Understanding the Matrix Equation and System of Linear Equations

The given problem asks for the solution of a matrix equation. A matrix equation of the form \(AX = B\) represents a system of linear equations. Here, \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix.

The given matrix equation is:

\[ \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 1&1&1\\ 1&{ - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9\\ 6\\ 2 \end{array}} \right] \]

This matrix equation can be written as the following system of linear equations:

1. \(2x - y + 3z = 9\)
2. \(x + y + z = 6\)
3. \(x - y + z = 2\)

Solving the System of Linear Equations

We can solve this system of linear equations using various methods, such as substitution, elimination, or matrix methods like Gaussian elimination or Cramer's rule. Let's use the elimination method as it's often straightforward for such systems.

Step 1: Eliminate a variable using two equations

Notice that equations (2) and (3) have opposite coefficients for the 'y' variable (+1 and -1). Adding these two equations will eliminate 'y'.

Add equation (2) and equation (3):

\[ (x + y + z) + (x - y + z) = 6 + 2 \]

\[ x + y + z + x - y + z = 8 \]

\[ (x+x) + (y-y) + (z+z) = 8 \]

\[ 2x + 0y + 2z = 8 \]

\[ 2x + 2z = 8 \]

Dividing by 2, we get a new equation:

\[ x + z = 4 \quad \text{(4)} \]

Now, let's eliminate 'y' again, perhaps using equations (2) and (3) in a different way, or using another pair. Subtracting equation (3) from equation (2) will also eliminate 'x' and 'z', giving us 'y' directly.

Subtract equation (3) from equation (2):

\[ (x + y + z) - (x - y + z) = 6 - 2 \]

\[ x + y + z - x + y - z = 4 \]

\[ (x-x) + (y+y) + (z-z) = 4 \]

\[ 0x + 2y + 0z = 4 \]

\[ 2y = 4 \]

Dividing by 2, we find the value of y:

\[ y = 2 \]

Step 2: Substitute the found variable into other equations

We found \(y = 2\). Substitute this value into equation (1):

\[ 2x - (2) + 3z = 9 \]

\[ 2x - 2 + 3z = 9 \]

Add 2 to both sides:

\[ 2x + 3z = 9 + 2 \]

\[ 2x + 3z = 11 \quad \text{(5)} \]

Step 3: Solve the new system with remaining variables

We now have a system of two linear equations with two variables (x and z):

4. \(x + z = 4\)
5. \(2x + 3z = 11\)

From equation (4), we can express x in terms of z:

\[ x = 4 - z \]

Substitute this expression for x into equation (5):

\[ 2(4 - z) + 3z = 11 \]

\[ 8 - 2z + 3z = 11 \]

Combine the 'z' terms:

\[ 8 + (3z - 2z) = 11 \]

\[ 8 + z = 11 \]

Subtract 8 from both sides to find the value of z:

\[ z = 11 - 8 \]

\[ z = 3 \]

Step 4: Find the value of the last variable

Now that we have \(z = 3\), substitute this value back into equation (4) (or \(x = 4 - z\)) to find the value of x:

\[ x = 4 - 3 \]

\[ x = 1 \]

Solution Verification

We found the solution to be \(x = 1\), \(y = 2\), and \(z = 3\). Let's check if these values satisfy all three original equations of the system derived from the matrix equation.

Equation 1: \(2x - y + 3z = 9\)

\[ 2(1) - (2) + 3(3) = 2 - 2 + 9 = 0 + 9 = 9 \quad (\text{Satisfied}) \]

Equation 2: \(x + y + z = 6\)

\[ 1 + 2 + 3 = 6 \quad (\text{Satisfied}) \]

Equation 3: \(x - y + z = 2\)

\[ 1 - 2 + 3 = -1 + 3 = 2 \quad (\text{Satisfied}) \]

Since all three equations are satisfied, the solution \(x=1, y=2, z=3\) is correct for the given matrix equation.

This step-by-step method clearly shows how to solve such a matrix equation by converting it into a system of linear equations.