Question

If A is a non-singular invertible square matrix of order n, then adj(adj(A)) is equal to
[a] $\det {{\left( A \right)}^{n}}A$
[b] $\det {{\left( A \right)}^{n-1}}A$
[c] $\det {{\left( A \right)}^{n-2}}A$
[d] None of these.

Answer 1

Hint: Use the property that adj(A)A= det(A)I where I is an identity matrix of the same dimensions as A.
Assume B = adj(adj(A)). Then multiply both sides by adj(A) and simplify. Again multiply both sides by A and simplify again to get the result. Use the property that if n is the order of the square matrix A, then $\det \left( adj\left( A \right) \right)=\det {{\left( A \right)}^{n-1}}$

Complete step-by-step answer:
Let B = adj(adj(A))
Pre multiplying both sides by adj(A), we get
adj(A)B = adj(A) adj (adj (A))
We know that adj(A)A= det(A)I
Hence, we get
adj(A)B = det( adj(A))I
Pre multiplying both sides by A, we get
Aadj(A)B = det(adj(A)) IA
We know that IA = AI = A and A adj(A) =det(A)I
Hence we get
det(A)IB = det(adj(A))A
Now, we know that $\det \left( adj\left( A \right) \right)={{\left( \det A \right)}^{n-1}}$
Hence, we have
$\det (A)B={{\left( \det \left( A \right) \right)}^{n-1}}A$
Since A is non-singular, dividing both sides by det(A), we get
$B=\det {{\left( A \right)}^{n-2}}A$
Hene option [c] is correct.

Note: [1] Alternative solution
We know that A adj(A) = det(A) I
Pre multiplying both sides by ${{A}^{-1}}$, we get
$adj\left( A \right)=\det \left( A \right){{A}^{-1}}$
Pre multiplying both sides by $adj{{\left( A \right)}^{-1}}$, we get
$A=\det \left( A \right)adj{{\left( A \right)}^{-1}}$
Hence we have
$adj\left( adj\left( A \right) \right)=\det \left( adj\left( A \right) \right)adj{{\left( A \right)}^{-1}}$
Now, we know that $\det \left( adj\left( A \right) \right)={{\left( \det A \right)}^{n-1}}$
Hence, we have
$adj\left( adj\left( A \right) \right)=\det {{\left( A \right)}^{n-1}}adj{{\left( A \right)}^{-1}}=\det {{\left( A \right)}^{n-2}}\det \left( A \right)adj{{\left( A \right)}^{-1}}$
Using $A=\det \left( A \right)adj{{\left( A \right)}^{-1}}$, we get
$adj\left( adj\left( A \right) \right)=\det {{\left( A \right)}^{n-2}}A$
Hence option [c] is correct.
[2] Note that pre multiplying by ${{A}^{-1}}$ is possible if and only if ${{A}^{-1}}$ exists, i.e. if and only if A is non-singular.
[3] If A is non-singular, then so is adj(A)