Is there a way I could factor out $B$?

Question

I need to know $\det M$

$$M = 3B + BT$$

where $T=\begin{pmatrix} k-3 & 2 & 0 \\ 2 & 0 & k \\ 0 & k & -3\end{pmatrix}$

and I know that $\det B = 5$

is there any way I could factor out $B$ on my original equation?

this are all $3 \times 3$ matrices, and I need to find out the value of $k$ so that $M$ is invertible

Answer 1

Yes. We have $$ 3B + BT = B(3I) + BT = B(3I + T) $$ where $I$ denotes the identity matrix.

Answer 2

$$M=3B+BT$$ yes you can take B common as below $$M=B(3I_n+T)$$ $$det(M)=det(B)\times det(3I_3+T)$$ M is invertible that means $$det(M)\neq 0$$ or $$det(3I_3+T)\neq 0$$ $$det(\begin{bmatrix}k-3&2&0\\2&0&k\\0&k&-3\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix})\neq 0$$ $$det(\begin{bmatrix}k&2&0\\2&3&k\\0&k&0\end{bmatrix})\neq 0$$ $$-k^3\neq 0$$ $$k\neq 0$$