Let A and B are square matrices of order 2 with real elements such that AB = $A^2B^2 - (AB)^2$

Question

A and B are 2 x 2 matrices with real elements and AB = $A^2B^2 - (AB)^2$ and |B| = 3,

Find the value of |A+2B| - |B+2A|.

Using the given relation, i managed to get |A| = 0 by the following;

AB = $A^2B^2 - (AB)^2$

A = $A^2B - ABA$

A(I - AB - BA) = 0 [ taking determinant on both sides ]

|A||I - AB - BA| = 0

It's here I was stuck on and later on seeing the solution it said this;

" For 2x2 matrices A and B, the following relationship holds true for some scalar x, |A+xB| = |A| + $x^2|B|$ + x[Tr(A)Tr(B)-Tr(AB)] "

Could anyone explain how this relationship was derived or any alternates to solve this problem.

Answer 1

Here is an alternative solution. You have already shown that $A$ is singular. Since $B$ is invertible, from $AB=A^2B^2-(AB)^2$ we obtain $A=A^2B-ABA$. Therefore $A$ has a zero trace. Since $A$ also has a zero determinant, its characteristic polynomial is $x^2$, i.e., it is nilpotent.

Clearly $0\subsetneq\ker(A)\subseteq\ker(BA)$. As the matrices are merely $2\times2$, this means $A$ and $BA$ are simultaneously triangulable. It follows that $|pI+qA+rBA|=|pI+rBA|$ for every real numbers $p,q$ and $r$.

Since $BA$ is singular, one of its eigenvalue is zero and the other is equal to $t=\operatorname{tr}(BA)$. So, we further obtain $$ |pA+qBA+rI|=|qBA+rI|=(qt+r)r=qrt+r^2.\tag{1} $$ Now let $b=\operatorname{tr}(B)$. By Cayley-Hamilton theorem, $B^2-bB+|B|I=0$. Therefore $B^{-1}=\frac{bI-B}{3}$ and $$ \begin{aligned} |xA+yB| &=|B||xB^{-1}A+yI|\\ &=3\left|x\left(\frac{bI-B}{3}\right)A+yI\right|\\ &=3\left|bxA-\frac{x}{3}BA+yI\right|\\ &=3\left(-\frac{xyt}{3}+y^2\right)\quad\text{by }(1)\\ &=3y^2-xyt. \end{aligned} $$ Thus $$ |A+2B|-|2A+B|=(12-2t)-(3-2t)=9. $$

Answer 2

Let $A=(y_1,y_2)$, where $y_1, y_2$ are column vectors, for example, $y_1=(a_{11},a_{21})^T$, $y_2=(a_{12},a_{22})^T$. Similar for $B=(\beta_1,\beta_2)$, where $\beta_1=(b_{11},b_{21})^T$, $\beta_2=(b_{12},b_{22})^T$.

$$\begin{align}|A+xB|&=|(y_1+x\beta_1,y_2+x\beta_2)|=|(y_1,y_2+x\beta_2)|+|(x\beta_1,y_2+x\beta_2)|\\ \\ &=|(y_1,y_2)|+|(y_1,x\beta_2)|+|(x\beta_1,y_2)|+|(x\beta_1,x\beta_2)|\\ \\ &=|A|+|(y_1,x\beta_2)|+|(x\beta_1,y_2)|+x^2|B|\\ \\ &=|A|+x(a_{11}b_{22}+a_{22}b_{11}-a_{12}b_{21}-a_{21}b_{12})+x^2|B|\\ \\ &=|A|+x(a_{11}b_{22}+a_{22}b_{11}+a_{11}b_{11}+a_{22}b_{22}-a_{11}b_{11}-a_{22}b_{22}-a_{12}b_{21}-a_{21}b_{12})+x^2|B|\\ \\ &=|A|+x\left[(a_{11}+a_{22})(b_{11}+b_{22})-(a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}+a_{22}b_{22})\right]+x^2|B|\\ \\ &=|A|+x\left[\text{Tr}(A)\text{Tr}(B)-\text{Tr}(AB)\right]+x^2|B|\end{align}$$