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Solve for \(x\). $$\left|\begin{array}{rrr} -1 & 0 & 2 \\ 0 & 5 & 3 \\ 0 & x & -2 \end{array}\right|=-2$$

Short Answer

Expert verified
The solution to the given equation is \(x = -4\).

Step by step solution

01

Calculate the determinant of the matrix

The determinant of a 3x3 matrix \(\left|\begin{array}{ccc}a & b & c \d & e & f \g & h & i \end{array}\right|\) can be calculated as follows: \(\text{det(A)}=a(ei−fh)−b(di−fg)+c(dh−eg)\). Here the values are a=-1, b=0, c=2, d=0, e=5, f=3, g=0, h=x, i=-2; Substituting these values gives \(-1*(5*-2 - 3*x) - 0 + 2*0\)
02

Simplify the equation

Simplify the obtained equation to isolate \(x\). The expression simplifies to \(10 + 3x = -2\)
03

Solve for \(x\)

Solve the equation for \(x\). Moving \(10\) to the right side we get \(3x = -2 - 10\), so \(3x = -12\), yielding \(x = -12/3 = -4\)

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Key Concepts

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

3x3 matrix
In mathematics, a 3x3 matrix is an arrangement consisting of three rows and three columns. Each element in this matrix can be a number, a variable, or a function. Such matrices are often used to solve systems of equations and to perform various transformations in space. Understanding their determinant is crucial, as it gives insight into the properties of the matrix.

The determinant of a 3x3 matrix is a special number calculated from its elements. It helps determine if the matrix has an inverse, or if the corresponding system of equations has a unique solution, no solution, or infinitely many solutions.
  • A 3x3 matrix, \ A = \(\begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\), has a determinant calculated as \(\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\).
  • This computation weighs the contributions of each element in the first row, adjusted by the determinants of smaller 2x2 matrices formed by the remaining rows and columns.
Understanding this process is vital since it lays the groundwork for solving complex mathematical problems that involve matrices.
linear equations
Linear equations are mathematical statements that describe a straight line when plotted on a graph. They take the general form \(ax + by + cz = d\). In the context of matrices, each linear equation can be seen as a row of the matrix. Breaking down complex systems into a matrix form can make solving them more manageable.

When dealing with a system of linear equations, particularly with three variables, a 3x3 matrix can represent the coefficients of these variables. The determinant then plays a vital role in determining the solution set of the system.
  • If the determinant is zero, the equations do not have a unique solution.
  • If non-zero, the solution can be found using various methods like Cramer's Rule, which relies on determinants.
Linear equations solved using matrices lead to elegant solutions, primarily when represented as a matrix equation, \(AX = B\), where \(A\) is the matrix of coefficients, \(X\) is the column matrix of variables, and \(B\) represents the constants.
precalculus problem solving
Precalculus is an essential branch of mathematics that prepares students for calculus by offering tools to handle more advanced problems. Concepts like matrices and determinants fall under this umbrella and aid in the transition from purely algebraic methods to more sophisticated approaches used in calculus.

Solving problems using matrices in precalculus teaches students logical thinking and systematic problem-solving skills. Understanding how to calculate a determinant and solve for variables like \(x\) in the initial exercise, refines analytical skills and prepares students for the complexities of calculus.
  • The process of operating with matrices involves various techniques: calculating determinants, applying matrix multiplication, and finding inverses when applicable.
  • These foundational skills are utilized when delving into calculus topics such as differentiating and integrating multivariable functions.
Mastery of precalculus topics strengthens mathematical foundations, making the journey through calculus a smoother experience.

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Most popular questions from this chapter

An airline charges 380 dollar for a round-trip flight from New York to Los Angeles if the ticket is purchased at least 7 days in advance of travel. Otherwise, the price is 700 dollar . If a total of 80 tickets are purchased at a total cost of 39,040 dollar, find the number of tickets sold at each price.

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. The athletic director of a local high school is ordering equipment for spring sports. He needs to order twice as many baseballs as softballs. The total number of balls he must order is \(300 .\) How many of each type should he order?

$$\begin{aligned}&\text { If } A=\left[\begin{array}{ll}0 & 1 \\\a & 0\end{array}\right] \text { and } B=\left[\begin{array}{ll}0 & a \\\1 & 0\end{array}\right], \text { for what value(s) of }\\\&a \operatorname{does} A B=\left[\begin{array}{ll}1 & 0 \\\0 & 1\end{array}\right] ?\end{aligned}$$

If \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]\) and \(B=\left[\begin{array}{cc}2 & 2 a+b \\ b-a & 6\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=B A ?\)

You wish to make a 1 -pound blend of two types of coffee, Kona and Java. The Kona costs \(\$ 8\) per pound and the Java costs \(\$ 5\) per pound. The blend will sell for \(\$ 7\) per pound. (a) Let \(k\) and \(j\) denote the amounts (in pounds) of Kona and Java, respectively, that go into making a 1 -pound blend. One equation that must be satisfied by \(k\) and \(j\) is $$k+j=1$$ Both \(k\) and \(j\) must be between 0 and \(1 .\) Why? (b) Using the variables \(k\) and \(j\), write an equation that expresses the fact that the total cost of 1 pound of the blend will be \(\$ 7\) (c) Solve the system of equations from parts (a) and (b), and interpret your solution. (d) To make a 1 -pound blend of Kona and Java that costs \(\$ 7.50\) per pound, which type of coffee would you use more of? Explain without solving any equations.

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