If A and B are two square matrices of order 3 | vedclass

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(A) Given $AB = A$ and $BA = B$.We know that $A^2 = A(A) = A(BA) = (AB)A = AA = A^2$. Wait,let us compute $A^2$ directly: $A^2 = A(A) = A(AB) = (AA)B$

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(A) Given $AB = A$ and $BA = B$.We know that $A^2 = A(A) = A(BA) = (AB)A = AA = A^2$. Wait,let us compute $A^2$ directly: $A^2 = A(A) = A(AB) = (AA)B$. Actually,$A^2 = A(AB) = (AA)B$ is not helpful. Let's use $A^2 = A \cdot A = A(AB) = (AA)B$. No,$A^2 = A \cdot A = (AB)A = A(BA) = AB = A$.Similarly,$B^2 = B \cdot B = (BA)B = B(AB) = BA = B$.Since $A^2 = A$,it follows that $A^n = A$ for all $n \in \mathbb{N}$.Similarly,since $B^2 = B$,it follows that $B^n = B$ for all $n \in \mathbb{N}$.Thus,$X = A^4 + B^4 = A + B$.And $Y = A^{10} + B^{10} = A + B$.Therefore,$X - Y = (A + B) - (A + B) = O$,where $O$ is the zero matrix.$A$ zero matrix is a singular matrix because its determinant is $0$.

If $A$ and $B$ are two square matrices of order $3$ such that $AB = A$ and $BA = B$,and matrices $X$ and $Y$ are defined as $X = A^4 + B^4$ and $Y = A^{10} + B^{10}$,then the matrix $X - Y$ is:

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