Chapter 8: Problem 33
Use matrix inversion to solve the system of equations. $$\left\\{\begin{array}{r}2 x-5 y=-7 \\\\-3 x+2 y=-6\end{array}\right.$$
Short Answer
Expert verified
\The solution to the given system of equations is x = 4 and y = -47/19
Step by step solution
01
Write the Coefficient Matrix and the Constants Matrix
The first step involves expressing the given system of equations in a matrix form. In the given system of equations, the coefficient matrix A = \[\[2, -5\], [-3, 2\]\] and the constants matrix b = \[-7, -6\]. Therefore, the system of equations can be written as Ax = b.
02
Compute the Inverse Matrix of A
The next step is to compute the inverse of matrix A if it exists. The inverse matrix A^-1 is given by 1/det(A) \* adj(A), where det(A) is the determinant of the matrix A, and adj(A) is the adjoint of A. For the given coefficient matrix A, its determinant is (2 * 2) - (5 * -3) = 4 + 15 = 19, so det(A) = 19. The adjoint matrix adj(A) is obtained by swapping the values of the main diagonal elements and changing the signs of the off-diagonal elements. Therefore, the adjoint of A, adj(A) = \[2, 3\], [5, 2]. Then, the inverse of A, A^-1 = 1/19 * \[2, 3\], [5, 2] is calculated.
03
Solve for x
After the inverse of A is computed, the solution to the system of equations can be found by multiplying A^-1 by b, so x = A^-1b. Therefore, the solution of the system of equations is x1 = 1/19 * (-14 + 18) = 4 and x2 = 1/19 * (-35 -12) = -47/19.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coefficient Matrix
The coefficient matrix plays a pivotal role in solving systems of linear equations using matrix inversion. Imagine it as the skeleton that holds all the coefficients from the algebraic expressions neatly in a grid where each row represents an equation and each column stands for a variable. For our system of equations, we have two equations and two variables, creating a 2x2 matrix.
Specifically, in the equation set \
2x - 5y = -7\
-3x + 2y = -6,
the coefficient matrix is represented as \[ A = \begin{bmatrix} 2 & -5 \ -3 & 2 \end{bmatrix} \].
The power of the coefficient matrix lies in its potential to simplify complex systems into a compact matrix form, facilitating operations like multiplication or inversion, which are central to solving the puzzle of what x and y could be.
Specifically, in the equation set \
2x - 5y = -7\
-3x + 2y = -6,
the coefficient matrix is represented as \[ A = \begin{bmatrix} 2 & -5 \ -3 & 2 \end{bmatrix} \].
The power of the coefficient matrix lies in its potential to simplify complex systems into a compact matrix form, facilitating operations like multiplication or inversion, which are central to solving the puzzle of what x and y could be.
Determinant
The determinant is a scalar value that is a property of square matrices, like our 2x2 coefficient matrix A. It serves as a multifaceted tool capable of telling us crucial information about the matrix. One of the key aspects of a determinant is that it can indicate if a matrix has an inverse — if the determinant is non-zero, the matrix is invertible.
To find the determinant of matrix A, we use the formula \[ \text{det}(A) = a_{11}a_{22} - a_{12}a_{21} \], where \( a_{ij} \) is the element in the ith row and jth column of A. For our matrix A, the determinant is \[ \text{det}(A) = (2)(2) - (-5)(-3) = 19 \], signifying that A is indeed invertible, and that we can proceed with the steps to solve the system using matrix inversion.
To find the determinant of matrix A, we use the formula \[ \text{det}(A) = a_{11}a_{22} - a_{12}a_{21} \], where \( a_{ij} \) is the element in the ith row and jth column of A. For our matrix A, the determinant is \[ \text{det}(A) = (2)(2) - (-5)(-3) = 19 \], signifying that A is indeed invertible, and that we can proceed with the steps to solve the system using matrix inversion.
Adjoint of a Matrix
When we deal with the inversion of a matrix, the adjoint plays a fundamental role in the calculation. It's crafted from our coefficient matrix, but with a twist. We create it by transposing the cofactor matrix of A, which involves both the determinants of certain submatrices and a checkerboard pattern of signs.
In simpler terms for our 2x2 matrix A, the process is straightforward: we swap the elements along the main diagonal and then change the signs of the off-diagonal elements. This leaves us with the adjoint matrix of A, represented as \[ \text{adj}(A) = \begin{bmatrix} 2 & 3 \ 5 & 2 \end{bmatrix} \].
The adjoint reflects the unique properties of matrix A, and it steps in as a key ingredient when we combine it with the determinant to unlock the inverse of matrix A.
In simpler terms for our 2x2 matrix A, the process is straightforward: we swap the elements along the main diagonal and then change the signs of the off-diagonal elements. This leaves us with the adjoint matrix of A, represented as \[ \text{adj}(A) = \begin{bmatrix} 2 & 3 \ 5 & 2 \end{bmatrix} \].
The adjoint reflects the unique properties of matrix A, and it steps in as a key ingredient when we combine it with the determinant to unlock the inverse of matrix A.
Inverse of a Matrix
Unlocking the inverse of a matrix is like finding a master key. Once we have it, we can solve our system of equations in a very elegant way. The inverse matrix essentially reverses the effect of the original matrix A. The formula to find the inverse of A is \( A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \), provided that the determinant of A is not zero.
With our given A, we have already determined that the determinant is 19 and that the adjoint is \[ \begin{bmatrix} 2 & 3 \ 5 & 2 \end{bmatrix} \]. Hence, the inverse matrix A^-1 is calculated as \[ A^{-1} = \frac{1}{19} \begin{bmatrix} 2 & 3 \ 5 & 2 \end{bmatrix} \].
This crucial matrix is the final piece we need to resolve the system, as we can apply it to transform the equation Ax = b into x = A^{-1}b, giving us the solution for our variables x and y.
With our given A, we have already determined that the determinant is 19 and that the adjoint is \[ \begin{bmatrix} 2 & 3 \ 5 & 2 \end{bmatrix} \]. Hence, the inverse matrix A^-1 is calculated as \[ A^{-1} = \frac{1}{19} \begin{bmatrix} 2 & 3 \ 5 & 2 \end{bmatrix} \].
This crucial matrix is the final piece we need to resolve the system, as we can apply it to transform the equation Ax = b into x = A^{-1}b, giving us the solution for our variables x and y.