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Use matrix inversion to solve the system of equations. $$\left\\{\begin{aligned}3 x+7 y &=-11 \\\x+2 y &=-3\end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = -17\) and \(y = -14\).

Step by step solution

01

Write the system of equations in matrix form

Let \(A\) be the coefficient matrix obtained from the coefficients of \(x\) and \(y\), \(X\) be the columna vector \(\[x, y\]\), and \(B\) be the column vector \(\[-11, -3\]\) on the right-hand side of the equations. Then the system of equations can be written in matrix form as \(AX = B\), where \[ A=\begin{bmatrix} 3 & 7 \\ 1 & 2 \end{bmatrix} \], \[ X=\begin{bmatrix} x \\ y \end{bmatrix} \], \[ B=\begin{bmatrix} -11 \\ -3 \end{bmatrix} \].
02

Find the inverse of the matrix \(A\)

The inverse of a 2x2 matrix \[ A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \]can be found using the formula \[ A^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \].Using this formula we can find the inverse of matrix \(A\) as \[ A^{-1}=\frac{1}{(3*2)- (7*1)}\begin{bmatrix} 2 & -7 \\ -1 & 3 \end{bmatrix} =\begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}\].
03

Multiply the inverse of \(A\) with \(B\)

We can then find the solution vector \(X\) as \(X = A^{-1}B\). So, multiply the inverse of \(A\) obtained from the previous step with \(B\) we get,\[ X=\begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} -11 \\ -3 \end{bmatrix} =\begin{bmatrix} 1*(-11) + 2*(-3) \\ 1*(-11) + 1*(-3) \end{bmatrix} =\begin{bmatrix} -17 \\ -14 \end{bmatrix}\]. \n\n This implies that \(x = -17\) and \(y = -14\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient Matrix
In the realm of linear algebra, a system of linear equations can be succinctly represented using matrices. The coefficient matrix plays a crucial role in this context. It is essentially a matrix constructed from the coefficients of the variables in the equations. For example, consider the system of equations:
\[ \begin{align*} 3x + 7y &= -11, \ x + 2y &= -3 \end{align*} \]
The coefficient matrix here would be:
\[ A = \begin{bmatrix} 3 & 7 \ 1 & 2 \end{bmatrix} \]
Each row in the matrix corresponds to an equation, and each column corresponds to a variable. This matrix is pivotal because it captures the essence of the system. By analyzing the coefficient matrix, we can understand the structure of the system and, with the help of additional matrix operations, solve for the unknown variables.
Inverse of a Matrix
Understanding the inverse of a matrix is critical when solving systems of linear equations. The
Matrix Multiplication
Matrix multiplication is a key operation in linear algebra that combines two matrices to produce a third matrix. Unlike element-wise multiplication, matrix multiplication follows a rule where the number of columns in the first matrix must match the number of rows in the second matrix. To calculate the product, we take the dot product of the rows of the first matrix with the columns of the second matrix. For matrices \( A \) and \( B \), the product \( AB \) will have the same number of rows as \( A \) and the same number of columns as \( B \). This method is crucial when solving a system of equations with matrix inversion. After finding the inverse of the coefficient matrix, we multiply it with the constant matrix to find the solution. For instance:
\[ A^{-1}B = \begin{bmatrix} 1 & 2 \ 1 & 1 \end{bmatrix} \begin{bmatrix} -11 \ -3 \end{bmatrix} = \begin{bmatrix} -17 \ -14 \end{bmatrix} \]
In the example provided, we used matrix multiplication to calculate the product of the inverse of the coefficient matrix, \( A^{-1} \), and the constant matrix \( B \) to arrive at the solutions for \( x \) and \( y \).
System of Linear Equations
A system of linear equations is a collection of two or more linear equations with the same set of variables. Solving such a system means finding the values of the unknowns that satisfy all the equations simultaneously. The system
\[3x + 7y = -11,
x + 2y = -3 \]
can be represented in matrix form as \( AX = B \), where \( A \) is the coefficient matrix, \( X \) is a column vector of variables, and \( B \) is the constant column vector. By using matrix inversion, we can isolate \( X \) and calculate its values. This is done by multiplying both sides of the equation \( AX = B \) by the inverse of matrix \( A \), denoted as \( A^{-1} \). When we do this, the equation becomes \( A^{-1}AX = A^{-1}B \), and since \( A^{-1}A \) is the identity matrix, we are left with \( X = A^{-1}B \). Correctly applying this process helps us find the values of the unknowns that make up the solutions to the system of linear equations.

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