Solve the following equations by Matrix Method: x + 2y + 5 = 10, x - y = -2, 2x + 3y - z = -11.

Steps to Solve

1. Rewrite equations in standard form

We need to rewrite the given equations to isolate all variable terms on one side.

From the equations:

1. ( x + 2y + 5 = 10 ) becomes ( x + 2y = 5 )
2. ( x - y = -2 ) stays the same.
3. ( 2x + 3y - z = -11 ) becomes ( 2x + 3y + z = -11 )

So the system of equations is: [ \begin{align*}

1. & \quad x + 2y + 0z = 5 \quad (1) \

2. & \quad x - y + 0z = -2 \quad (2) \

3. & \quad 2x + 3y + z = -11 \quad (3) \ \end{align*} ]

4. Set up the Matrix Equation

We can now express the system in matrix form ( AX = B ), where ( A ) is the coefficient matrix, ( X ) is the variable matrix, and ( B ) is the constant matrix.

The coefficient matrix ( A ) is: [ A = \begin{bmatrix} 1 & 2 & 0 \ 1 & -1 & 0 \ 2 & 3 & 1 \end{bmatrix} ]

The variable matrix ( X ) is: [ X = \begin{bmatrix} x \ y \ z \end{bmatrix} ]

The constant matrix ( B ) is: [ B = \begin{bmatrix} 5 \ -2 \ -11 \end{bmatrix} ]

So the matrix equation is: [ AX = B ]

3. Find the Inverse of Matrix ( A )

To solve for ( X ), we need to find the inverse of ( A ) (if it exists) and multiply it by ( B ): [ X = A^{-1}B ]

Calculate the determinant of ( A ): [ \text{det}(A) = 1 \cdot [-1 \cdot 1 - 0 \cdot 3] - 2 \cdot [1 \cdot 1 - 0 \cdot 2] + 0 = -1 - 2 = -3 ] Since the determinant is non-zero, ( A ) is invertible.

4. Calculate the Inverse of Matrix ( A )

To find ( A^{-1} ), we can use the formula for the inverse of a 3x3 matrix. For simplicity, approaching using minors and cofactors is an option, or we can use a computational tool to find: [ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) ] Here, we calculate adjugate and use -3 to finalize ( A^{-1} ).

5. Solve for Variables

Once we get ( A^{-1} ), we can calculate: [ X = A^{-1}B ] This will yield the values of ( x, y, ) and ( z ).