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Chapter 1: Problem 8

Let \(A=\left[\begin{array}{cc}1 & \alpha \\ \alpha & 3 \alpha\end{array}\right]\). a. For which numbers \(\alpha\) will \(A\) be singular? b. For all numbers \(\alpha\) not on your list in part \(a\), we can solve \(A \mathbf{x}=\mathbf{b}\) for every vector \(\mathbf{b} \in \mathbb{R}^{2}\). For each of the numbers \(\alpha\) on your list, give the vectors \(\mathbf{b}\) for which we can solve \(A \mathbf{x}=\mathbf{b}\).

Short Answer

Expert verified
The matrix A is singular for \(\alpha=0\) and \(\alpha=3\). When \(\alpha=0\), we can solve the system \(A\mathbf{x}=\mathbf{b}\) for any vector \(\mathbf{b}\) with \(b_2=0\). When \(\alpha=3\), we can solve the system for any vector \(\mathbf{b}\) with \(b_2=3b_1\).

Step by step solution

01

Find the determinant of A

To find the determinant of the given 2x2 matrix A, use the formula: \(\det(A)=ad - bc\) Here, a=1, b=\(\alpha\), c=\(\alpha\), and d=3\(\alpha\). So, \(\det(A)=(1)(3\alpha) - (\alpha)(\alpha) = 3\alpha - \alpha^2\)
02

Find values of \(\alpha\) that make A singular

A matrix is singular if its determinant is equal to zero. So, equate the determinant to zero and solve for \(\alpha\): \(3\alpha - \alpha^2 = 0\) Factor out \(\alpha\): \(\alpha(3-\alpha) = 0\) Thus, the two values of \(\alpha\) that make A singular are \(\alpha = 0\) and \(\alpha = 3\).
03

Solve \(A\mathbf{x}=\mathbf{b}\) for each singular \(\alpha\)

Now we'll find the vectors \(\mathbf{b}\) for which we can solve the system \(A\mathbf{x}=\mathbf{b}\) when \(\alpha\) is equal to 0 or 3 (the singular values). a) \(\alpha = 0\): The matrix A becomes: \(A = \left[\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right]\) Write down the augmented matrix for the system: \(\left[\begin{array}{cc|c} 1 & 0 & b_1 \\ 0 & 0 & b_2 \end{array}\right]\) For this system to be consistent, we need: \(b_2 = 0\) The vectors \(\mathbf{b}\) for which we can solve \(A\mathbf{x}=\mathbf{b}\) when \(\alpha = 0\) are given by: \(\mathbf{b} = \begin{bmatrix} b_1 \\ 0 \end{bmatrix}\), for any value of \(b_1\). b) \(\alpha = 3\): The matrix A becomes: \(A = \left[\begin{array}{cc} 1 & 3 \\ 3 & 9 \end{array}\right]\) Write down the augmented matrix for the system: \(\left[\begin{array}{cc|c} 1 & 3 & b_1 \\ 3 & 9 & b_2 \end{array}\right]\) Use Row Echelon form: 1. Multiply first row by 3 and subtract it from the second row: \(\left[\begin{array}{cc|c} 1 & 3 & b_1 \\ 0 & 0 & b_2 - 3b_1 \end{array}\right]\) For this system to be consistent, we need: \(b_2 - 3b_1 = 0 \Rightarrow b_2 = 3b_1\) The vectors \(\mathbf{b}\) for which we can solve \(A\mathbf{x}=\mathbf{b}\) when \(\alpha = 3\) are given by: \(\mathbf{b} = \begin{bmatrix} b_1 \\ 3b_1 \end{bmatrix}\), for any value of \(b_1\). In conclusion, the matrix A is singular for \(\alpha=0\) and \(\alpha=3\). When \(\alpha=0\), we can solve the system \(A\mathbf{x}=\mathbf{b}\) for any vector \(\mathbf{b}\) with \(b_2=0\). When \(\alpha=3\), we can solve the system for any vector \(\mathbf{b}\) with \(b_2=3b_1\).

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Key Concepts

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

Matrix Determinant
The determinant of a matrix is a special number that can help us understand some important properties of the matrix. For a 2x2 matrix, like the one in our example, the determinant can be calculated using the formula:
  • \( ext{det}(A) = ad - bc \)
The letters \( a, b, c, \) and \( d \) refer to the specific entries in the matrix. Calculating the determinant tells us if a matrix is singular. A singular matrix has a determinant of zero. When a matrix is singular, it means we cannot uniquely solve the system of linear equations related to this matrix for every possible input vector. In simpler terms, if you end up with a determinant of zero, your matrix is not invertible. It doesn't have all the information needed to find a unique solution for certain problems.
Systems of Linear Equations
Systems of linear equations involve finding values for variables that satisfy multiple equations at the same time. Imagine trying to find a point where two lines meet on a graph. That's what solving a system of equations is often like. A matrix can represent a system of linear equations, with each row corresponding to an equation and each column to a variable. Here's where it gets interesting: the singularity of a matrix (when the determinant is zero) affects the system of equations it represents. If the determinant is zero, some of the equations might be redundant or dependent on each other. This can make solutions impossible to find uniquely, or sometimes at all. When solving systems, we note the conditions of vectors \( \mathbf{b} \) that allow solutions in cases where the determinant equals zero. For different values of \( \alpha \), the vector \( \mathbf{b} \) may need to meet certain criteria to make the system solvable.
Row Echelon Form
Row echelon form is a special way to simplify matrices to make solving systems of linear equations easier. When you have a matrix, you can perform row operations to transform it into a row echelon form. This involves using three basic operations: swapping rows, multiplying a row by a non-zero constant, and adding or subtracting rows. The goal is to get the matrix to a point where it's easier to see what solutions might exist.
  • In row echelon form, any rows with all zeros are at the bottom.
  • The leading entry in each non-zero row is 1, and it appears to the right of the leading entry in the previous row.
Converting a matrix to row echelon form helps us solve the system of equations by making it easier to apply methods like back substitution. Given this simplification, it's straightforward to determine which vectors \( \mathbf{b} \) can satisfy the equation when the matrix is singular, provided certain conditions are met.

