If A = \(\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \), prove that A⁻¹ = A^T.

If A = \(\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \), verify that A(adj A) = (adj A)A = |A|I₂.

If A = \(\left[ \begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix} \right] \) and B = \(\left[ \begin{matrix} -1 & -3 \\ 5 & 2 \end{matrix} \right] \), verify that (AB)^-1 = B^-1A^-1

Decrypt the received encoded message \(\left[ \begin{matrix} 2 & -3 \end{matrix} \right] \left[ \begin{matrix} 20 & 4 \end{matrix} \right] \) with the encryption matrix \(\left[ \begin{matrix} -1 & -1 \\ 2 & 1 \end{matrix} \right] \) and the decryption matrix as its inverse, where the system of codes are described by the numbers 1 - 26 to the letters A - Z respectively, and the number 0 to a blank space.

Find the rank of the following matrices by row reduction method:
\(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \)

Solve: 2x + 3y = 10, x + 6y = 4 using Cramer's rule.

For what value of t will the system tx +3y - z = 1, x + 2y + z = 2, -tx + y + 2z = -1 fail to have unique solution?

Solve: 3x+ay = 4, 2x + ay = 2, a ≠ 0 by Cramer's rule.

Solve: x + y + 3z = 4, 2x + 2y + 6z = 7, 2x + y +  z = 10.

If the rank of the matrix \(\left[ \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right] \) is 2, then find ⋋.