Question

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{r} x-y+z=-4 \\ 2 x-3 y+4 z=-15 \\ 5 x+y-2 z=12 \end{array}\right. $$

Short answer

The solution is \( x = 1, y = 3, z = -2 \).

Step-by-step solution

- Write the system of equations in matrix form

First, represent the system of equations as an augmented matrix: \[ \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 2 & -3 & 4 & | & -15 \ 5 & 1 & -2 & | & 12 \ \ \ \end{pmatrix} \]

- Perform row operations to get a row-echelon form

Use row operations to pivot and simplify the matrix. Start by making the first row's leading coefficient (first element) 1, which it already is, then eliminate the first variable from the other rows.1. Subtract 2 times the first row from the second row: \[ R2 = R2 - 2R1 \] Resulting matrix: \[ \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & -1 & 2 & | & -7 \ 5 & 1 & -2 & | & 12 \ \ \end{pmatrix} \]2. Subtract 5 times the first row from the third row: \[ R3 = R3 - 5R1 \] Resulting matrix: \[ \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & -1 & 2 & | & -7 \ 0 & 6 & -7 & | & 32 \ \ \end{pmatrix} \]

- Make the pivot in the second row to 1 and simplify

Replace the second row with its negation to make the pivot a positive 1: \[ R2 = -R2 \] Resulting matrix: \[ \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & 1 & -2 & | & 7 \ 0 & 6 & -7 & | & 32 \ \ \end{pmatrix} \]Next, eliminate the second variable from the third row by subtracting 6 times the second row from the third row: \[ R3 = R3 - 6R2 \] Resulting matrix: \[ \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & 1 & -2 & | & 7 \ 0 & 0 & 5 & | & -10 \ \ \end{pmatrix} \]

- Back substitution to find the solutions

From the simplified matrix: \[ \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & 1 & -2 & | & 7 \ 0 & 0 & 5 & | & -10 \ \ \end{pmatrix} \]We can write the equations as: 1. \( z = -2 \)2. \( y - 2z = 7 \)3. \( x - y + z = -4 \)Substitute \( z = -2 \) into the second equation: \( y - 2(-2) = 7 \)\( y + 4 = 7 \)\( y = 3 \)Then substitute \( y = 3 \) and \( z = -2 \) into the first equation: \( x - 3 + (-2) = -4 \)\( x - 5 = -4 \)\( x = 1 \)

Key concepts

Matrix Representation

Understanding systems of equations becomes easier when using matrices. Matrices allow us to organize and simplify the process of finding solutions. For our given system of equations: \ 1. \( x - y + z = -4 \)\ 2. \( 2x - 3y + 4z = -15 \)\ 3. \( 5x + y - 2z = 12 \)\ We can represent this system as an augmented matrix: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 2 & -3 & 4 & | & -15 \ 5 & 1 & -2 & | & 12 \ \end{pmatrix} \ \)This matrix captures the coefficients of each variable and the constants from each equation in a structured form. The vertical bar separates the coefficients on the left from the constants on the right.

Row Operations

Row operations are the steps we take to simplify our matrix. They help us eliminate variables systematically. There are three key row operations: \

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting the multiple of one row to/from another

For our augmented matrix: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 2 & -3 & 4 & | & -15 \ 5 & 1 & -2 & | & 12 \ \end{pmatrix} \ \), we perform these steps: \

• Subtract 2 times the first row from the second row \( R2 = R2 - 2R1 \)

Updating the matrix: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & -1 & 2 & | & -7 \ 5 & 1 & -2 & | & 12 \ \end{pmatrix} \ \)\ Similarly, subtracting 5 times the first row from the third results in: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & -1 & 2 & | & -7 \ 0 & 6 & -7 & | & 32 \ \end{pmatrix} \ \). These operations help in forming our next key step: Row-Echelon Form.

Row-Echelon Form

Row-Echelon Form (REF) is all about simplifying the matrix to a state where it is easier to solve. In REF, each leading entry (the first non-zero number from the left in a row) is 1, and it is to the right of the leading entry in the row above. Also, all entries below a leading entry are zeros. Continuing from our current matrix: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & -1 & 2 & | & -7 \ 0 & 6 & -7 & | & 32 \ \end{pmatrix} \ \), we make the pivot in the second row positive by multiplying it by -1: \ \( R2 = -R2 \). This gives us: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & 1 & -2 & | & 7 \ 0 & 6 & -7 & | & 32 \ \end{pmatrix} \ \).\ Next, we want all elements below the second row's pivot to be zeros, so we subtract 6 times the second row from the third row: \ \( R3 = R3 - 6R2 \). The matrix now looks like this: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & 1 & -2 & | & 7 \ 0 & 0 & 5 & | & -10 \ \end{pmatrix} \ \). The matrix is closer to its final form, making it easier to solve using back substitution.

Back Substitution

Back substitution is the final step in solving a system of equations using matrices. It involves solving from the last row upwards. With our matrix now in row-echelon form: \ \( \begin{pmatrix} 1 & -1 & 1 & | & -4 \ 0 & 1 & -2 & | & 7 \ 0 & 0 & 5 & | & -10 \ \end{pmatrix} \ \), we can rewrite it to start solving: \

• From the third row: \ \( 5z = -10 \)\ Hence, \( z = \frac{-10}{5} = -2 \)
\
• Second row: \ \( y - 2z = 7 \)\ Substituting \( z = -2 \): \ \( y - 2(-2) = 7 \)\ \( y + 4 = 7 \)\ Hence, \( y = 3 \)
\
• First row: \ \( x - y + z = -4 \)\ Substituting \( y = 3 \) and \( z = -2 \): \ \( x - 3 + (-2) = -4 \)\ \( x - 5 = -4 \)\ Hence, \( x = 1 \)

\ Therefore, the solution to the system of equations is \( x = 1, y = 3, z = -2 \).