Question

If A is a non-singular matrix satisfying $AB-BA=A$, then
(a) $\det \left( B+I \right)=\det \left( B-I \right)$
(b) $\det \left( B+I \right)=-\det \left( B-I \right)$
(c) $\det \left( B+I \right)=\det \left( {{B}^{2}}-I \right)$
(d) None of these

Answer 1

Hint: We can add $BA$ on both the sides of the given equation to get the equation $AB=A\left( I+B \right)$. On taking the determinant on both the sides, we will get an equation for $\det \left( B+I \right)$. Similarly, on subtracting A and then adding $BA$ on both the sides of the equation, we will get $A\left( B-I \right)=BA$. On taking determinant on both the sides, we will get the equation for $\det \left( B-I \right)$. From the two equations obtained we will get the final answer.

Complete step-by-step answer:
The matrix equation given in the above question is
$\Rightarrow AB-BA=A.......\left( i \right)$
Let us add $BA$ on both the sides of the above equation to get
\[\begin{align}
  & \Rightarrow AB-BA+BA=A+BA \\
 & \Rightarrow AB=A+BA \\
\end{align}\]
Writing \[A=IA\] in the RHS of the above equation, we get
\[\Rightarrow AB=IA+BA\]
Taking $A$ common on the RHS, we get
\[\Rightarrow AB=\left( I+B \right)A\]
Now, let us take the determinant on both the sides of the above equation to get
\[\Rightarrow \det \left( AB \right)=\det \left[ \left( I+B \right)A \right]\]
From the properties of determinant of a matrix, we know that the determinant of the product of matrices is equal to the product of the determinant of the individual matrices. Therefore, we can write the above equation as
\[\Rightarrow \det \left( A \right)\det \left( B \right)=\det \left( I+B \right)\det \left( A \right)\]
Subtracting \[\det \left( I+B \right)\det \left( A \right)\] from both the sides, we get
\[\begin{align}
  & \Rightarrow \det \left( A \right)\det \left( B \right)-\det \left( I+B \right)\det \left( A \right)=\det \left( I+B \right)\det \left( A \right)-\det \left( I+B \right)\det \left( A \right) \\
 & \Rightarrow \det \left( A \right)\det \left( B \right)-\det \left( I+B \right)\det \left( A \right)=0 \\
\end{align}\]
Taking $\det \left( A \right)$ common, we get
$\Rightarrow \det \left( A \right)\left[ \det \left( B \right)-\det \left( I+B \right) \right]=0$
Now, since the matrix A is non-singular, $\det \left( A \right)\ne 0$. Therefore, we can divide the above equation by $\det \left( A \right)$ to get
\[\begin{align}
  & \Rightarrow \dfrac{\det \left( A \right)\left[ \det \left( B \right)-\det \left( I+B \right) \right]}{\det \left( A \right)}=\dfrac{0}{\det \left( A \right)} \\
 & \Rightarrow \det \left( B \right)-\det \left( I+B \right)=0 \\
\end{align}\]
Adding \[\det \left( I+B \right)\] on both sides we get
\[\begin{align}
  & \Rightarrow \det \left( B \right)-\det \left( I+B \right)+\det \left( I+B \right)=0+\det \left( I+B \right) \\
 & \Rightarrow \det \left( B \right)=\det \left( I+B \right) \\
 & \Rightarrow \det \left( B \right)=\det \left( B+I \right).......\left( ii \right) \\
\end{align}\]
Now, we subtract A from both the sides of the equation (i) to get
$\begin{align}
  & \Rightarrow AB-BA-A=A-A \\
 & \Rightarrow AB-BA-A=0 \\
\end{align}$
Now, we add $BA$ on both the sides of the above equation to write
$\begin{align}
  & \Rightarrow AB-BA-A+BA=0+BA \\
 & \Rightarrow AB-A=BA \\
\end{align}$
Writing \[A=AI\] in the LHS of the above equation we get
\[\Rightarrow AB-AI=BA\]
Taking A common on the LHS, we get
\[\Rightarrow A\left( B-I \right)=BA\]
Taking determinant on both the sides, we get
$\Rightarrow \det \left( A \right)\det \left( B-I \right)=\det \left( B \right)\det \left( A \right)$
Subtracting $\det \left( B \right)\det \left( A \right)$ from both the sides, we get
\[\begin{align}
  & \Rightarrow \det \left( A \right)\det \left( B-I \right)-\det \left( B \right)\det \left( A \right)=\det \left( B \right)\det \left( A \right)-\det \left( B \right)\det \left( A \right) \\
 & \Rightarrow \det \left( A \right)\det \left( B-I \right)-\det \left( B \right)\det \left( A \right)=0 \\
\end{align}\]
Taking \[\det \left( A \right)\] common, we get
\[\Rightarrow \det \left( A \right)\left[ \det \left( B-I \right)-\det \left( B \right) \right]=0\]
Dividing the above equation by $\det \left( A \right)$ we get
\[\Rightarrow \det \left( B-I \right)-\det \left( B \right)=0\]
Adding \[\det \left( B \right)\] both sides, we get
\[\begin{align}
  & \Rightarrow \det \left( B-I \right)-\det \left( B \right)+\det \left( B \right)=0+\det \left( B \right) \\
 & \Rightarrow \det \left( B-I \right)=\det \left( B \right)......\left( iii \right) \\
\end{align}\]
Finally, form the equations (ii) and (iii) we can write
\[\Rightarrow \det \left( B+I \right)=\det \left( B+I \right)\]

So, the correct answer is “Option (a)”.

Note: We must note that we cannot treat matrix equations just like the algebraic equations, even if they appear to be alike. This is because the matrix multiplication does not obey the commutative law, which is valid in case of the algebraic multiplication.