Let A be a 2 times 2 matrix with non-zero ent | vedclass

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(D) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Given $A^2 = I$,we have:$\begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} a & b

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Statement $-1$ is true,Statement $-2$ is false.

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(D) Let $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Given $A^2 = I$,we have:$\begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$\begin{bmatrix} a^2 + bc & b(a+d) \ c(a+d) & bc + d^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$From $b(a+d) = 0$,since $b 
eq 0$,we get $a+d = 0$,which implies $d = -a$.Thus,$tr(A) = a + d = a - a = 0$. So,Statement $-1$ is true.Now,$|A| = ad - bc = a(-a) - bc = -(a^2 + bc)$.From the matrix multiplication,$a^2 + bc = 1$,so $|A| = -1$.Therefore,Statement $-2$ is false.

Let $A$ be a $2 \times 2$ matrix with non-zero entries and let $A^2 = I$,where $I$ is the $2 \times 2$ identity matrix. Define $tr(A) = \text{sum of diagonal elements of } A$ and $|A| = \text{determinant of matrix } A$.
Statement $-1: tr(A) = 0$
Statement $-2: \det(A) = 1$

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Let $A$ be a $2 \times 2$ matrix with non-zero entries and let $A^2 = I$,where $I$ is the $2 \times 2$ identity matrix. Define $tr(A) = \text{sum of diagonal elements of } A$ and $|A| = \text{determinant of matrix } A$.Statement $-1: tr(A) = 0$Statement $-2: \det(A) = 1$