$A$ and $B$ are $3\times 3$ matrices ($A,B \in \mathcal{M}_{3 \times 3}(\mathbb{R})$). There are two equations:
$$A^2+3BA=I$$
$$A^2=AB$$
I want to prove that $A$ does not have an inverse.
I tried to substitute $AB$ at the first equation but could not get the solution. I know that if the matrix is invertible, then $AA^{-1}=I$.
$\endgroup$ 32 Answers
$\begingroup$Take $A=B=\frac{1}{2}I$, with $I=I_3$ the identity matrix. Then we have $$ A^2+3BA=(\frac{1}{4}+\frac{3}{4})I=I, $$ and also $A^2=AB$. However, both $A$ and $B$ are invertible.
$\endgroup$ 0 $\begingroup$Your first equation says :
$$(A+3B)A=I$$
Taking determinant both the sides, you get
$$\det(A+3B) \cdot \det (A)=1$$
This implies $\det (A)$ can never be $0$, hence $A$ is invertible.
Moreover, you can easily see that the inverse of $A$ is $A+3B$, since both multiply to identity matrix.
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