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(D) Given the equation: $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$. Expanding the squares,we get: $(4 A^2+2 A B+2 B A+B^2)+(A^2-3 A B-3 B A+9 B^2)=5 A^2

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$A^2 B^2$

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(D) Given the equation: $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$. Expanding the squares,we get: $(4 A^2+2 A B+2 B A+B^2)+(A^2-3 A B-3 B A+9 B^2)=5 A^2-2 A B+10 B^2$. Combining like terms: $(4 A^2+A^2)+(B^2+9 B^2)+(2 A B-3 A B)+(2 B A-3 B A)=5 A^2-2 A B+10 B^2$. This simplifies to: $5 A^2+10 B^2-A B-B A=5 A^2-2 A B+10 B^2$. Subtracting $5 A^2+10 B^2$ from both sides,we get: $-A B-B A=-2 A B$. Adding $A B$ to both sides,we get: $-B A=-A B$,which implies $A B=B A$. Since $A$ and $B$ commute,$A B A B=A(B A) B=A(A B) B=(A A)(B B)=A^2 B^2$.

If $A$ and $B$ are $n \times n$ square matrices such that $(2 A+B)^2+(A-3 B)^2=5 A^2-2 A B+10 B^2$,then $A B A B=$

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