Question

If A is a square matrix such that, \[AadjA = diag\left( {k.k.k} \right)\], then \[\left| {adjA} \right| = \]
A. \[k\]
B. \[{k^2}\]
C. \[{k^3}\]
D. \[{k^4}\]

Answer 1

Hint: Let \[A = \left[ {{a_{ij}}} \right]\] be a square matrix. Then the transpose of a matrix of the cofactors of \[A\] is called adjoint of \[A\] and is denoted by \[adj(A)\] . So, Adjoint is the transpose of a matrix whose \[(i,j)\] entry is the \[{a_{ij}}\] cofactor.
* Any matrix of order \[m \times n\]is said to be a square matrix when \[m = n\], that is, the number of rows in the matrix is equal to the number of columns.
* A diagonal matrix is the matrix that has non zero diagonal elements and other elements as 0 and is represented as \[diag\left( {{a_1}, \ldots ,{a_n}} \right)\].
* The determinant of a diagonal matrix is the product of all the diagonal elements.
* \[A\left( {adjA} \right) = \left( {adjA} \right)A = \left| A \right|{I_n}\] , where \[n\]is the order of matrix and \[{I_n}\] represents the identity matrix of order \[n \times n\]

Complete step by step solution:
Given, \[AadjA = diag\left( {k.k.k} \right)\]
From the property of adjoint of a matrix \[A\left( {adjA} \right) = \left| A \right|{I_n}\]. Also \[AadjA = diag\left( {k.k.k} \right)\].
Therefore, \[\left| A \right|{I_n} = diag\left( {k.k.k} \right)\].
From the property of diagonal of diagonal matrix \[\left| {diag\left( {k.k.k} \right)} \right| = {k^3}\].
And, determinant of \[\left| A \right|{I_n}\]is equal to \[{\left| A \right|^n}\].
It can be observed that the order of matrices is 3, that is \[n = 3\]therefore determinant of \[\left| A \right|{I_n}\] is \[{\left| A \right|^3}\].
This implies that \[{\left| A \right|^3} = {k^3} \Rightarrow \left| A \right| = k\].
Substitute \[k\] for \[\left| A \right|\] and 3 for \[n\] into the formula for determinant of adjoint of a matrix, that is \[\left| {adjA} \right| = {\left| A \right|^{n - 1}}\] and solve.
\[\begin{gathered}
  \left| {adjA} \right| = {\left( k \right)^{3 - 1}} \\
   = {\left( k \right)^2} \\
   = {k^2} \\
\end{gathered} \]

Therefore, Option B is correct.

Note:
The order of the diagonal matrix can be found by the number of elements in the diagonal. In these types of questions determinant of diagonal matrix and adjoint matrix is very important and must be calculated cautiously. Also, the order of the adjoint matrix and diagonal matrix must be equal. Whenever operating two matrices, always keep in mind the order of the matrices because only matrices with the same order can be added or subtracted. Also, when multiplying two matrices, the value of \[n\] for the first matrix should be equal to value of \[m\] for the second matrix, only then they can be multiplied and the product matrix has order \[m\] (of first matrix) \[ \times n\] (of second matrix).
Some other properties of Adjoint of a matrix are:
1) \[\left| {adjA} \right| = {\left| A \right|^{n - 1}}\]
2) \[adj(adjA) = {\left| A \right|^{n - 2}}.A\]
3) \[adj(AB) = adj(B).adj(A)\]
4) \[adj(kA) = {k^{n - 1}}.adj(A)\]
5) \[{(adjA)^T} = adj({A^T})\]