Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{aligned} 3 x+y-z &=\frac{2}{3} \\ 2 x-y+z &=1 \\ 4 x+2 y &=\frac{8}{3} \end{aligned}\right. $$

Short answer

The solution is \(x = \frac{1}{45}\), \(y = \frac{1}{3}\), \(z = \frac{8}{15}\).

Step-by-step solution

Write the augmented matrix

Represent the system of equations as an augmented matrix:\[\begin{bmatrix}3 & 1 & -1 & \frac{2}{3} \2 & -1 & 1 & 1 \4 & 2 & 0 & \frac{8}{3}\end{bmatrix}\]

Normalize the first row

Divide the first row by 3 to get a leading 1 in the first position:\[R1 \rightarrow \frac{1}{3} R1\begin{bmatrix}1 & \frac{1}{3} & -\frac{1}{3} & \frac{2}{9} \2 & -1 & 1 & 1 \4 & 2 & 0 & \frac{8}{3}\end{bmatrix}\]

Eliminate first column's entries below the leading 1

Subtract 2 times the first row from the second row and 4 times the first row from the third row to form zeros in the first column below the leading 1:\[R2 \rightarrow R2 - 2R1\R3 \rightarrow R3 - 4R1\begin{bmatrix}1 & \frac{1}{3} & -\frac{1}{3} & \frac{2}{9} \0 & -\frac{5}{3} & \frac{5}{3} & \frac{7}{9} \0 & \frac{4}{3} & \frac{4}{3} & \frac{20}{9}\end{bmatrix}\]

Normalize the second row

Divide the second row by \(-\frac{5}{3}\) to get a leading 1 in the second row:\[R2 \rightarrow -\frac{3}{5} R2\begin{bmatrix}1 & \frac{1}{3} & -\frac{1}{3} & \frac{2}{9} \0 & 1 & -1 & -\frac{7}{15} \0 & \frac{4}{3} & \frac{4}{3} & \frac{20}{9}\end{bmatrix}\]

Eliminate the second column's entry below the leading 1

Subtract \(\frac{4}{3}\) times the second row from the third row:\[R3 \rightarrow R3 - \frac{4}{3}R2\begin{bmatrix}1 & \frac{1}{3} & -\frac{1}{3} & \frac{2}{9} \0 & 1 & -1 & -\frac{7}{15} \0 & 0 & \frac{8}{3} & \frac{64}{45}\end{bmatrix}\]

Normalize the third row

Divide the third row by \(\frac{8}{3}\) to get a leading 1 in the third row:\[R3 \rightarrow \frac{3}{8}R3\begin{bmatrix}1 & \frac{1}{3} & -\frac{1}{3} & \frac{2}{9} \0 & 1 & -1 & -\frac{7}{15} \0 & 0 & 1 & \frac{8}{15}\end{bmatrix}\]

Back-substitute to get upper triangular form

Eliminate the above-diagonal entries in the third column:\[R2 \rightarrow R2 + R3\R1 \rightarrow R1 + \frac{1}{3}R3\begin{bmatrix}1 & \frac{1}{3} & 0 & \frac{10}{45} \0 & 1 & 0 & \frac{1}{3} \0 & 0 & 1 & \frac{8}{15}\end{bmatrix}\]

Simplify to get the final solution

Now, simplify the matrix back to its simplest form to read the solutions:\[\begin{bmatrix}1 & 0 & 0 & \frac{1}{45} \0 & 1 & 0 & \frac{1}{3} \0 & 0 & 1 & \frac{8}{15}\end{bmatrix}\]So the solution is \(x = \frac{1}{45}\), \(y = \frac{1}{3}\), \(z = \frac{8}{15}\)

Key concepts

Row Operations

Understanding row operations is essential for solving systems of equations using matrices. Row operations include three main types: swapping rows, multiplying a row by a constant, and adding or subtracting one row from another. These operations help in transforming the matrix into a simpler form while preserving the solutions of the system. For instance, when solving a system, you might need to create zeros below a pivot (a leading 1) to simplify the matrix. By performing appropriate row operations, such as subtracting multiples of one row from another, you can systematically reduce the matrix to row-echelon form and eventually to reduced row-echelon form to find the solution.

Augmented Matrix

An augmented matrix is a convenient and compact way to represent a system of linear equations. It combines the coefficients of the system and the constants into a single matrix. To form an augmented matrix from a system of equations:

• Write all the coefficients of the variables in order as rows.
• Add a vertical bar to separate the constants from the coefficients.

For example, the system:\[\begin{aligned} 3x + y - z &= \frac{2}{3} \ 2x - y + z &= 1 \ 4x + 2y &= \frac{8}{3} \end{aligned}\] is represented as:\[\begin{bmatrix} 3 & 1 & -1 & \frac{2}{3} \ 2 & -1 & 1 & 1 \ 4 & 2 & 0 & \frac{8}{3} \end{bmatrix}\]. This matrix includes all the essential information needed to solve the system using techniques like Gaussian elimination.

Gaussian Elimination

Gaussian elimination is a method used to solve systems of linear equations. It transforms the augmented matrix to row-echelon form using a series of row operations. This form makes it easy to solve the system by back-substitution. The steps involved in Gaussian elimination include:

• Form the augmented matrix.
• Use row operations to create zeros below each pivot (leading 1).
• Continue the process until you have an upper triangular matrix.

Once you have achieved the upper triangular form, you can solve the system by starting from the bottom row and working upwards, substituting known values to find the unknowns. This method is highly systematic and breaks down complex systems into manageable steps.

Matrix Normalization

Matrix normalization refers to the process of scaling the rows of the matrix to make the leading coefficient (pivot) in each row equal to 1. This step simplifies further row operations and ensures the matrix is in row-echelon form or reduced row-echelon form. The key steps to normalize a matrix include:

• Start with the first row and scale it so that the leading coefficient becomes 1.
• Proceed to the next row, adjusting it similarly by subtracting multiples of previous rows to create zeros.
• Normalize each subsequent row by scaling and performing necessary row operations.

Normalizing the matrix makes it straightforward to identify the solutions of the system by simplifying the augmented matrix to a form where the solutions are easily readable, as seen in the final steps of Gaussian elimination.