Suppose A = begin{bmatrix} a & b c & d end{b | vedclass

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(A) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $A^2 = A$.Calculating $A^2$:$A^2 = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begi

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(A) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ and $A^2 = A$.Calculating $A^2$:$A^2 = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd \ ac + cd & bc + d^2 \end{bmatrix}$.Since $A^2 = A$,we have:$\begin{bmatrix} a^2 + bc & ab + bd \ ac + cd & bc + d^2 \end{bmatrix} = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Comparing the elements,we get $ab + bd = b$,which implies $b(a + d) = b$.Since the entries are non-zero,$b 
eq 0$,so we can divide by $b$ to get $a + d = 1$.

Suppose $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is a real matrix with non-zero entries,$ad - bc = 0$ and $A^2 = A$. Then,$a + d$ equals

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