/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Find an upper triangular matrix ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 24

Find an upper triangular matrix \(A\) such that \(A^{3}=\left[\begin{array}{rr}8 & -57 \\ 0 & 27\end{array}\right]\).

Short Answer

Expert verified
The upper triangular matrix A that satisfies the given condition is: \[A = \left[\begin{array}{rr}2 & -\frac{73}{9} \\ 0 & 3\end{array}\right]\]

Step by step solution

01

Multiply the given matrix A three times

As A is an upper triangular matrix, let's assume it has the form: \[A = \left[\begin{array}{rr}a & b \\ 0 & c\end{array}\right]\] First, we need to multiply A by itself three times to find A cubed (A^3): \[A^3 = A \cdot A \cdot A\]
02

Calculate A^2

Start by finding A^2 (A squared): \[A^2 = A \cdot A = \left[\begin{array}{rr}a & b \\ 0 & c\end{array}\right] \cdot \left[\begin{array}{rr}a & b \\ 0 & c\end{array}\right]\] Perform the multiplication: \(A^2 = \left[\begin{array}{rr}a^2 & ab+bc \\ 0 & c^2\end{array}\right]\)
03

Calculate A^3

Next, find A^3 by multiplying A^2 and A: \[A^3 = A^2 \cdot A = \left[\begin{array}{rr}a^2 & ab+bc \\ 0 & c^2\end{array}\right] \cdot \left[\begin{array}{rr}a & b \\ 0 & c\end{array}\right]\] Perform the multiplication: \(A^3 = \left[\begin{array}{rr}a^3 & a(ab+bc)+b(c^2) \\ 0 & (ab+bc)(c)\end{array}\right]\)
04

Equate A^3 with the given matrix

Now equate A^3 with the given matrix: \[\left[\begin{array}{rr}a^3 & a(ab+bc)+b(c^2) \\ 0 & (ab+bc)(c)\end{array}\right] = \left[\begin{array}{rr}8 & -57 \\ 0 & 27\end{array}\right]\] This results in a system of linear equations: \(a^3 = 8\) \(a(ab+bc)+b(c^2) = -57\) \((ab+bc)(c) = 27\)
05

Solve for a, b, and c

Start by solving for a in the first equation: \(a^3 = 8\) \(a = 2\) Then, solve for c in the third equation using the value of a: \((ab+bc)(c) = 27\) \((2b+b2)(c)=27\) (Since a=2) \(2bc+b^2c=27\) Now, we have \(2bc+b^2c=27\) and since we know \(a^{3} = 8\). We can say that \(c^{3} = 27\), so \(c = 3\). Plug the values of a and c back into the second equation: \(a(ab+bc)+b(c^2) = -57\) \(2(2(2)+3(2))+b(3)^2 = -57\) (Since a=2 and c=3) \(16 + 9b = -57\) \(9b = -73\) \(b = -\frac{73}{9}\)
06

Construct the matrix A

Finally, construct the matrix A using the values of a, b, and c: \[A = \left[\begin{array}{rr}a & b \\ 0 & c\end{array}\right] = \left[\begin{array}{rr}2 & -\frac{73}{9} \\ 0 & 3\end{array}\right]\] So, the upper triangular matrix A is: \[A = \left[\begin{array}{rr}2 & -\frac{73}{9} \\ 0 & 3\end{array}\right]\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that \(A=\left[\begin{array}{cc}\frac{1}{3}-\frac{2}{3} i & \frac{2}{3} i \\\ -\frac{2}{3} i & -\frac{1}{3}-\frac{2}{3} i\end{array}\right]\) is unitary.

Let \(A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\) and \(B=\left[\begin{array}{ccc}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right],\) (a) \(\quad\) Find \(A^{n} \cdot(b) \quad\) Find \(B^{n}\).

Let \(D_{k}=k I,\) the scalar matrix belonging to the scalar \(k .\) Show (a) \(D_{k} A=k A\) (b) \(\quad B D_{k}=k B\) (c) \(\quad D_{k}+D_{k^{\prime}}=D_{k+k^{\prime}}\) (d) \(\quad D_{k} D_{k^{\prime}}=D_{k k^{\prime}}\)

Let \(c_{0}, c_{1}, \ldots, c_{n}\) be distinct scalars from an infinite field \(F\). Define \(\mathrm{T}: \mathrm{P}_{n}(F) \rightarrow \mathrm{F}^{n+1}\) by $\mathrm{T}(f)=\left(f\left(c_{0}\right), f\left(c_{1}\right), \ldots, f\left(c_{n}\right)\right)\(. Prove that \)\mathrm{T}$ is an isomorphism. Hint: Use the Lagrange polynomials associated with \(c_{0}, c_{1}, \ldots, c_{n}\).

Refer to the following matrices: $$A=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 3 & 4 \end{array}\right], \quad B=\left[\begin{array}{rrr} 4 & 0 & -3 \\ -1 & -2 & 3 \end{array}\right], \quad C=\left[\begin{array}{rrrr} 2 & -3 & 0 & 1 \\ 5 & -1 & -4 & 2 \\ -1 & 0 & 0 & 3 \end{array}\right], \quad D=\left[\begin{array}{rr} 2 \\ -1 \\ 3 \end{array}\right].$$ Find (a) \(3 A-4 B,\) (b) \(A C,\) (c) \(B C,\) (d) \(A D,\) (e) \(B D,\) (f) \(C D.\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.