Problem 47
Question
Complete the following. (A) Write the system in the form \(A X=B\). (B) Solve the system by finding \(A^{-1}\) and then using the equation \(\boldsymbol{X}=\boldsymbol{A}^{-1} \boldsymbol{B}\). (Hint: Some of your answers from Exercises \(15-28\) may be helpful.) $$ \begin{aligned} &x+2 y=3\\\ &x+3 y=6 \end{aligned} $$
Step-by-Step Solution
Verified Answer
The solution is \(x = -3\), \(y = 3\).
1Step 1: Write System in Matrix Form
First, write the given system of equations in matrix form as \(AX = B\). The system is: \(x + 2y = 3\) and \(x + 3y = 6\). This translates to:\[A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 3 \ 6 \end{bmatrix}\] Thus, the system is expressed as \(AX = B\).
2Step 2: Find the Inverse of Matrix A
Now, find the inverse of matrix \(A\). The formula for the inverse of a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is \(\frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). For \(A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}\), calculate the determinant: \( \text{det}(A) = 1 \cdot 3 - 2 \cdot 1 = 1 \). Since the determinant is 1, the inverse is:\[A^{-1} = \begin{bmatrix} 3 & -2 \ -1 & 1 \end{bmatrix}\].
3Step 3: Solve for X using A Inverse
Use the equation \(X = A^{-1}B\) to solve for \(X\). Multiply \(A^{-1}\) by \(B\): \(X = \begin{bmatrix} 3 & -2 \ -1 & 1 \end{bmatrix} \begin{bmatrix} 3 \ 6 \end{bmatrix}\)Perform the multiplication:\[X = \begin{bmatrix} 3\times 3 + (-2)\times 6 \ -1\times 3 + 1\times 6 \end{bmatrix} = \begin{bmatrix} 9 - 12 \ -3 + 6 \end{bmatrix} = \begin{bmatrix} -3 \ 3 \end{bmatrix}\]Thus, \(x = -3\) and \(y = 3\).
Key Concepts
Systems of Linear EquationsMatrix InversionDeterminants
Systems of Linear Equations
Understanding systems of linear equations is key to solving various mathematical problems. A system of linear equations consists of two or more equations, each representing a straight line, which can be solved together to find a common solution. In simple terms, it's about finding the point(s) where these lines intersect.
In our example, we have two equations: \(x + 2y = 3\) and \(x + 3y = 6\). When writing this system in matrix form, each equation represents a line in a two-dimensional space. The task is to find \(x\) and \(y\) such that both equations hold true simultaneously.
This can be represented in matrix terms as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. This representation allows us to use powerful matrix operations to find the solution.
In our example, we have two equations: \(x + 2y = 3\) and \(x + 3y = 6\). When writing this system in matrix form, each equation represents a line in a two-dimensional space. The task is to find \(x\) and \(y\) such that both equations hold true simultaneously.
This can be represented in matrix terms as \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix. This representation allows us to use powerful matrix operations to find the solution.
Matrix Inversion
Matrix inversion is a crucial concept for solving systems of equations that are expressed in matrix form. If a matrix \(A\) is invertible, we can find a matrix called \(A^{-1}\), such that when it is multiplied by \(A\), it yields the identity matrix. This inverted matrix helps in finding the solution of the equation \(AX = B\) by transforming it into \(X = A^{-1}B\).
For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), its inverse can be calculated if the determinant \( \text{det}(A) = ad-bc eq 0 \). The formula to find the inverse is:\[ A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]
For our matrix \(A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}\), the determinant is 1, which is non-zero, making \(A\) invertible. By applying the formula, we find \(A^{-1} = \begin{bmatrix} 3 & -2 \ -1 & 1 \end{bmatrix}\). Matrix inversion transforms the problem into a simple multiplication, bringing us closer to the solution.
For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), its inverse can be calculated if the determinant \( \text{det}(A) = ad-bc eq 0 \). The formula to find the inverse is:\[ A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]
For our matrix \(A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}\), the determinant is 1, which is non-zero, making \(A\) invertible. By applying the formula, we find \(A^{-1} = \begin{bmatrix} 3 & -2 \ -1 & 1 \end{bmatrix}\). Matrix inversion transforms the problem into a simple multiplication, bringing us closer to the solution.
Determinants
Determinants are mathematical expressions that provide important information about a matrix, such as whether a matrix is invertible. For a 2x2 matrix, the determinant is calculated as \(ad-bc\). If this value is zero, the matrix cannot be inverted, meaning the system of equations may not have a unique solution.
In our problem, we calculated the determinant of matrix \(A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}\) as 1. The fact that the determinant is non-zero confirms that \(A\) is invertible, allowing us to use its inverse to solve the system of equations.
Determinants are invaluable for understanding the nature of a matrix and ensuring that calculations such as inversion are valid. Always verify the determinant before attempting to invert a matrix in any problem involving matrix equations, as this will guide whether the matrix inversion approach can be applied.
In our problem, we calculated the determinant of matrix \(A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}\) as 1. The fact that the determinant is non-zero confirms that \(A\) is invertible, allowing us to use its inverse to solve the system of equations.
Determinants are invaluable for understanding the nature of a matrix and ensuring that calculations such as inversion are valid. Always verify the determinant before attempting to invert a matrix in any problem involving matrix equations, as this will guide whether the matrix inversion approach can be applied.
Other exercises in this chapter
Problem 46
If possible, solve the system of linear equations and check your answer. $$ \begin{aligned} &3 x+6 y=0\\\ &4 x-2 y=-5 \end{aligned} $$
View solution Problem 47
If possible, find \(A B\) and \(B A\). $$A=\left[\begin{array}{rrr}1 & 0 & -2 \\\3 & -4 & 1 \\\2 & 0 & 5\end{array}\right], \quad B=\left[\begin{array}{r}1 \\\\
View solution Problem 47
Solve the system, if possible. $$ \begin{array}{r} 5 x & +4 z=7 \\ 2 x-4 y=6 \\ 3 y+3 z=3 \end{array} $$
View solution Problem 47
If possible, solve the system of linear equations and check your answer. $$ \begin{aligned} \frac{1}{2} x-y &=-5 \\ x+\frac{1}{2} y &=10 \end{aligned} $$
View solution