Problem 56 - Algebra and Trigonometry Real Mathematics, Real People

Question

Matrix Multiplication Use matrix multiplication to determine whether each matrix is a solution of the system of equations. Use a graphing utility to verify your results. \(\left\\{\begin{aligned} 6 x+2 y &=0 \\\\-x+5 y &=16 \end{aligned}\right.\) (a) \(\left[\begin{array}{r}-1 \\ 3\end{array}\right]\) (b) \(\left[\begin{array}{r}2 \\ -6\end{array}\right]\) (c) \(\left[\begin{array}{r}3 \\ -9\end{array}\right]\) (d) \(\left[\begin{array}{r}-3 \\ 9\end{array}\right]\)

Step-by-Step Solution

Answer

The solutions of the given system of equations are vectors (a) and (d).

Step 1: Represent the system of equations in matrix form

First, rewrite the system of equations into matrix form. The system of equations can be represented as \[ \begin{bmatrix} 6 & 2 \\-1 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 16 \end{bmatrix} \]

Step 2: Multiply each vector with the coefficients matrix

Now, for each provided vector, perform matrix multiplication with the coefficients matrix. Then, compare the resultant matrix with the right-hand side matrix. If it equates, the vector is a solution, else it's not.

Step 3: Verifying solution (a) [-1, 3]

Multiplying coefficient matrix by vector (a) we get \[ \begin{bmatrix} 6 & 2 \\-1 & 5 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 16 \end{bmatrix} \] Thus, vector (a) is a solution.

Step 4: Verifying solution (b) [2, -6]

Multiply coefficient matrix by vector (b) \[ \begin{bmatrix} 6 & 2 \\-1 & 5 \end{bmatrix} \begin{bmatrix} 2 \\ -6 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 16 \end{bmatrix} \] Thus, vector (b) is not a solution.

Step 5: Verifying solution (c) [3, -9]

Multiply coefficient matrix by vector (c) \[ \begin{bmatrix} 6 & 2 \\-1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ -9 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 16 \end{bmatrix} \] Thus, vector (c) is not a solution.

Step 6: Verifying solution (d) [-3, 9]

Multiply coefficient matrix by vector (d) \[ \begin{bmatrix} 6 & 2 \\-1 & 5 \end{bmatrix} \begin{bmatrix} -3 \\ 9 \end{bmatrix} = \begin{bmatrix} 0 \\ 16 \end{bmatrix} \] Thus, vector (d) is a solution.

Key Concepts

Systems of Equations

A system of equations refers to a set of two or more equations with the same set of unknowns. You can solve systems of equations using various methods, such as substitution, elimination, or matrix methods. In the context of matrices, these methods offer a structured way to handle multiple equations simultaneously. Each equation in the system should be satisfied for a solution to be valid. For instance, in our exercise, the system consists of two linear equations with two variables, given as:

• 6x + 2y = 0
• -x + 5y = 16

These equations need to be solved together to find a pair of values for x and y that satisfy both equations. Utilizing matrices can greatly simplify this process, especially with larger systems.

Matrix Form

To solve systems of equations using matrices, we first convert the system into what is known as matrix form. This involves representing the coefficients of the variables in a matrix, while the variables themselves form another column matrix. Similarly, the constants on the right-hand side form another matrix. For example, with our given system:

1. Coefficient matrix: \[ \begin{bmatrix} 6 & 2 \ -1 & 5 \end{bmatrix} \]
2. Variable matrix: \[ \begin{bmatrix} x \ y \end{bmatrix} \]
3. Constant matrix: \[ \begin{bmatrix} 0 \ 16 \end{bmatrix} \]

With these, the system in matrix form is: \[ \begin{bmatrix} 6 & 2 \ -1 & 5 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 16 \end{bmatrix} \] This matrix equation allows us to apply matrix algebra to find solutions, if any.

Matrix-Vector Multiplication

Matrix-vector multiplication involves the process of multiplying a matrix by a column vector. Each element in the resulting vector is the dot product of a row from the matrix and the input vector. For our exercise, we multiply the given coefficient matrix by each vector representing a potential solution:

• For \(\begin{bmatrix} -1 \ 3 \end{bmatrix}\), multiplying givesthe correct result: the dot products match the constants in our matrix-form equation.
• For other vectors like \(\begin{bmatrix} 2 \ -6 \end{bmatrix}\), the resulting vector does not match \(\begin{bmatrix} 0 \ 16 \end{bmatrix}\).

This multiplication process is crucial in verifying or discarding candidate solutions. Matrix-vector multiplication helps us solve systems efficiently and with precision.

Solution Verification

Solution verification is the final step in determining which, if any, of the candidate vectors we have are actual solutions to the system of equations. After performing the matrix-vector multiplication, we compare the resulting vector with the constant vector from our matrix equation.

• If they match, as with vector \(\begin{bmatrix} -1 \ 3 \end{bmatrix}\), then we can be confident that the vector is indeed a solution.
• When the results do not match, as in the case of vector \(\begin{bmatrix} 2 \ -6 \end{bmatrix}\), we can conclude it is not a solution.
• This method of verification not only provides an exact check but also allows for consistency across different systems of equations, no matter the complexity.

By systematically applying matrix calculations and comparing results, you can accurately verify solutions for any system of equations you encounter.