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Question: If A is a square matrix such that, \[AadjA = diag\left( {k.k.k} \right)\], then \[\left| {adjA} \rig...

If A is a square matrix such that, AadjA=diag(k.k.k)AadjA = diag\left( {k.k.k} \right), then adjA=\left| {adjA} \right| =
A. kk
B. k2{k^2}
C. k3{k^3}
D. k4{k^4}

Explanation

Solution

Let A=[aij]A = \left[ {{a_{ij}}} \right] be a square matrix. Then the transpose of a matrix of the cofactors of AA is called adjoint of AA and is denoted by adj(A)adj(A) . So, Adjoint is the transpose of a matrix whose (i,j)(i,j) entry is the aij{a_{ij}} cofactor.

  • Any matrix of order m×nm \times nis said to be a square matrix when m=nm = 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(a1,,an)diag\left( {{a_1}, \ldots ,{a_n}} \right).
  • The determinant of a diagonal matrix is the product of all the diagonal elements.
  • A(adjA)=(adjA)A=AInA\left( {adjA} \right) = \left( {adjA} \right)A = \left| A \right|{I_n} , where nnis the order of matrix and In{I_n} represents the identity matrix of order n×nn \times n

Complete step by step solution:
Given, AadjA=diag(k.k.k)AadjA = diag\left( {k.k.k} \right)
From the property of adjoint of a matrix A(adjA)=AInA\left( {adjA} \right) = \left| A \right|{I_n}. Also AadjA=diag(k.k.k)AadjA = diag\left( {k.k.k} \right).
Therefore, AIn=diag(k.k.k)\left| A \right|{I_n} = diag\left( {k.k.k} \right).
From the property of diagonal of diagonal matrix diag(k.k.k)=k3\left| {diag\left( {k.k.k} \right)} \right| = {k^3}.
And, determinant of AIn\left| A \right|{I_n}is equal to An{\left| A \right|^n}.
It can be observed that the order of matrices is 3, that is n=3n = 3therefore determinant of AIn\left| A \right|{I_n} is A3{\left| A \right|^3}.
This implies that A3=k3A=k{\left| A \right|^3} = {k^3} \Rightarrow \left| A \right| = k.
Substitute kk for A\left| A \right| and 3 for nn into the formula for determinant of adjoint of a matrix, that is adjA=An1\left| {adjA} \right| = {\left| A \right|^{n - 1}} and solve.

\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})$$