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$2\left(\left[\begin{array}{cc}{3 x} & {-1} \\ {8} & {5}\end{array}\right]+\left[\begin{array}{cc}{4} & {1} \\ {-2} & {-y}\end{array}\right]\right)=\left[\begin{array}{cc}{26} & {0} \\ {12} & {8}\end{array}\right]$

$2\left(\begin{array}{cc}{3 x+4} & {0} \\ {6} & {5 - y}\end{array}\right)=\left(\begin{array}{cc}{26} & {6} \\ {12} & {8}\end{array}\right)$

$\left(\begin{array}{cc}{6 x+8} & {0} \\ {12} & {10 - 2 y}\end{array}\right)=\left(\begin{array}{cc}{26} & {0} \\ {12} & {8}\end{array}\right)$

$6 x+8=26$
$6 x=18$
$x=3$

$10-2 y=8$
$-2 y=-2$
$y=1$

$A=\left[\begin{array}{lll}{1} & {2} & {3} \\ {4} & {5} & {6}\end{array}\right] \quad B=\left[\begin{array}{ll}{1} & {2} \\ {2} & {3} \\ {3} & {2}\end{array}\right]$

$A B=\left(\begin{array}{ll}{(1 \times 1)+(2 \times 2)+(3 \times 3)} & {(1 \times 2)+(2 \times 3)+(3 \times 2)} \\ {(4 \times 1)+(5 \times 2)+(6 \times 3)} & {(4 \times 2)+(5 \times 3)(6 \times 2)}\end{array}\right)$

$=\left(\begin{array}{ll}{1+4+9} & {2+6+6} \\ {4+10+18} & {8+15+12}\end{array}\right)$

$=\left(\begin{array}{ll}{14} & {14} \\ {32} & {35}\end{array}\right)$

show that $A C=B C$, but $A \neq B$

$A=\left[\begin{array}{ll}{1} & {3} \\ {0} & {1}\end{array}\right] \quad B=\left[\begin{array}{ll}{2} & {4} \\ {2} & {3}\end{array}\right] \quad C=\left[\begin{array}{cc}{1} & {-2} \\ {-1} & {2}\end{array}\right]$

$A C=\left(\begin{array}{ll}{1} & {3} \\ {0} & {1}\end{array}\right)\left(\begin{array}{ll}{1} & {-2} \\ {-1} & {2}\end{array}\right)$

$=\left(\begin{array}{cc}{-2} & {4} \\ {-1} & {2}\end{array}\right)$

$B C=\left(\begin{array}{ll}{2} & {4} \\ {2} & {3}\end{array}\right)\left(\begin{array}{cc}{1} & {-2} \\ {-1} & {2}\end{array}\right)$

$=\left(\begin{array}{cc}{-2} & {4} \\ {-1} & {2}\end{array}\right)$

\$A C=B C$

but $A \neq B$

Is $(A+B)^{2}=A^{2}+2 A B+B^{2} ? ?$

$(A+B)^{2}=(A+B)(A+B)$

$=A A+A B+B A+B B$
$=A^{2}+A B+B A+B^{2}$

\$(A+B)^{2} \neq A^{2}+2 A B+B^{2}$

Because $A B \neq B A$ in general