# 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 $\displaystyle A$ such that $\displaystyle A^2\neq A$ and $\displaystyle B=I_n$. Then $\displaystyle ABA=A^2\neq BAB=A$.
But $\displaystyle A^2B=A^2$ and $\displaystyle BAB=A$ 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 $\displaystyle A$ and $\displaystyle B$ are $\displaystyle n\times n$ matrices, and $\displaystyle ABA \,\neq \,BAB$,

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

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

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

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

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