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Solve for \(u, x, y\), and \(z\) in the given matrix equation. $$ \left[\begin{array}{rr} 1 & 2 \\ 3 & 4 \\ x & -1 \end{array}\right]-3\left[\begin{array}{rr} y-1 & 2 \\ 1 & 2 \\ 4 & 2 z+1 \end{array}\right]=2\left[\begin{array}{rr} -4 & -u \\ 0 & -1 \\ 4 & 4 \end{array}\right] $$

Short Answer

Expert verified
The values for the variables are: \(u = -4\), \(x = -4\), \(y = -\frac{7}{3}\), and \(z = 2\).

Step by step solution

01

Multiply the second (A) and third (B) matrices by their respective constants

Before performing any matrix operations, we need to multiply the second matrix by -3 and the third matrix by 2. $$ -3\left[\begin{array}{rr} y-1 & 2 \\\ 1 & 2 \\\ 4 & 2 z+1 \end{array}\right] = \left[\begin{array}{rr} -3(y-1) & -6 \\\ -3 & -6 \\\ -12 & -3(2 z+1) \end{array}\right] $$ and $$ 2\left[\begin{array}{rr} -4 & -u \\\ 0 & -1 \\\ 4 & 4 \end{array}\right] = \left[\begin{array}{rr} -8 & -2u \\\ 0 & -2 \\\ 8 & 8 \end{array}\right] $$
02

Subtract the right-hand side matrix from the left-hand side matrix

Now we need to subtract the right-hand side matrix from the left-hand side matrix, including the initial matrix with variable 'x'. $$ \left[\begin{array}{rr} 1 & 2 \\\ 3 & 4 \\\ x & -1 \end{array}\right] - \left[\begin{array}{rr} -3(y-1) & -6 \\\ -3 & -6 \\\ -12 & -3(2 z+1) \end{array}\right] = \left[\begin{array}{rr} -8 & -2u \\\ 0 & -2 \\\ 8 & 8 \end{array}\right] $$ After performing subtraction between the matrices we get the following: $$ \left[\begin{array}{rr} 1+3(y-1) & 2+6 \\\ 3+3 & 4+6 \\\ x-(-12) & -1-(-3(2 z+1)) \end{array}\right] = \left[\begin{array}{rr} -8 & -2u \\\ 0 & -2 \\\ 8 & 8 \end{array}\right] $$
03

Equate the corresponding entries and solve for the variables

To find the values of the variables, we equate the corresponding entries in the two matrices. We get the following equations: $$ 1+3(y-1) = -8 \\ 2+6 = -2u \\ 3+3 = 0 \\ 4+6 = -2 \\ x-(-12) = 8 \\ -1-(-3(2 z+1)) = 8 $$ Now we can solve each equation for the variable: $$ y = -\frac{7}{3} \\ u = -4 \\ x = -4 \\ z = 2 $$ So, the values for u, x, y, and z are \(u = -4, x = -4, y = -\frac{7}{3},\) and \(z = 2\).

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Key Concepts

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

Linear Algebra
Linear algebra is a foundational branch of mathematics that focuses on studying vectors, vector spaces, linear mappings, and systems of linear equations. In the context of our exercise, the application of linear algebra is evident through the use of matrices to represent and solve systems of linear equations.

Matrices are rectangular arrays of numbers, which can compactly represent complex linear systems. Each element in a matrix corresponds to a coefficient in a linear equation. The method of solving matrix equations, as seen in the exercise, involves performing operations such as matrix addition, subtraction, multiplication, and sometimes inversion to find the values of unknown variables that satisfy the equation. These operations carry many similarities to arithmetic but must follow rules specific to matrices.
Matrix Subtraction
Matrix subtraction, a critical operation in our exercise, is the process of taking two matrices of the same dimensions and subtracting the corresponding entries. This is similar to element-wise subtraction in two-dimensional arrays. It's essential to ensure that matrices are of the same size; otherwise, the operation is undefined.

Important conditions to remember when performing matrix subtraction include, maintaining the order of subtraction as it is not commutative; this means that A - B is not the same as B - A. Additionally, matrix subtraction, like addition, is associative, allowing grouping without changing the result: (A - B) - C = A - (B + C). In our case, we used matrix subtraction to combine the simplified matrices to form a series of linear equations that could be solved for the variables.
Systems of Equations
Systems of equations are collections of two or more equations with a common set of variables. In linear algebra, these systems can be elegantly represented using matrices. Solving a system of linear equations typically means finding the set of values for the variables that makes all the equations true simultaneously.

There are various methods to solve systems of equations, such as substitution, elimination, and matrix operations—including the use of inverse matrices when applicable. In our exercise, we treated the matrix equation as a system of linear equations. By equating the matrices on both sides after performing the necessary operations, we established a set of scalar equations corresponding to each element of the matrix. Solving these equations, we found the values for the variables u, x, y, and z. Students should understand that matrix operations like subtraction can simplify many cumbersome steps involved in solving a system, leading to a more efficient and organized method, especially when dealing with large-scale systems.

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Most popular questions from this chapter

Matrix \(A\) is an input-output matrix associated with an economy, and matrix \(D\) (units in millions of dollars) is a demand vector. In each problem,find the final outputs of each industry such that the demands of industry and the consumer sector are met. $$ A=\left[\begin{array}{lll} 0.2 & 0.4 & 0.1 \\ 0.3 & 0.2 & 0.1 \\ 0.1 & 0.2 & 0.2 \end{array}\right] \text { and } D=\left[\begin{array}{r} 6 \\ 8 \\ 10 \end{array}\right] $$

K & R Builders build three models of houses, \(M_{1}, M_{2}\), and \(M_{3}\), in three subdivisions I, II, and III located in three different areas of a city. The prices of the houses (in thousands of dollars) are given in matrix \(A\) : K\& R Builders has decided to raise the price of each house by \(3 \%\) next year. Write a matrix \(B\) giving the new prices of the houses.

Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{aligned} -x_{2}+x_{3} &=2 \\ 4 x_{1}-3 x_{2}+2 x_{3} &=16 \\ 3 x_{1}+2 x_{2}+x_{3} &=11 \end{aligned} $$

Let $$ A=\left[\begin{array}{rr} 6 & -4 \\ -4 & 3 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 3 & -5 \\ 4 & -7 \end{array}\right] $$ a. Find \(A B, A^{-1}\), and \(B^{-1}\). b. Show that \((A B)^{-1}=B^{-1} A^{-1}\).

Mr. and Mrs. Garcia have a total of $$\$ 100,000$$ to be invested in stocks, bonds, and a money market account. The stocks have a rate of return of \(12 \% /\) year, while the bonds and the money market account pay \(8 \% /\) year and \(4 \%\) year, respectively. The Garcias have stipulated that the amount invested in the money market account should be equal to the sum of \(20 \%\) of the amount invested in stocks and \(10 \%\) of the amount invested in bonds. How should the Garcias allocate their resources if they require an annual income of $$\$$ 10,000$ from their investments?

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