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Either compute the inverse of the given matrix, or else show that it is singular. \(\left(\begin{array}{rr}{1} & {4} \\ {-2} & {3}\end{array}\right)\)

Short Answer

Expert verified
Answer: The inverse of matrix A is A⁻¹ = ( [3/11, -4/11], [2/11, 1/11] ).

Step by step solution

01

Find the determinant

First, let's find the determinant of the given matrix A: \(\det(A) = a_{11}a_{22} - a_{12}a_{21} = (1)(3) - (4)(-2) = 3 + 8 = 11\) Since \(\det(A) \neq 0\), the matrix A is invertible.
02

Calculate the inverse matrix

Now, we'll find the inverse of matrix A using the formula mentioned earlier: \(A^{-1} = \frac{1}{\det(A)} \left(\begin{array}{rr}{a_{22}} & {-a_{12}} \\\ {-a_{21}} & {a_{11}}\end{array}\right) = \frac{1}{11} \left(\begin{array}{rr}{3} & {-4} \\\ {2} & {1}\end{array}\right)\) So the inverse of the given matrix is: \(A^{-1}=\left(\begin{array}{rr}{\frac{3}{11}} & {\frac{-4}{11}} \\\ {\frac{2}{11}} & {\frac{1}{11}}\end{array}\right)\)

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Key Concepts

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

Determinant of a Matrix
The determinant of a matrix is a special scalar value that provides key insights into the matrix's properties. It can help to determine whether a matrix has an inverse, the volume distortion factor during a linear transformation, and solutions to systems of linear equations. To compute the determinant of a 2x2 matrix like the one in the exercise \( \left(\begin{array}{rr}{1} & {4} \ {-2} & {3}\end{array}\right) \), you take the product of the entries on the main diagonal and subtract the product of the entries on the off-diagonal, yielding \( \det(A) = a_{11}a_{22} - a_{12}a_{21} \).

Understanding determinants is crucial because only matrices with non-zero determinants can be inverted, which is a fundamental operation in solving linear equations. When a determinant is zero, it indicates that the matrix is singular, which leads us to the next section.
Singular Matrix
A singular matrix is one that does not have an inverse, which occurs if and only if its determinant is zero. This has significant implications in linear algebra because singular matrices do not allow for certain computations, such as solving linear systems using the inverse.

In the context of the exercise, the determinant computed was 11, not zero, thus confirming that the matrix is not singular. If it had been zero, it would have meant that the matrix could not be used to find unique solutions for a system of linear equations, nor could it represent an invertible linear transformation in geometry.
Inverse of a Matrix
The inverse of a matrix is analog to reciprocal of a number, where multiplying the matrix by its inverse yields the identity matrix. The operation of finding a matrix's inverse is fundamental in various areas, such as solving systems of equations, in algorithms, and analyzing electrical circuits.

For a 2x2 matrix \(A\), the inverse \(A^{-1}\) can be calculated using the algebraic formula \(A^{-1} = \frac{1}{\det(A)} \left(\begin{array}{rr}{a_{22}} & {-a_{12}} \ {-a_{21}} & {a_{11}}\end{array}\right)\). It's important to note that the existence of an inverse relies heavily on the determinant. As shown in the exercise, the inverse exists and is explicitly computed when the determinant is not zero.
Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces, linear mappings, and systems of linear equations. It provides the language and framework for various areas including physics, computer science, statistics, and more.

Understanding concepts like matrix inversion and determinants are central to the study of linear algebra, as they are used to describe linear transformations, which are the main actions in this field. The ability to compute matrix inverses, as demonstrated in the exercise, is one of the many powerful techniques in linear algebra that allow for the analysis and solution of real-world problems involving linear systems.

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Most popular questions from this chapter

Show that all solutions of the system $$ x^{\prime}=\left(\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right) \mathbf{x} $$ approach zero as \(t \rightarrow \infty\) if and only if \(a+d<0\) and \(a d-b c>0 .\) Compare this result with that of Problem 38 in Section \(3.5 .\)

The method of successive approximations (see Section \(2.8)\) can also be applied to systems of equations. For example, consider the initial value problem $$ \mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}, \quad \mathbf{x}(0)=\mathbf{x}^{0} $$ where \(\mathbf{A}\) is a constant matrix and \(\mathbf{x}^{0}\) a prescribed vector. (a) Assuming that a solution \(\mathbf{x}=\Phi(t)\) exists, show that it must satisfy the integral equation $$ \Phi(t)=\mathbf{x}^{0}+\int_{0}^{t} \mathbf{A} \phi(s) d s $$ (b) Start with the initial approximation \(\Phi^{(0)}(t)=\mathbf{x}^{0} .\) Substitute this expression for \(\Phi(s)\) in the right side of Eq. (ii) and obtain a new approximation \(\Phi^{(1)}(t) .\) Show that $$ \phi^{(1)}(t)=(1+\mathbf{A} t) \mathbf{x}^{0} $$ (c) Reppeat this process and thereby obtain a sequence of approximations \(\phi^{(0)}, \phi^{(1)}\), \(\phi^{(2)}, \ldots, \phi^{(n)}, \ldots\) Use an inductive argument to show that $$ \phi^{(n)}(t)=\left(1+A t+A^{2} \frac{2}{2 !}+\cdots+A^{x} \frac{r^{2}}{n !}\right) x^{0} $$ (d) Let \(n \rightarrow \infty\) and show that the solution of the initial value problem (i) is $$ \phi(t)=\exp (\mathbf{A} t) \mathbf{x}^{0} $$

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\ {-8} & {-5} & {-3}\end{array}\right) \mathbf{x} $$

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{ccc}{1} & {0} & {0} \\ {2} & {1} & {-2} \\ {3} & {2} & {1}\end{array}\right) $$

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{3} & {-2} \\ {2} & {-2}\end{array}\right) \mathbf{x} $$

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