Chapter 8: Problem 43
In Problems \(43-50\), use an inverse matrix to solve the given system of equations. $$ \begin{aligned} x_{1}+x_{2} &=4 \\ 2 x_{1}-x_{2} &=14 \end{aligned} $$
Short Answer
Expert verified
The solution is \(x_1 = 6\) and \(x_2 = -2\).
Step by step solution
01
Express the System as a Matrix Equation
The given system of equations can be expressed in matrix form as follows:\[\begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix} \begin{bmatrix} x_{1} \ x_{2} \end{bmatrix} = \begin{bmatrix} 4 \ 14 \end{bmatrix}\]Here, the coefficient matrix is \(A = \begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix}\), the variable matrix is \(X = \begin{bmatrix} x_{1} \ x_{2} \end{bmatrix}\), and the constant matrix is \(B = \begin{bmatrix} 4 \ 14 \end{bmatrix}\).
02
Find the Inverse of the Coefficient Matrix
First, we need to calculate the inverse of the coefficient matrix \(A\). The inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our matrix \(A = \begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix}\), we calculate:\[\det(A) = (1)(-1) - (1)(2) = -3\]Thus, the inverse is:\[A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 & -1 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}\]
03
Multiply the Inverse by the Constant Matrix
Now, to find \(X\), multiply the inverse matrix \(A^{-1}\) with the constant matrix \(B\):\[X = A^{-1}B = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \begin{bmatrix} 4 \ 14 \end{bmatrix}\]Carrying out the matrix multiplication, we get:\[X = \begin{bmatrix} \frac{1}{3}(4) + \frac{1}{3}(14) \ \frac{2}{3}(4) - \frac{1}{3}(14) \end{bmatrix} = \begin{bmatrix} \frac{18}{3} \ \frac{8}{3} - \frac{14}{3} \end{bmatrix} = \begin{bmatrix} 6 \ -2 \end{bmatrix}\]
04
Interpret the Solution
The result of the matrix multiplication gives the values for \(x_1\) and \(x_2\). Hence, the solution to the system of equations is:\[x_1 = 6, \quad x_2 = -2\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Matrix Equations
Matrix equations are a compact way to represent and solve linear systems of equations. For example, the system of equations given in the problem:\[ \begin{aligned} x_{1}+x_{2} &=4 \ 2 x_{1}-x_{2} &=14 \end{aligned} \]is transformed into a matrix equation: \[\begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix} \begin{bmatrix} x_{1} \ x_{2} \end{bmatrix} = \begin{bmatrix} 4 \ 14 \end{bmatrix}\]Here:- The **coefficient matrix** is the array \(A = \begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix} \).- The **variable matrix** \(X = \begin{bmatrix} x_{1} \ x_{2} \end{bmatrix} \) contains the unknowns.- The **constant matrix** \(B = \begin{bmatrix} 4 \ 14 \end{bmatrix} \) represents the constants from the original equations.The goal is to solve for \(X\) by manipulating the equation \(AX = B\). Understanding how matrix multiplication works is key to solving these equations effectively.
The Properties of a 2x2 Matrix
A 2x2 matrix is a simple square matrix that contains exactly two rows and two columns. It is often used in handling systems of two linear equations. For instance, in the given exercise, we have:\[ A = \begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix} \]Key points to remember about a 2x2 matrix include:- Two rows and two columns- Four elements in total- The position of elements matters significantly for calculationThis type of matrix can be inverted (if it is non-singular) which means we can find another matrix, called the inverse, that will help solve systems of equations using the equation \(A^{-1}AX = A^{-1}B = X\). Understanding these basics makes it easier to work with matrices to solve systems of equations.
Exploring the Determinant
The determinant of a matrix is a special value that can tell us if the matrix is invertible. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated as:\[\det(A) = ad - bc\]In our example, with matrix \(A = \begin{bmatrix} 1 & 1 \ 2 & -1 \end{bmatrix}\), the determinant is:\[\det(A) = (1)(-1) - (1)(2) = -3\]Why is the determinant important?- It tells us whether a matrix is invertible (non-zero determinant)- Influences the formula for the inverse of a 2x2 matrixIf \(\det(A)\) is zero, the matrix does not have an inverse and cannot be used to solve systems of equations via matrix methods. Thus, calculating the determinant is a critical step in confirming that a solution path exists.
Solving Systems of Equations Using Matrices
Solving a system of linear equations with matrices relies heavily on understanding and applying inverse matrices. The primary equation used is:\[ AX = B \]where \(A\) is a known matrix, \(X\) is the matrix of variables, and \(B\) is the matrix of constants. To find \(X\), we use:\[ X = A^{-1}B \]Let’s look at an example from our solution:- As found, \(A^{-1} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \ \frac{2}{3} & -\frac{1}{3} \end{bmatrix} \)- The multiplication \(A^{-1}B = \begin{bmatrix} 6 \ -2 \end{bmatrix} \)This means the values of the variables are \(x_1 = 6\) and \(x_2 = -2\). Understanding matrix operations simplifies complex equations and provides an efficient method for finding solutions.