​$$\text { Let } A=\left(\begin{array}{cc}\alpha & 1 \\0 & -1\end{array}\right) \text { and } B=\left(\begin{array}{ll}4 & 1 \\0 & 1\end{array}\right) \text {, such that } A^2=B \text {, then the value of } \alpha \text { is: }$$​​

The sum of the eigenvalues of the matrix $$\begin{pmatrix} 5 & 4 \\1 & 2\end{pmatrix}$$ is​

If $$A = \begin{pmatrix} \alpha & 0 \\1 & 1\end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\5 & 1\end{pmatrix}$$and $$A^2=B$$, then value of $$\alpha$$ is​

if $$a, b, c$$ are three unequal numbers and $$\left| \begin{array}{ccc}0 & x - a & x - b \\x + a & 0 & x - c \\x + b & x + c & 0 \\\end{array}\right| = 0$$ then $$x$$ is​

​$$\text { Let } A=\left(\begin{array}{cc}\alpha & 1 \\0 & -1\end{array}\right) \text { and } B=\left(\begin{array}{ll}4 & 1 \\0 & 1\end{array}\right) \text {, such that } A^2=B \text {, then the value of } \alpha \text { is: }$$​​

​$$\text{If } A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{bmatrix}, \text{ then } \det(A + B) = ?$$​​

The product of the cofactors of 3 and -2 in the matrix $$\begin{bmatrix}1 & 0 & -2 \\3 & -1 & 2 \\4 & 5 & 6\end{bmatrix}$$​ is: