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(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. $$ \begin{array}{l} 3 x-2 y=b_{1} \\ 4 x+3 y=b_{2} \\ \text { where } & \text { (i) } b_{1}=-6, b_{2}=10 \\ \text { and } & \text { (ii) } b_{1}=3, b_{2}=-2 \end{array} $$

Short Answer

Expert verified
For the given systems of equations with cases (i) and (ii), we first wrote them as matrix equations in the form \(AX=B\). Then, we computed the inverse of the coefficient matrix, \(A^{-1}\). Finally, we solved the systems of equations by multiplying \(A^{-1}\) with \(B\). For case (i), the solution is \(x = 2\) and \(y = -2\). For case (ii), the solution is \(x = 1\) and \(y = -1\).

Step by step solution

01

Case (i) - Write the system of equations as a matrix equation

For this case, we are given \(b_1 = -6\) and \(b_2 = 10\). Write the system of equations as a matrix equation. \( AX = B \), with \( A = \begin{bmatrix} 3 & -2 \\ 4 & 3 \end{bmatrix} \), \(X = \begin{bmatrix} x \\ y \end{bmatrix} \), and \(B = \begin{bmatrix} -6 \\ 10 \end{bmatrix} \).
02

Case (i) - Compute the inverse of the coefficient matrix

Now compute the inverse of the coefficient matrix, \(A^{-1}\), by using the formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] where \(A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) and \(\text{det}(A)=ad-bc\). Calculate the determinant, \(\text{det}(A) = (3)(3)-(-2)(4) = 9+8=17\) Calculate the inverse: \(A^{-1} = \frac{1}{17} \begin{bmatrix} 3 & 2 \\ -4 & 3 \end{bmatrix}\)
03

Case (i) - Solve the system of equations

Now multiply the inverse of the coefficient matrix, \(A^{-1}\), with the matrix containing the constants, \(B\), to find the solution, \(X\). \[ X = A^{-1}B \] \(X = \frac{1}{17} \begin{bmatrix} 3 & 2 \\ -4 & 3 \end{bmatrix}\begin{bmatrix} -6 \\ 10 \end{bmatrix}\) \(X = \begin{bmatrix} 2 \\ -2 \end{bmatrix}\) So for case (i), the solution is \(x = 2\) and \(y = -2\).
04

Case (ii) - Write the system of equations as a matrix equation

For this case, we are given \(b_1 = 3\) and \(b_2 = -2\). Write the system of equations as a matrix equation. \( AX = B \), with \( A = \begin{bmatrix} 3 & -2 \\ 4 & 3 \end{bmatrix} \), \(X = \begin{bmatrix} x \\ y \end{bmatrix} \), and \(B = \begin{bmatrix} 3 \\ -2 \end{bmatrix} \).
05

Case (ii) - Compute the inverse of the coefficient matrix

The matrix \(A\) and its inverse \(A^{-1}\) are the same as in case (i), so we don't need to compute it again.
06

Case (ii) - Solve the system of equations

Multiply the inverse of the coefficient matrix, \(A^{-1}\), with the matrix containing the constants, \(B\), to find the solution, \(X\). \[ X = A^{-1}B \] \(X = \frac{1}{17} \begin{bmatrix} 3 & 2 \\ -4 & 3 \end{bmatrix}\begin{bmatrix} 3 \\ -2 \end{bmatrix}\) \(X = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\) So for case (ii), the solution is \(x = 1\) and \(y = -1\).

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Key Concepts

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

Systems of Linear Equations
Systems of linear equations are collections of two or more equations with two or more variables. Each equation provides a line, and the solution to the system is the intersection point of these lines. For example, in our exercise, the two equations are:
  • \( 3x - 2y = b_1 \)
  • \( 4x + 3y = b_2 \)
Here, \( x \) and \( y \) are the variables. More often, we need to solve these systems because they model real-world situations where different conditions must be satisfied simultaneously. The visual solution of a system can be represented where the two lines intersect in a coordinate plane. However, solving large systems manually can become complex. Luckily, using matrices helps simplify the solution process. Matrices systematically handle equations which supports finding a solution through computational methods.
Thus, when placed into a matrix form, the system becomes something that computers can process quickly, making them invaluable for solving problems with many variables.
Inverse of a Matrix
The inverse of a matrix is a crucial component in the matrix equation approach to solve systems of linear equations. An inverse matrix, often denoted as \( A^{-1} \), is a matrix that, when multiplied with the original matrix \( A \), yields the identity matrix. The identity matrix is a special square matrix with ones on the diagonal and zeros elsewhere, symbolically represented as \( I \). The fundamental property of an inverse matrix is:
  • \( A \times A^{-1} = I \)
In our exercise example, to find the solution \( X \), we compute \( X = A^{-1}B \) where \( B \) is a matrix containing constants from the system.
Finding an inverse involves using the formula:
  • \( A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \)
This method requires calculating the determinant of the matrix \( A \), and if the determinant is zero, the inverse does not exist.
Determinants
Determinants are values computed from square matrices and play a significant role in matrix algebra. They are used in determining whether a matrix has an inverse and play a fundamental role in solving systems of linear equations. The determinant of a matrix, often abbreviated as \( \text{det}(A) \), provides essential information about the matrix. For a 2x2 matrix, as in our exercise:
  • \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \)
  • The determinant is computed as: \( \text{det}(A) = ad-bc \)
If \( \text{det}(A) \) is zero, the matrix does not have an inverse, meaning the system of equations has no unique solution. This wasn't the case in our provided example where we calculated:
  • \( \text{det}(A) = 9+8=17 \)
A non-zero determinant allowed us to find an inverse and consequently solve the system of equations using the matrix equation approach. Thus, understanding determinants is key in verifying if a system can be uniquely solved.

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