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Chapter 8: Problem 101

For what values of \(x\) does the inverse of \(A\) not exist, given \(A=\left[\begin{array}{ll}x & 6 \\ 3 & 2\end{array}\right] ?\)

Short Answer

Expert verified
The inverse of matrix \(A\) does not exist when \(x = 9\).

Step by step solution

01

Understand the Condition for Invertibility

To determine when the inverse of matrix \(A\) does not exist, we need to know when the determinant of \(A\) is zero. Invertibility is violated when the determinant equals zero, meaning no inverse exists.
02

Calculate the Determinant of Matrix A

Given matrix \(A = \begin{bmatrix} x & 6 \ 3 & 2 \end{bmatrix}\), the determinant is calculated as follows:\[\text{det}(A) = (x)(2) - (3)(6) = 2x - 18\]
03

Solve for Zero Determinant

To find the values of \(x\) for which the determinant is zero, set the determinant expression equal to zero:\[2x - 18 = 0\]Solve for \(x\):\[2x = 18\]\[x = 9\]
04

Conclusion from the Calculation

The determinant of \(A\) becomes zero when \(x = 9\). Therefore, the inverse of \(A\) does not exist when \(x = 9\).

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Key Concepts

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

Matrix Inverse
The concept of a matrix inverse is a crucial piece of understanding in linear algebra. If you have a square matrix, like matrix \( A \) in our exercise, you can find its inverse if it exists. The inverse of a matrix \( A \), denoted \( A^{-1} \), is essentially the matrix you multiply \( A \) by to get the identity matrix, \( I \). The identity matrix acts like the number "1" in matrix world; multiplying any matrix by the identity matrix leaves the original matrix unchanged. For matrix \( A \) to have an inverse, its determinant must be non-zero.

Why is the inverse important? It allows us to solve linear equations efficiently just like we solve simple algebraic equations. Knowing when a matrix has an inverse helps us identify when such a solution is possible. Understanding this can be a game changer when dealing with systems of linear equations as well as in many applied fields.
Linear Algebra
Linear algebra is the study of vectors, vector spaces, and linear transformations. Central to this field are matrices, which are used to represent linear transformations. When dealing with systems of linear equations, matrices provide a structured way to handle various transformations and solve these systems.

In our exercise, matrix \( A \) is used to represent a set of linear transformations. Each element in the matrix corresponds to specific coefficients in the linear expressions. The determinant of the matrix provides valuable information about these transformations, particularly related to invertibility.
  • Matrices can be thought of as a way to encode information.
  • Linear transformations can bring about rotations, scalings, and shearing in vector space.
  • The concept of span, basis, rank are related to linear algebra's application and influence on matrices.
Grasping linear algebra is essential for deeper studies in both mathematics and applied fields like engineering, computer science, and natural sciences.
Invertibility
Invertibility, often a key property in linear systems, determines whether a matrix has an inverse. When a matrix is invertible, it implies that every linear transformation encoded by the matrix is reversible, meaning that you can find a solution to the system of equations it represents.

The magic number here is the determinant: - If the determinant is zero, the matrix is non-invertible, often referred to as singular.- A non-zero determinant signals the matrix is invertible, or non-singular.For our specific exercise, matrix \( A \) is non-invertible when \( x = 9 \), because it makes the determinant zero (\( 2x - 18 = 0\)).

This concept of invertibility is especially important in real-world applications, such as cryptography, computer graphics, and many optimization problems where we need to solve linear systems quickly and reliably. Understanding when a matrix is invertible allows us to handle these applications more effectively.

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Most popular questions from this chapter

Involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will eventually reach a maximum height and then fall to the ground. The equation that relates the height \(h\) of a projectile \(t\) seconds after it is projected upward is given by $$h=\frac{1}{2} a t^{2}+v_{0} t+h_{0}$$ where \(a\) is the acceleration due to gravity, \(h_{0}\) is the initial height of the object at time \(t=0,\) and \(v_{0}\) is the initial velocity of the object at time \(t=0 .\) Note that a projectile follows the path of a parabola opening down, so \(a<0\). An object is thrown upward, and the table below depicts the height of the ball \(t\) seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. $$\begin{array}{|c|c|} \hline t \text { (seconos) } & \text { HeiGHT (FEET) } \\ \hline 1 & 34 \\ \hline 2 & 36 \\ \hline 3 & 6 \\ \hline \end{array}$$

Apply a graphing utility to perform the indicated matrix operations. $$A=\left[\begin{array}{rrrr}1 & 7 & 9 & 2 \\\\-3 & -6 & 15 & 11 \\\0 & 3 & 2 & 5 \\\9 & 8 & -4 & 1\end{array}\right]$$ Find \(A A^{-1}\)

Orange juice producers use three varieties of oranges: Hamlin, Valencia, and navel. They want to make a juice mixture to sell at \(\$ 3.00\) per gallon. The price per gallon of each variety of juice is \(\$ 2.50, \$ 3.40,\) and \(\$ 2.80,\) respectively. To maintain their quality standards, they use the same amount of Valencia and navel oranges. Determine the quantity of each juice used to produce 1 gallon of mixture.

Maximize the objective function \(z=2 x+y\) subject to the conditions, where \(a>2\) $$\begin{aligned}a x+y & \geq-a \\\\-a x+y & \leq a \\\a x+y & \leq a \\\\-a x+y & \geq-a\end{aligned}$$

Solve the system of linear equations. $$ \begin{array}{l} 4 x-6 y=0 \\ 4 x+6 y=4 \end{array} $$ Solution: Set up the determinants. $$ D=\left|\begin{array}{rr} 4 & -6 \\ 4 & 6 \end{array}\right|, D_{x}=\left|\begin{array}{rr} 0 & -6 \\ 4 & 6 \end{array}\right|, \text { and } D_{y}=\left|\begin{array}{ll} 4 & 0 \\ 4 & 4 \end{array}\right| $$ Evaluate the determinants. $$ \begin{array}{c} D=48, D_{x}=24, \text { and } D_{y}=16 \\ x=\frac{D}{D_{x}}=\frac{48}{24}=2 \text { and } y=\frac{D_{y}}{D}=\frac{48}{16}=3 \end{array} $$ Solve for \(x\) and \(y\) \(x=2, y=3\) is incorrect. What mistake was made?
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