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

$G = \left\{ \begin{bmatrix} x & x \\ x & x \end{bmatrix} : x \in \mathbb{R} \setminus \{0\} \right\}$ is a group with respect to matrix multiplication. In this group,the inverse of $\begin{bmatrix} 1/3 & 1/3 \\ 1/3 & 1/3 \end{bmatrix}$ is

If $A = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} y \\ x \\ 1 \end{bmatrix}$ are such that $AB = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$,then:

Let $M$ and $N$ be two matrices over $\mathbb{R}$ of order $2$. Then,$MN = NM$ if .......

If $A+2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$ and $2A-B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$,then $\operatorname{tr}(A)-\operatorname{tr}(B) =$

Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$,$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$,and $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$. Find $A + B$.