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I recently took an algebra test, and one of the questions was like this (apologise upfront because I don't remember the exact matrix elements, but the idea is essentially the same):

Let $A$ and $B$ be two square matrices. We know that: $$AB=\begin{bmatrix}1 & 2 \\ 3&4\end{bmatrix}, \quad BA=\begin{bmatrix}3 & a \\b &2\end{bmatrix}$$ What can be said about values $a$ and $b$? What possible values can they have?

The real exam problem gave me only 4 options and only 1 was correct (it was a multiple-choice test question). The first 3 of the answers suggested possible values for $a$ and $b$, whereas the last answer stated $\text{(d)Nothing can be said about a and b}$.

How would you proceed? I spent like 10 minutes trying several things to no avail. No further conditions were given on A nor B. We only know they are square and 2x2.

What I tried:

I stablished: $$AB=X, \quad BA=Y$$ Then I multiplied by the inverse of $B$ (assuming it has an inverse), to reach: $$A=XB^{-1}=B^{-1}Y$$ $$B=A^{-1}X=YA^{-1}$$ But I really don't know where this will lead me.

José Carlos Santos
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Jose Lopez Garcia
  • 948

2 Answers2

4

Nothing can be said besides the fact that $a\neq0$ and that $b=\frac 8a$ (which means that both matrices $\left(\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}3&a\\b&2\end{smallmatrix}\right)$ have the same determinant). Indeed, if you take$$A=\begin{pmatrix}0&\frac a4\\1&\frac a8\end{pmatrix}\text{ and }B=\begin{pmatrix}\frac52&3\\\frac4a&\frac8a\end{pmatrix},$$then$$AB=\begin{pmatrix}1&2\\3&4\end{pmatrix}\text{ and }BA=\begin{pmatrix}3&a\\\frac8a&2\end{pmatrix}.$$

José Carlos Santos
  • 427,504
1

Answer on the same lines, typed it too slowly...

First of all we have the two conditions $$ \begin{aligned} \det(AB) &=\det(BA)\ ,\\ \operatorname{tr}(AB) &=\operatorname{tr}(BA)\ , \end{aligned} $$ the trace condition is fulfilled, the determinant condition becomes $$ab=8\ .$$ We consider only such values of $a,b$, show that each matrix can be realized (in many ways). For this, we search for a general $B$ in the form $$ B=\begin{bmatrix}x&y\\z&t\end{bmatrix}\ . $$ Then the condition $AB=BA$ can be equivalently rewritten $ (AB)^{-1}=(BA)^{-1}$, so we get both $$ \begin{aligned} A^{-1} &= B(B^{-1}A^{-1})=\frac 1{-2} \begin{bmatrix}x&y\\z&t\end{bmatrix} \begin{bmatrix}4&-2\\-3&1\end{bmatrix}\ , \\ A^{-1} &= (A^{-1}B^{-1})B=\frac 1{-2} \begin{bmatrix}2&-a\\-b&3\end{bmatrix} \begin{bmatrix}x&y\\z&t\end{bmatrix} \ . \end{aligned} $$ This leads to an algebraic system in $x,y,z,t$ with parameters $a,b$, $$ \left\{ \begin{aligned} 4x-3y &= 2x-az\\ -2x+y &= 2y-at\\ 4z-3t &= -bx+3z\\ -2z+t &= -by+3t\\ ab &= 8 \end{aligned} \right. $$ and the solution is simplest (found and) typed using a computer algebra system, here sage:

sage: var( 'x,y,z,t,b' );
sage: a = 8/b
sage: AB = matrix( 2, 2, [1,2,3,4] ) 
sage: BA = matrix( 2, 2, [3,a,b,2] )
sage: B  = matrix( 2, 2, [x,y,z,t] )
sage: Ainverse1 = B * AB.inverse() 
sage: Ainverse2 = BA.inverse() * B
sage: solve( [ Ainverse1[j,k] == Ainverse2[j,k] for j in [0,1] for k in [0,1] ], [x,y,z,t] ) 
[[x == r4, y == r3, z == 1/8*b*(3*r3 - 2*r4), t == 1/8*b*(r3 + 2*r4)]]

so any choice of r3, r4 above leads to a solution. My choice is $y=r_3=0$, $x=r_4=4$, so $$ B = \begin{bmatrix}4&0\\-b&b\end{bmatrix} \ , \qquad A = \begin{bmatrix}3/4&2/b\\7/4&4/b\end{bmatrix} \ . $$ Code and check:

sage: B = matrix( 2, 2, [4,0,-b,b] )
sage: B
[ 4  0]
[-b  b]
sage: A = AB * B.inverse()
sage: A
[3/4 2/b]
[7/4 4/b]
sage: A*B
[1 2]
[3 4]
sage: B*A
[  3 8/b]
[  b   2]
dan_fulea
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