
If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is
A. $d^n$
B. $d^{n-1}$
C. $d^{n+1}$
D. $d$
Answer
164.1k+ views
Hint:
In this case, we must know the property of the inverse and adjoint matrices. According to the property $|(Adj A)| = |A|^{n-1}$ if $A$ is a square, non-singular matrix of order $n \times n$.
Formula Used:
For non-singular matrix A of order $n \times n$ we have:
$|(Adj A)| = |A|^{n-1}$
Complete step-by-step solution:
Given that, $d$ is the determinant of a square matrix $A$ of order $n$
We know that, $|(Adj A)| = |A|^{n-1}$
where $A$ is a square, non-singular matrix of order $n$.
Also, $|A| = d$
$\therefore |(Adj A)| = d ^{n-1}$
So, option B is correct.
Note:
Property: The determinant of adjoint A is equal to the determinant of A power (n-1) where A is invertible n x n square matrix.
In this case, we must know the property of the inverse and adjoint matrices. According to the property $|(Adj A)| = |A|^{n-1}$ if $A$ is a square, non-singular matrix of order $n \times n$.
Formula Used:
For non-singular matrix A of order $n \times n$ we have:
$|(Adj A)| = |A|^{n-1}$
Complete step-by-step solution:
Given that, $d$ is the determinant of a square matrix $A$ of order $n$
We know that, $|(Adj A)| = |A|^{n-1}$
where $A$ is a square, non-singular matrix of order $n$.
Also, $|A| = d$
$\therefore |(Adj A)| = d ^{n-1}$
So, option B is correct.
Note:
Property: The determinant of adjoint A is equal to the determinant of A power (n-1) where A is invertible n x n square matrix.
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