Question

If A be any square matrix of order 3×3 and |A| = 5, then find the value of |adj.(adj.A)|​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us first derive an expression for |adj(adjA)|

Let us consider a square matrix A of order n × n.

We know that,

$\rm :\longmapsto\:A(adjA) = |A| I_n$

Replace A by adjA, we get

$\rm :\longmapsto\:adjA\bigg(adj(adjA) \bigg) = |adjA| I_n$

$\rm :\longmapsto\:adjA\bigg(adj(adjA) \bigg) = { |A| }^{n - 1} I_n$

$\: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{ \blue{ \bf \: |adjA| = { |A| }^{n - 1}}}$

Pre-multiply by A on both sides, we get

$\rm :\longmapsto\:AadjA\bigg(adj(adjA) \bigg) = A{ |A| }^{n - 1} I_n$

$\rm :\longmapsto\: |A|I_n \bigg(adj(adjA) \bigg) ={ |A| }^{n - 1}A$

$\rm :\longmapsto\: \bigg(adj(adjA) \bigg) ={ |A| }^{n - 2}A$

So, by taking determinant on both sides, we get

$\rm :\longmapsto\: |adj(adjA)| = | { |A| }^{n - 2}A |$

$\rm :\longmapsto\: |adj(adjA)| = { |A| }^{n(n - 2)} |A|$

$\: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{ \blue{ \bf \: |k \: A| = {k}^{n} \: |A| }}$

$\rm :\longmapsto\: |adj(adjA)| = { |A| }^{n(n - 2) + 1}$

$\rm :\longmapsto\: |adj(adjA)| = { |A| }^{ {n}^{2} - 2n + 1}$

$\rm :\longmapsto\: |adj(adjA)| = { |A| }^{ {(n - 1)}^{2}}$

$\rm :\implies\:\boxed{ \red{ \bf \:\: |adj(adjA)| = { |A| }^{ {(n - 1)}^{2}}}}$

According to statement, we have

• Order of matrix, n = 3

and

• |A| = 5

So,

$\rm :\longmapsto\:{ \bf \:\: |adj(adjA)| = {5}^{ {(3 - 1)}^{2}} = {5}^{4} = 625}$

Additional Information :-

$\boxed{ \blue{ \bf \: |I| = 1}}$

$\boxed{ \blue{ \bf \: |adjA| = { |A| }^{n - 1} }}$

$\boxed{ \blue{ \bf \: |kA| = {k}^{n} |A| }}$

$\boxed{ \blue{ \bf \: |AB| = |A| |B| }}$

$\boxed{ \blue{ \bf \: | {A}^{T} | = |A|}}$

$\boxed{ \blue{ \bf \: | {A}^{ - 1} | = \dfrac{1}{ |A| }}}$

$\boxed{ \blue{ \bf \: |A {A}^{ - 1} | = 1}}$

$\boxed{ \blue{ \bf \: | {A}^{T} {A}^{ - 1} | = 1}}$