# Matrices - given $AB$, how to find determinant of $BA$ ??

Let A$$_{3×2}$$ and B$$_{2×3}$$ be matrices such that their product $$AB$$ is $$AB=\begin{pmatrix} 8&2 & -2\\ 2&5&4 \\ -2&4&5 \\ \end{pmatrix}$$ And $$BA$$ is nonsingular

Find the determinant of $$BA$$.

I have no idea , how to solve this type of question. All I could notice is that $$|AB| = 0$$ and it's a symmetric matrix. I tried assuming a general matrix , but I get simply too many unknowns and very few equations.

• Try Cauchy-Binet, or compute the characteristic polynomial... BA and AB have the same charpolys except AB has an extra zero in its polynomial for dimension reasons – user8675309 Feb 14 at 10:03
• I dont know what Cauchy Binet is ... But I do have a vague idea of characteristic polynomials.... If you say so , that the property is true, how to decipher, which of the root of AB , is bot part of BA?? – RandomAspirant Feb 14 at 11:54
• my read is: the point of this exercise is you are supposed to learn to relate the characteristic polynomials of AB and BA... I infer you are working in $\mathbb R$ here -- another choice is to compute $\text{trace}\big(AB\big)$ --which must be $= \text{trace}\big(BA\big)$ and compute $\text{trace}\big((AB)^2\big)$-- --which must be $= \text{trace}\big((BA)^2\big)$-- and Newton's Identities give you the result – user8675309 Feb 14 at 18:59

A direct calculation is also possible in case one wants to find out the answer easily. If we denote the coefficients of $$A$$ by $$a_i$$ and the ones of $$B$$ by $$b_j$$, and the given matrix by $$C$$, then the matrix equation $$AB=C$$ is equivalent to equations in $$a_i,b_j$$. We can solve them case by case. The first equation is $$a_1b_1 + a_2b_4 =8$$. For $$a_1=0$$ we obtain $$a_2\neq 0$$ and $$b_4=\frac{8}{a_2}, b_5= \frac{2}{a_2}, b_6=-\frac{2}{a_2}, a_4= \frac{a_2a_3b_3 - 4a_2}{2}, b_3= -\frac{1}{a_3(a_3b_2 - 9}, b_2= \frac{1}{(4a_3)(a_3b_1 + 18)}, a_6= -\frac{a_2a_5b_1 + 2a_2}{8}, a_5=a_3.$$ Then we obtain $$BA=\begin{pmatrix}9 & 0 \cr 0 & 9\end{pmatrix}.$$ The other case $$a_1\neq 0$$ is similar. Note that $$\det(AB)\neq \det(BA)$$ in general, but $$tr(AB)=tr(BA)$$ is true in general. So the determinant is $$81$$.
• We don't assume values. We just do both cases, first assuming $a_1=0$ and then $a_1\neq 0$. This covers everything. The second case has the advantage that we can use the first equation $a_1b_1 + a_2b_4 =8$ to eliminate $b_1$. I have edited my post. – Dietrich Burde Feb 14 at 15:09
The characteristic polynomial of $$AB$$ is $$p(x)=\left| \begin{array}{rrr} 8-x & 2 & -2 \\ 2 & 5-x & 4 \\ -2 & 4 & 5-x \end{array} \right|=\cdots =-x^3+18x^2-81x=-x(x-9)^2$$ Cayley-Hamilton Theorem implies that $$p(AB)=0$$, i.e., $$-(AB)^3+18(AB)^2-81AB=0$$ In fact, since $$AB$$ is symmetric, it is diagonalisable and hence its minimal polynomial is $$m(x)=x(x-9)$$.
Hence $$(AB)^2-9AB=0 \quad\Longrightarrow\quad B(AB)^2A-9BABA=0 \quad\Longrightarrow\quad (BA)^3-9(BA)^2=0. \quad\Longrightarrow\quad (BA)^2\big((BA)-9\big)=0.$$ Clearly, rank$$(AB)=2$$, and also rank$$((AB)^2)=2$$ (as $$(AB)^2$$ is also symmetric with eigenvalues $$81,81,0$$) and hence rank$$(BA)=2$$, and thus $$BA$$ is invertible and thus $$(BA)^2\big((BA)-9\big)=0\quad\Longrightarrow\quad (BA)-9=0,$$ i.e., $$BA=\left(\begin{array}{cc} 9&0 \\ 0&9\end{array}\right)$$ and hence det$$(BA)=81$$.