Question regarding matrices and determinants.

Question

The question states : If $A$ is a non-singular square matrix satisfying $AB-BA=A$, then prove that $|B+I|=|B-I|$.

Note :

$1.$ Here $I$ is the Identity matrix.

$2.$ Modulus sign means determinant. $|A|=\det{A}$.

I have never encountered questions where there was addition or subtraction inside a determinant. My first thought was to use the property $|A|.|B|=|A.B|$.

So, I wrote $$|B+I| = \frac{1}{|A|}.|BA+A| $$ (should it be $|BA+A|$ or $|AB+A|$ or am I wrong altogether?)

Anyway, that leads to $$|B+I|=\frac{1}{|A|}.|AB-A+A|$$

or $$|B+I|=|B|.$$

Obviously, this is not what the question asked. So I must have made a mistake. Where did I make a mistake and what is a good way to approach this question?

Answer 1

This is what I get: $$ B-I = A^{-1}A(B-I)=A^{-1}(AB-A)=A^{-1}BA. $$ Taking determinants, $$ |B-I|= |B|, $$ and joining it with what you got, you are done. Notice that what you did is correct, since $$ B+I=(B+I)AA^{-1}=(BA+A)A^{-1}=ABA^{-1}. $$ The product of matrices is distributive respect to the sum of matrices, so the only thing you have to be careful is on which side you multiply by the matrix $A$;

Answer 2

Multiply $A^{-1}$ on both sides so that we have two equations : $$ ABA^{-1} - B =I $$ and $$ B-A^{-1} BA = I $$

Hence $$ |I +B | = |ABA^{-1}|=|A^{-1} BA | = |I-B |$$