - 1} & 1  \
   { - 1} & 0 & 2  \
   1 & { - 2} & { - 1}  \

\end{pmatrix} } \right]$$

$$B = adj\,(adj\,A)$$

$= |A{|^{n - 2}}A$

$= |A{|^{3 - 2}}.A$ [As here n = 3]

$= |A|.A$ .....(1)

Now, $$|adj\,A| = \left[ {\begin{pmatrix} 2 & { - 1} & 1  \
   { - 1} & 0 & 2  \
   1 & { - 2} & { - 1}  \

\end{pmatrix} } \right]$$

$=  + 2(0 + 4) + 1(1 - 2) + 1(2 - 0)$

$= 8 - 1 + 2$

$= 9$

Also we know, |adj A| = |A|n-1

$$ \therefore $$ Here |adj A| = |A|3 - 1 = |A|2

$$ \therefore $$ |A|2 = 9

$$ \Rightarrow $$ |A| = $$ \pm $$3

Given, |A| = $$\lambda $$

$$ \therefore $$ $$\lambda $$ = $$ \pm $$3

$$ \Rightarrow $$ |$$\lambda $$| = 3

From (1)

B = ($$ \pm $$3)A

Given, |(B-1)T| = $$\mu $$

$$ \Rightarrow $$ |(BT)-1| = $$\mu $$

$$ \Rightarrow $$ $${1 \over {|{B^T}|}} = \mu $$

$$ \Rightarrow $$ $${1 \over {|B|}} = \mu $$ [As |BT| = |B|]

$$ \Rightarrow $$ $${1 \over {| \pm 3A|}} = \mu $$

$$ \Rightarrow $$ $${1 \over {{{\left( { \pm 3} \right)}^3}\left| A \right|}} = \mu $$ [As |KA| = Kn |A|]

$$ \Rightarrow {1 \over {( \pm 27)|A|}} = \mu $$

$$ \Rightarrow \mu  = {1 \over {( \pm 27)( \pm 3)}} =  \pm {1 \over {81}}$$

$$ \therefore $$ $$(|\lambda |,\,\mu ) = \left( {3,\, \pm {1 \over {81}}} \right)$$

Here correct option will be $$\left( {3,\,{1 \over {81}}} \right)$$

Solutions · 1