Question

If A is a square matrix of order n*n , then adj(adj A) is equal to
1.${\left| A \right|^{n}}A$
2.${\left| A \right|^{n-1}}A$
3.${\left| A \right|^{n-2}}A$
4.${\left| A \right|^{n-3}}A$

Answer 1

Hint: we know the inverse formula of a matrix is ${A^{ - 1}} = \dfrac{1}{{\det A}}adjA$ and using the property $A{A^{ - 1}} = I$ and equating them we get $A(adjA) = \left| A \right|$ and by replacing A by adj A and the property $\left| {adjA} \right| = {\left| A \right|^{n - 1}}$ we get the required value.

Complete step-by-step answer:
We know that the inverse of a matrix is given by
$ \Rightarrow {A^{ - 1}} = \dfrac{1}{{\det A}}adjA$………(1)
And we also know that
$ \Rightarrow A{A^{ - 1}} = I$
From this,
$ \Rightarrow {A^{ - 1}} = \dfrac{I}{A}$ …………(2)
Now equating (1) and (2)
$ \Rightarrow \dfrac{{adjA}}{{\det A}} = \dfrac{I}{A}$
$ \Rightarrow A(adjA) = \left| A \right|$ ……..(3)
Now let's replace A by adj A in (3)
 $
   \Rightarrow adjA(adj(adjA)) = \left| {adjA} \right| \\
   \Rightarrow (adj(adjA)) = \dfrac{{\left| {adjA} \right|}}{{adjA}} \\
$
From (3) we get $adjA = \dfrac{{\left| A \right|}}{A}$
Substituting in the above equation
$ \Rightarrow (adj(adjA)) = \dfrac{{\left| {adjA} \right|}}{{\left| A \right|}}*A$
We know that $\left| {adjA} \right| = {\left| A \right|^{n - 1}}$ as A is a matrix of order n
$
   \Rightarrow (adj(adjA)) = \dfrac{{{{\left| A \right|}^{n - 1}}}}{{\left| A \right|}}*A \\
   \Rightarrow (adj(adjA)) = {\left| A \right|^{n - 2}}*A \\
$
Hence the correct option is c.

Note: Determinant evaluated across any row or column is the same.
If all the elements of a row (or column) are zeros, then the value of the determinant is zero.
Determinant of an Identity matrix (${I_n}$ ) is 1.
If rows and columns are interchanged then the value of the determinant remains the same (value does not change). Therefore, det(A) = det(${A^T}$ ), here ${A^T}$ is the transpose of matrix A.
If any two rows (or two columns) of a determinant are interchanged the value of the determinant is multiplied by -1.
If all elements of a row (or column) of a determinant are multiplied by some scalar number k, the value of the new determinant is k times the given determinant. Therefore, If A be an n-rowed square matrix and K be any scalar. Then |KA| = ${K^n}$|A| .