# matrix products

• Aug 24th 2007, 09:12 AM
kamaksh_ice
matrix products
if A and B are n* n matrix, and AB is not equal to BA . and ABA is equal to A^2B; BAB is equal to B^2 A; then prove that A+B is not invertible.
• Aug 24th 2007, 10:33 AM
red_dog
I think is not true.
Let $A$ such that $A^2\neq A$ and $B=I_n$. Then $ABA=A^2\neq BAB=A$.
But $A^2B=A^2$ and $BAB=A$ and there are not equal.
• Aug 24th 2007, 10:38 AM
Soroban
Hello, kamaksh_ice!

Is there a typo in the problem?
. . The statement is not true . . .

Quote:

If $A$ and $B$ are $n\times n$ matrices, and $ABA \,\neq \,BAB$,

. . prove that: . $A^2B \:=\:B^2A$

Let: . $A \:=\:\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad B \:=\:\begin{pmatrix}2&0\\0&2\end{pmatrix}$

$\begin{array}{cc}\text{Then:} & ABA \:=\:\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pm atrix}2&0\\0&2\end{pmatrix}\begin{pmatrix}1&0\\0&1 \end{pmatrix}\;=\;\begin{pmatrix}2&0\\0&2\end{pmat rix} \\ \\
\text{And:} & BAB \:=\:\begin{pmatrix}2&0\\0&2\end{pmatrix}\begin{pm atrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&2 \end{pmatrix}\:=\:\begin{pmatrix}4&0\\0&4\end{pmat rix}\end{array}$
. . $\text{Hence: }\;ABA \:\neq\:BAB$

$\begin{array}{cc}\text{But:} & A^2B \:=\:\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pm atrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&2 \end{pmatrix}\:=\:\begin{pmatrix}2&0\\0&2\end{pmat rix} \\ \\
\text{And: }& B^2A \:=\:\begin{pmatrix}2&0\\0&2\end{pmatrix}\begin{pm atrix}2&0\\0&2\end{pmatrix}\begin{pmatrix}1&0\\0&1 \end{pmatrix}\;=\;\begin{pmatrix}4&0\\0&4\end{pmat rix} \end{array}$
. . $\text{. . . and: }\;A^2B \:\neq \:B^2A$

Ha! red_dog beat me to it ... and explained it better!
.