The above system of linear equations can be written in the form of AX = B where A is the matrix of coefficients, X is the matrix of unknown variables and B is the matrix of constants.
A = \begin{bmatrix} 2 & 1 \\ 2 & 3 \end{bmatrix}
X = \begin{bmatrix} x \\ y \end{bmatrix}
B = \begin{bmatrix} 3 \\ 6 \end{bmatrix}
First find out the |A|
as you can see |A| = 4 ≠0. Hence, the system of equations is consistent and will possess an unique solution and the solution can be found out using X = A-1B
A^{-1} = \frac{adj A}{|A|}
A-1 = \frac{1}{4}\begin{bmatrix} 3 & -1 \\ -2 & 2 \end{bmatrix}
X = A-1B
\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 3 & -1 \\ -2 & 2 \end{bmatrix} \begin{bmatrix} 3 \\ 6 \end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 9-6 \\ -6+12 \end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 3 \\ 6 \end{bmatrix}
From here, you can conclude that
x = 3/4
and y = 6/4
The above system of equations can be written in the form of AX = B, where
A = \begin{bmatrix} 1 & 3\\ 2 & 6 \end{bmatrix}
X = \begin{bmatrix} x \\ y \end{bmatrix}
B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}
Now check determinant of A
|A| = 6 - 6 = 0g
For checking consistency of the system, you have to check (adj A)B
adj A = \begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}
(adj A)B = \begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix}
(adj A)B = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix}
(adj A)B = \begin{bmatrix} 6 \\ -2 \end{bmatrix}
Since (adj A)B ≠0, the solution of the given system of linear equations doesn't exist. Hence, the system of equations is inconsistent.
The above system of equations can be written in the form of AX = B, where
A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}
X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}
B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}
Now check determinant of A
|A| = 1 * (12 - 5) + 1 * (9 + 10) + 2 * (-3 - 8)
|A| = 7 + 19 - 22 = 4
|A| ≠0
Hence, its inverse exists and hence there exists a unique solution that can be found out by X = A-1B
A^{-1} = \frac{adj A}{|A|}
A-1 = \frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11\\ -11 & -1 & 7 \end{bmatrix}
X = A-1B
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11\\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}
From here, you can see that x =2, y = 1, and z = 3.
Hence, x = 2, y = 1, and z = 3.