Problem 8 - Elementary and Intermediate Algebra

Question

Solve using matrices. $$ \begin{aligned} x+3 y &=16 \\ 6 x+y &=11 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \) and \( y = 5 \).

Step 1: Write the system of equations in matrix form

First, represent the system of equations as a matrix equation of the form \textbf{A} \textbf{x} = \textbf{b}. The given system is: \[ \begin{aligned} x + 3y &= 16 \ 6x + y &= 11 \end{aligned} \] This can be written in matrix form as: \[ \begin{pmatrix} 1 & 3 \ 6 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 16 \ 11 \end{pmatrix} \]

Step 2: Form the augmented matrix

Combine the coefficient matrix and the constant matrix into an augmented matrix: \[ \left( \begin{array}{cc|c} 1 & 3 & 16 \ 6 & 1 & 11 \end{array} \right) \]

Step 3: Apply row operations to achieve row echelon form

Step-by-step row operations are needed to transform the augmented matrix into row echelon form. First, make the entry in the first row, first column a 1 (if it isn't already): \[ R1 = R1/1 \rightarrow \left( \begin{array}{cc|c} 1 & 3 & 16 \ 6 & 1 & 11 \end{array} \right) \] Next, eliminate the first column entry of the second row: \[ R2 = R2 - 6R1 \rightarrow \left( \begin{array}{cc|c} 1 & 3 & 16 \ 0 & -17 & -85 \end{array} \right) \] Lastly, simplify the second row: \[ R2 = R2 / -17 \rightarrow \left( \begin{array}{cc|c} 1 & 3 & 16 \ 0 & 1 & 5 \end{array} \right) \]

Step 4: Back substitution

Starting from the second row, solve for \( y \): \[ y = 5 \] Substitute \( y \) back into the first equation to find \( x \): \[ x + 3(5) = 16 \rightarrow x + 15 = 16 \rightarrow x = 1 \]

Step 5: Conclusion

The solution to the system of equations is \( x = 1 \) and \( y = 5 \).

Key Concepts

matrix form

Solving a system of equations using matrices begins by converting the given equations into a matrix form. This involves writing the coefficients of the variables as a matrix. For example, for the system of equations:

\[ \begin{aligned} x + 3y &= 16 \ 6x + y &= 11 \end{aligned} \]
We represent it in matrix form by writing the coefficients and constants in separate matrices:
\[ \begin{pmatrix} 1 & 3 \ 6 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 16 \ 11 \end{pmatrix} \]
Here, the first matrix contains coefficients, the second matrix contains the variables, and the third contains the constants. Now we are ready to combine these matrices into one augmented matrix.

row operations

Once we have the augmented matrix, we need to perform row operations to transform it into a simpler form. Row operations help us to systematically simplify the matrix. There are three types of row operations:

• Swapping two rows (if necessary)
• Multiplying a row by a non-zero constant
• Adding or subtracting a multiple of one row to another row

Using these operations, we aim to achieve a row echelon form. For example, starting from:
\[ \left( \begin{array}{cc|c} 1 & 3 & 16 \ 6 & 1 & 11 \end{array} \right) \]
We may need to make one entry a pivot (like turning a 6 into 0 by subtraction). Each operation moves us closer to solving the system.

augmented matrix

An augmented matrix is a combination of the coefficient matrix and the constants from a system of linear equations. This matrix consolidates all the information we need to perform row operations efficiently. For our example, the system:
\[ \begin{aligned} x + 3y &= 16 \ 6x + y &= 11 \end{aligned} \]
Transforms into the augmented matrix:
\[ \left( \begin{array}{cc|c} 1 & 3 & 16 \ 6 & 1 & 11 \end{array} \right) \]
Here, the vertical bar separates the coefficient part from the constants part. This format makes it easier to apply row operations and see the impact on the constants simultaneously. The goal is to convert this augmented matrix into a simpler form, usually the row echelon form.

row echelon form

The row echelon form of a matrix is achieved when all non-zero rows are above rows with all zero entries, and the leading coefficient (the first non-zero number from the left, also known as a pivot) of a non-zero row is always to the right of the leading coefficient of the row above it. Consider our ongoing example:
Initially, we have:
\[ \left( \begin{array}{cc|c} 1 & 3 & 16 \ 6 & 1 & 11 \end{array} \right) \]
First, we make the first element of the second row zero by subtracting 6 times the first row from the second row:
\[ R2 = R2 - 6R1 \rightarrow \left( \begin{array}{cc|c} 1 & 3 & 16 \ 0 & -17 & -85 \end{array} \right) \]
Next, we scale the second row:
\[ R2 = R2 / -17 \rightarrow \left( \begin{array}{cc|c} 1 & 3 & 16 \ 0 & 1 & 5 \end{array} \right) \]
Now we have the row echelon form. It directly helps us to back-substitute and find the values of the variables. In our example, this form immediately tells us that:
\[ y = 5 \]
From there, substituting back, we find that:
\[ x = 1 \].