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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.
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.
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.
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.

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Most popular questions from this chapter

Involve the use of matrix multiplication to transform one or more points. This technique, which can be applied to any set of points, is used extensively in computer graphics. Let \(A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{r}2 \\ -1\end{array}\right]\) (a) Calculate the product matrix \(A B\) (b) On a single coordinate system, plot the point (2,-1) and the point whose coordinates \((x, y)\) are the entries of the product matrix found in part (a). Explain geometrically what the matrix multiplication did to the point (2,-1) (c) How would you undo the multiplication in part (a)?

A financial advisor offers three specific investment instruments: a stock- based mutual fund, a high-yield bond, and a certificate of deposit (CD). Risk factors for individual instruments can be quantified on a scale of 1 to \(5,\) with 1 being the most risky. The risk factors associated with these particular instruments are summarized in the following table.$$\begin{array}{lc} \text { Type of Investment } & \text { Risk Factor } \\ \text { Stock-based mutual fund } & 3 \\\\\text { High-yield bond } & 1 \\\\\text { CD } & 5\end{array}$$.One of the advisor's clients can tolerate an overall risk level of \(3.5 .\) In addition, the client stipulates that the amount of money invested in the mutual fund must equal the sum of the amounts invested in the high-yield bond and the CD. To satisfy the client's requirements, what percentage of the total investment should be allocated to each instrument?

Consider the following system of equations. $$\left\\{\begin{aligned} x^{2}+y^{2} &=r^{2} \\ (x-h)^{2}+y^{2} &=r^{2} \end{aligned}\right.$$ Let \(r\) be a (fixed) positive number. For what value(s) of \(h\) does this system have (a) exactly one real solution? (b) exactly two real solutions? (c) infinitely many real solutions? (d) no real solution? (Hint: Visualize the graphs of the two equations.)

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. An electronics store carries two brands of video cameras. For a certain week, the number of Brand A video cameras sold was 10 less than twice the number of Brand B cameras sold. Brand A cameras cost \(\$ 200\) and Brand \(B\) cameras cost \(\$ 350 .\) If the total revenue generated that week from the sale of both types of cameras was \(\$ 16,750,\) how many of each type were sold?

Apply elementary row operations to a matrix to solve the system of equations. If there is no solution, state that the system is inconsistent. $$\left\\{\begin{array}{r}3 x+4 y-8 z=10 \\ -6 x-8 y+16 z=20\end{array}\right.$$

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