Problem 24 - Algebra and Trigonometry

Question

Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example 6. Use the inverses from Exercises 7–10, 15, 16, 19, and 21. \(\left\\{\begin{array}{l}{3 x+4 y=10} \\ {7 x+9 y=20}\end{array}\right.\)

Step-by-Step Solution

Answer

The solution is \( x = 10 \) and \( y = -10 \).

Step 1: Write the System as a Matrix Equation

The system of equations can be expressed in the form of a matrix equation: \( A\mathbf{x} = \mathbf{b} \).Here, \( A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \), \( \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix} \), and \( \mathbf{b} = \begin{bmatrix} 10 \ 20 \end{bmatrix} \). So the matrix equation is:\[ \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 10 \ 20 \end{bmatrix} \]

Step 2: Find the Inverse of the Coefficient Matrix

The inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \].For \( A = \begin{bmatrix} 3 & 4 \ 7 & 9 \end{bmatrix} \), compute:- Determinant: \( (3)(9) - (4)(7) = 27 - 28 = -1 \)- Thus,\[ A^{-1} = \begin{bmatrix} 9 & -4 \ -7 & 3 \end{bmatrix} \]

Step 3: Multiply by the Inverse

Multiply the inverse of the coefficient matrix \( A^{-1} \) by \( \mathbf{b} \) to solve for \( \mathbf{x} \):\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} 9 & -4 \ -7 & 3 \end{bmatrix} \begin{bmatrix} 10 \ 20 \end{bmatrix} \]Perform multiplying:\[ \begin{bmatrix} (9)(10) + (-4)(20) \ (-7)(10) + (3)(20) \end{bmatrix} = \begin{bmatrix} 90 - 80 \ -70 + 60 \end{bmatrix} = \begin{bmatrix} 10 \ -10 \end{bmatrix} \]

Step 4: Write Solution for Variables

From the matrix \( \mathbf{x} = \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 10 \ -10 \end{bmatrix} \), we conclude:- \( x = 10 \)- \( y = -10 \)

Key Concepts

Inverse of a Matrix

The inverse of a matrix is akin to the reciprocal of a number. For a number, multiplying it by its reciprocal results in one, such as: \( 5 \times \frac{1}{5} = 1 \). In matrix terms, if \( A \) is your matrix and \( A^{-1} \) is its inverse, then their product \( AA^{-1} \) results in the identity matrix, which acts like "1".
To find the inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), we follow a formula: swap \( a \) and \( d \), negate \( b \) and \( c \), and divide by the determinant \( ad-bc \). This gives us:

• Inverse: \( A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \)

Remember, if the determinant is zero, the inverse does not exist. Having this tool allows us to solve matrix equations by transforming them into something manageable.

System of Equations

Solving systems of equations is a core skill in algebra, letting us find unknown values satisfying multiple conditions. These systems can appear in different ways: linear, where you might see straight-line graphs, or otherwise more complex cases.
For each unknown, you typically need at least one equation to solve a system accurately. Representing a system in matrix format provides a powerful method, simplifying the complexity by working with numbers in grid form. In our exercise, the system of equations is expressed as a matrix equation, \( A\mathbf{x} = \mathbf{b} \), where each element in the matrix corresponds to coefficients in the system. Here, the goal is to see it mathematically, solve it practically, and understand how all parts interlink.Using matrices to solve a system can help visualize relations between equations and often provide computational efficiency.

Determinant of a Matrix

The determinant is a special value that helps us understand certain properties of a square matrix. It's particularly useful in finding the inverse of a matrix, as previously mentioned.
For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as:

• \( \det(A) = ad - bc \)

This snippet of mathematics shows its true worth when ensuring that a matrix can be inverted. If the determinant is not zero, the inverse exists. But if the determinant is zero, the matrix is singular and can't be inverted.
By determining \( ad-bc \), we unlock a simple check to see if the matrix can help with solving equations efficiently or needs an alternative method.

Matrix Multiplication

Matrix multiplication might look complex at first glance, but when broken down, it’s quite systematic. It allows us to transform matrices and calculate new outcomes.
When we multiply two matrices together, we take each row from the first matrix and multiply it by each column in the second matrix, summing the result for positioning in the resulting matrix.
For instance, to multiply a 2x2 matrix by a 2x1 matrix:

• \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \times \begin{bmatrix} e \ f \end{bmatrix} = \begin{bmatrix} ae+bf \ ce+df \end{bmatrix} \)

In the context of solving systems of equations with matrices, multiplication is the step where we find the solution by applying the inverse matrix to the constants matrix. This gives us the final values. Remember, alignment regarding dimensions is crucial when performing matrix multiplication to ensure valid results.