Problem 2 - Algebra and Trigonometry

Question

(a) Write the following system as a matrix equation \(A X=B\) System $$5 x+3 y=4$$ $$3 x+2 y=3$$ Matrix equation $$A \quad X=B$$ (b) The inverse of \(A\) is \(A^{-1}=\)(c) The solution of the matrix equation is \(X=A^{-1} B\) . $$ X=A^{-1} \quad B $$ $$\left[\begin{array}{l}{x} \\ {y}\end{array}\right]=$$ (d) The solution of the system is \(x=\) _______, \(y=\) ______

Step-by-Step Solution

Answer

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

Step 1: Write the system as a matrix equation

Let's express the given system of equations as a matrix equation.Given the system:\[\begin{align*}5x + 3y &= 4 \3x + 2y &= 3\end{align*}\]We can rewrite this as a matrix equation:\[ A \begin{bmatrix} x \ y \end{bmatrix} = B \]where \( A = \begin{bmatrix} 5 & 3 \ 3 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 4 \ 3 \end{bmatrix} \).

Step 2: Determine the inverse of matrix A

To find the inverse of matrix \( A = \begin{bmatrix} 5 & 3 \ 3 & 2 \end{bmatrix} \), we use the formula for the inverse of a 2x2 matrix:\[A^{-1} = \frac{1}{det(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]where \( det(A) = ad - bc \).For matrix \( A \), we have:\[det(A) = (5)(2) - (3)(3) = 10 - 9 = 1\]So, the inverse is:\[ A^{-1} = \begin{bmatrix} 2 & -3 \ -3 & 5 \end{bmatrix} \].

Step 3: Solve the matrix equation for X

Now, we solve for \( X \) by multiplying the inverse of \( A \) with \( B \):\[X = A^{-1} B = \begin{bmatrix} 2 & -3 \ -3 & 5 \end{bmatrix} \begin{bmatrix} 4 \ 3 \end{bmatrix}\]Calculating the multiplication:\[\begin{bmatrix} 2 & -3 \ -3 & 5 \end{bmatrix} \begin{bmatrix} 4 \ 3 \end{bmatrix} = \begin{bmatrix} (2)(4) + (-3)(3) \ (-3)(4) + (5)(3) \end{bmatrix} = \begin{bmatrix} 8 - 9 \ -12 + 15 \end{bmatrix} = \begin{bmatrix} -1 \ 3 \end{bmatrix}\]Thus, \( X = \begin{bmatrix} -1 \ 3 \end{bmatrix} \).

Step 4: Write the solution of the system

The vector \( X = \begin{bmatrix} -1 \ 3 \end{bmatrix} \) represents the solution of the system, where \( x = -1 \) and \( y = 3 \). Thus, the solution to the system of equations is \( x = -1 \) and \( y = 3 \).

Key Concepts

Inverse Matrix

Understanding the inverse of a matrix is crucial when solving matrix equations. An inverse matrix is like the "opposite" of a given square matrix that, when multiplied with the original matrix, yields the identity matrix. If matrix A has an inverse, it's denoted as \(A^{-1}\), and it satisfies the equation \(A \times A^{-1} = I\), where \(I\) is the identity matrix. The identity matrix has ones on the diagonal and zeros elsewhere.
To compute the inverse of a 2x2 matrix, use the formula:

• Given a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \)
• The determinant \(det(A) = ad - bc\)
• The inverse \(A^{-1} = \frac{1}{det(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \)

It's crucial to note, if the determinant is zero, the matrix does not have an inverse. In our example, the determinant was 1, allowing us to find the inverse of matrix \(A\) as \(\begin{bmatrix} 2 & -3 \ -3 & 5 \end{bmatrix}\).

Systems of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. These systems can often be solved using matrices, which provide a structured way to find the values of the unknowns.
For example, consider the system of linear equations:

• \(5x + 3y = 4\)
• \(3x + 2y = 3\)

We aim to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Using matrices, these equations can be expressed as a single matrix equation \(AX = B\), where \(A\) represents the coefficients of the variables, \(X\) is a column matrix of the variables, and \(B\) is a column matrix of the equations' constants. Solving this matrix equation involves finding the matrix \(X\) that satisfies the equation, which leads us to compute the inverse of \(A\) if possible.

2x2 Matrix

A 2x2 matrix is a simple yet fundamental structure in linear algebra comprised of two rows and two columns. Each entry in the matrix is typically a number or a variable. For instance, a typical 2x2 matrix looks like this:\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]In our problem, the matrix \(A\) of system coefficients is depicted as \(\begin{bmatrix} 5 & 3 \ 3 & 2 \end{bmatrix}\). This matrix compactly represents the system of equations, taking the "x" coefficients from the first column and the "y" coefficients from the second.
Determining the inverse of a 2x2 matrix is manageable due to its simple structure, especially when compared to larger matrices. For solving equations, we usually represent the coefficients in a 2x2 matrix, find its inverse if its determinant is non-zero, and then solve the system using this inverse. These principles are foundational for solving algebraic problems efficiently using matrices.