\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left( \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right) \)
adj A =\(\left( \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right) \)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)-(14-9) \\ +(3-4)-(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

The matrix form of the system is AX = B , where A = \(\left[ \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] \)
We find |A| = \(\left| \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right| \) = 10 - 6 = 4 ≠ 0. So, A⁻¹ exists and A⁻¹ = \(\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \)
Then, applying the formula X = A⁻¹B, we get
\(\left[ \begin{matrix} x \\ y \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} -4 \\ 16 \end{matrix} \right] =\left[ \begin{matrix} \frac { -4 }{ 4 } \\ \frac { 16 }{ 4 } \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 4 \end{matrix} \right] \).
So the solution is (x = −1, y = 4).

Let A be the matrix. Performing elementary row operations, we get
A = \(\left[ \begin{matrix} 2 \\ \begin{matrix} -3 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -1 \\ 7 \end{matrix} \end{matrix} \right] \)\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} -6 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 8 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -4 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -2 \\ 7 \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -13 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -2 \end{matrix} \end{matrix} \right] \).
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -45 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -30 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -15 \right) }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ 2 \end{matrix} \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3.

\(\left[ \begin{matrix} 0 \\ -1 \\ 4 \end{matrix}\begin{matrix} 3 \\ 0 \\ 2 \end{matrix}\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}\begin{matrix} 6 \\ 5 \\ 0 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 4 & 2 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 0 & 0 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 2 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 8 & 20 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 2 }{ 3 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ \frac { 22 }{ 3 } & 16 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 3R_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 0 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 2 & 5 \\ 1 & 6 \\ 22 & 48 \end{matrix} \right] \)