Prove that $ \det(A+2B)=3\det(B)$ under the condition that $\det(A)$ $=\det(A+B)$ $=\det(A-B)$ $=0$

Question

Let $A, B\in M_{3\times 3}(R)$ such that $$\det(A)=\det(A+B)=\det(A-B)=0$$ Show that $ \det(A+2B)=3\det(B)$

Observe that $$\det(A+0B)=\det(A+B)=\det(A-B)=0$$ and hence the $3\!-\!\mathrm{degree}$ polynomial of$$\det(A+tB)$$ has roots $t=0,1,-1$

But unfortunately there is nothing more I can tell.

Could anyone give me some hint please?

Answer 1

$$\det(A+tB)=c\cdot t(t-1)(t+1)\tag{1}$$

If $t\neq0$, rewrite (1) it as

$$\det\left(\frac1tA+B\right)=c\cdot \left(1-\frac{1}{t^2}\right)$$

Let $t\to\infty$, we get

$$\det B=c$$

Let $t=2$ in eq.(1)

$$\det(A+2B)=6c=6\det B$$

Remarks:

The OP is incorrect, take an example:

$$A=diag(0,1,1),~~~~B=diag(1,1,-1)$$

which satisfies $$\det A=\det(A+B)=\det(A-B)=0$$

Hence, $A+2B=diag(2,3,-1)$, and easy to check

$$\det(A+2B)=-6=6\det B$$

Answer 2

Letting the polynomial be $f(t)=\det(A+tB)$, its leading coefficient is nothing but $$\lim_{t\rightarrow\infty}(f(t)/t^3)=\det(B).$$ Therefore, you have shown that $f(t)=(\det(B))(t^3-t)$, so $\det(A+2B)=f(2)=6\det(B)\ (!?)$

Answer 3

To prove that $\det(A + 2B) = 3\det(B)$ under the conditions $\det(A) = \det(A + B) = \det(A - B) = 0$, we can use the properties of determinants. Let's consider the determinant of $(A + 2B)$ and let's expand this determinant using the properties of determinants:

$\det(A + 2B) = \det(A) + \det(2B) + 2(\det(A)\det(2B))$

Since $\det(A) = 0$, we can simplify the expression:

$\det(A + 2B) = 0 + \det(2B) + 2(0)(\det(2B))$

$\det(A + 2B) = \det(2B)$

Now, let's consider the determinant of $B$. Since det(A + B) = 0, we can write:

$\det(A + B) = 0 = \det(A) + \det(B) + \det(A)\det(B)$

$0 = 0 + \det(B) + 0(\det(B))$

$0 = \det(B)$

Therefore, we have shown that $\det(A + 2B) = \det(2B) and \det(B) = 0$.

Now:

$\det(A + 2B) = \det(2B) = 3(\det(B))$ $\square$