Question

If $$B$$ is a nonsingular matrix and $$A$$ is a square matrix then $$det\left(B^{-1}AB\right)$$ is equal to
A. $$det\left(A\right)$$
B. $$det\left(B\right)$$
C. $$det\left(B^{-1}\right)$$
D. $$det\left(A^{-1}\right)$$

Answer 1

- Hint: First of all, split the matrices inside the determinants by using the multiplicative properties of determinants. Then make the product of two matrices equal to a unit matrix to obtain the required answer.

Complete step-by-step solution -

Given $$B$$ is a nonsingular matrix and $$A$$ is a square matrix.
Now, consider $$det\left(B^{-1}AB\right)$$
We know that $$det\left(ABC\right)=det\left(A\right)det\left(B\right)det\left(C\right)$$, so we have
$$det\left(B^{-1}AB\right)=det\left(B^{-1}\right)det\left(A\right)det\left(B\right)$$
As determinants obeys commutative property of multiplication, we have
$$det\left(B^{-1}AB\right)=det\left(B^{-1}\right)det\left(B\right)det\left(A\right)$$
We know that, $$det\left(AB\right)=det\left(A\right)det\left(B\right)$$
$$det\left(B^{-1}AB\right)=det\left(B^{-1}B\right)det\left(A\right)$$
As $$B^{-1}B=I$$, we have
$$det\left(B^{-1}AB\right)=det\left(I\right)det\left(A\right)$$
We know that, $$det\left(I\right)=1$$
$$\therefore det\left(B^{-1}AB\right)=det\left(A\right)$$
Thus, the correct option is A. $$det\left(A\right)$$
Note: The multiplicative property of determinants tells us that the det of product of the given matrices is equal to their product of individual det of matrices. The product of a matrix and its inverse matrix gives us a unit matrix.