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Most popular questions from this chapter

Suppose a living organism that can live to a maximum age of 3 years has Leslie matrix $$ A=\left[\begin{array}{ccc} 0 & 0 & 8 \\ \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{4} & 0 \end{array}\right] $$ Find a stable age distribution vector \(\mathbf{x}\), i.e., a vector \(\mathbf{x} \in \mathbb{R}^{3}\) with \(A \mathbf{x}=\mathbf{x}\).

Find a normal vector to the given hyperplane and use it to find the distance from the origin to the hyperplane. a. \(\mathbf{x}=(-1,2)+t(3,2)\) b. The plane in \(\mathbb{R}^{3}\) given by the equation \(2 x_{1}+x_{2}-x_{3}=5\) "c. The plane passing through \((1,2,2)\) and orthogonal to the line \(\mathbf{x}=(3,1,-1)+\) \(t(-1,1,-1)\) d. The plane passing through \((2,-1,1)\) and orthogonal to the line \(\mathbf{x}=(3,1,1)+\) \(t(-1,2,1)\) *e. The plane spanned by \((1,1,4)\) and \((2,1,0)\) and passing through \((1,1,2)\) f. The plane spanned by \((1,1,1)\) and \((2,1,0)\) and passing through \((3,0,2)\) g. The hyperplane in \(\mathbb{R}^{4}\) spanned by \((1,-1,1,-1),(1,1,-1,-1)\), and \((1,-1,-1,1)\) and passing through \((2,1,0,1)\)

The equation \(2 x_{1}-3 x_{2}=5\) defines a line in \(\mathbb{R}^{2}\). a. Give a normal vector a to the line. b. Find the distance from the origin to the line by using projection. c. Find the point on the line closest to the origin by using the parametric equation of the line through \(\mathbf{0}\) with direction vector a. Double- check your answer to part \(b\). d. Find the distance from the point \(\mathbf{w}=(3,1)\) to the line by using projection. e. Find the point on the line closest to \(w\) by using the parametric equation of the line through \(\mathbf{w}\) with direction vector a. Double-check your answer to part \(d\).

Suppose \(\mathbf{x}=\mathbf{x}_{0}+t \mathbf{v}\) and \(\mathbf{y}=\mathbf{y}_{0}+s \mathbf{w}\) are two parametric representations of the same line \(\ell\) in \(\mathbb{R}^{n}\). a. Show that there is a scalar \(t_{0}\) so that \(\mathbf{y}_{0}=\mathbf{x}_{0}+t_{0} \mathbf{v}\). b. Show that \(\mathbf{v}\) and \(\mathbf{w}\) are parallel.

One might need to find solutions of \(A \mathbf{x}=\mathbf{b}\) for several different \(\mathbf{b}\) 's, say \(\mathbf{b}_{1}, \ldots, \mathbf{b}_{k}\). In this event, one can augment the matrix \(A\) with all the b's simultaneously, forming the "multi-augmented" matrix \(\left[A \mid \mathbf{b}_{1} \mathbf{b}_{2} \cdots \mathbf{b}_{k}\right]\). One can then read off the various solutions from the reduced echelon form of the multi-augmented matrix. Use this method to solve \(A \mathbf{x}=\mathbf{b}_{j}\) for the given matrices \(A\) and vectors \(\mathbf{b}_{j}\). a. \(A=\left[\begin{array}{rrr}1 & 0 & -1 \\ 2 & 1 & -1 \\ -1 & 2 & 2\end{array}\right], \quad \mathbf{b}_{1}=\left[\begin{array}{r}-1 \\ 1 \\\ 5\end{array}\right], \quad \mathbf{b}_{2}=\left[\begin{array}{l}1 \\ 3 \\\ 2\end{array}\right]\) b. \(A=\left[\begin{array}{rrrr}1 & 2 & -1 & 0 \\ 2 & 3 & 2 & 1\end{array}\right], \quad \mathbf{b}_{1}=\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad \mathbf{b}_{2}=\left[\begin{array}{l}0 \\\ 1\end{array}\right]\) c. \(A=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 1\end{array}\right], \quad \mathbf{b}_{1}=\left[\begin{array}{l}1 \\ 0 \\\ 0\end{array}\right], \quad \mathbf{b}_{2}=\left[\begin{array}{l}0 \\ 1 \\\ 0\end{array}\right], \quad \mathbf{b}_{3}=\left[\begin{array}{l}0 \\ 0 \\\ 1\end{array}\right]\)
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