The equations can be written in matrix form as
\(\left( \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 4 \end{matrix} \right) \) ⇒ AX = B where
A =\(\left( \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right) ,X=\left( \begin{matrix} x \\ y \end{matrix} \right) ,B=\left( \begin{matrix} 3 \\ 4 \end{matrix} \right) \)
∴ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -1 \\ 5 & 1 \end{matrix} \right| \) = 2+5 =7 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 7 } \left[ \begin{matrix} 1 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ X = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 1 & 1 \\ -5 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 4 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 3+4 \\ -15+8 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 7 \\ -7 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ -1 \end{matrix} \right] \)
∴ Solution set is {1, -1}

The matrix form of the system is
\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where
A=\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
|A| =\(\left| \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right| =3\left| \begin{matrix} 2 & 12 \\ 1 & 7 \end{matrix} \right| -1\left| \begin{matrix} 3 & 12 \\ 2 & 7 \end{matrix} \right| +9\left| \begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix} \right| \)
= 3 (14 - 12) - 1 (21 -24) + 9 (3 -4)
= 3 (2) -1 (-3) + 9 (-1)
= 6 + 3 - 9 = 9 - 9 = 0
Since |A| = 0, the homogeneous system of equations have non-trivial solutions also.

Let A =\(\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{matrix} \right] \)
Now |A| = \(\left| \begin{matrix} 6 & -5 \\ -2 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & -5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & 6 \\ 5 & -2 \end{matrix} \right| \)
= 3 (12 - 10) + 1 (-30 + 25) + 1 (30 - 30)
= 3(2) + 1 (-5) + 0 = 6 - 5 = 1 ≠ 0
∴ Rank of A is 3.

Given A³ = I ⇒ IA³I = |I|
⇒ |A.A.A| = 1
⇒ |A|. |A|·|A| = 1
⇒ |A|³ = 1
∴ |A| ≠ 0
Hence, A is non-singular.

Given A (adj A) =\(\left[ \begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix} \right] =10\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)...(1)
We know A (adj A) = (adj A) A = |A|. I₂ ...(2)
Comparing (1) and (2), we get |A| = 10