/*! 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 291 if $$ A=\left|\begin{array}{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 291

if $$ A=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right| $$ then \(\mathrm{A}^{3}=\) (a) 0 (b) \(\mathrm{A}^{\mathrm{T}}\) (c) I (d) \(\mathrm{A}^{-1}\)

Short Answer

Expert verified
The matrix \(A^3 = A\), which implies that \(A^2\) is the inverse of \(A\), i.e., \(A^2 = A^{-1}\). Therefore, the correct answer is option (d).

Step by step solution

01

Find A^2

First, we need to calculate A^2. To do this, we will multiply matrix A with itself: $$ A^2 = A \times A $$
02

Perform the matrix multiplication for A^2

To perform the matrix multiplication for A^2, follow the normal matrix multiplication procedure: $$ A^2=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right]\times\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] $$ $$ A^2= \left[\begin{array}{ccc} 1(1) + (-1)(2) + 1(1) & 1(-1) + (-1)(-1) + 1(0) & 1(1) + (-1)(0) + 1(0) \\ 2(1) + (-1)(2) + 0(1) & 2(-1) + (-1)(-1) + 0(0) & 2(1) + (-1)(0) + 0(0) \\ 1(1) + 0(2) + 0(1) & 1(-1) + 0(-1) + 0(0) & 1(1) + 0(0) + 0(0) \end{array}\right] $$ $$ A^2=\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & -1 & 1 \end{array}\right] $$
03

Find A^3

Now that we have found A^2, we can find A^3 by multiplying A^2 by A: $$ A^3 = A^2 \times A $$
04

Perform the matrix multiplication for A^3

To perform the matrix multiplication, follow the normal matrix multiplication procedure: $$ A^3=\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 1 & -1 & 1 \end{array}\right]\times\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] $$ $$ A^3= \left[\begin{array}{ccc} 0(1) + 1(2) + 1(1) & 0(-1) + 1(-1) + 1(0) & 0(1) + 1(0) + 1(0) \\ 0(1) + 1(2) + 2(1) & 0(-1) + 1(-1) + 2(0) & 0(1) + 1(0) + 2(0) \\ 1(1) + (-1)(2) + 1(1) & 1(-1) + (-1)(-1) + 1(0) & 1(1) + (-1)(0) + 1(0) \end{array}\right] $$ $$ A^3=\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right] $$ Now, we can compare the obtained matrix A^3 with the given options: (a) 0: The matrix A^3 is not a zero matrix. (b) \(A^T\): The transpose of A is: $$ A^T = \left[\begin{matrix} 1 & 2 & 1 \\ -1 & -1 & 0 \\ 1 & 0 & 0 \end{matrix}\right] $$ which is not equal to A^3. (c) I (identity matrix): The identity matrix is $$ I = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $$ A^3 is not equal to the identity matrix. (d) \(A^{-1}\) (inverse of A): It can be observed that A^3 = A, which implies that A^2 is the inverse of A (i.e., \(A^2 = A^{-1}\)). Therefore, the correct answer is option (d).

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!

Key Concepts

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

Determinants
The concept of a determinant is crucial in understanding many matrix operations, including whether or not a matrix is invertible. For a square matrix, the determinant is a scalar value that provides insight into certain properties of the matrix. Specifically, if a matrix has a determinant of 0, it is deemed singular, meaning it does not have an inverse. Calculating the determinant for a 3x3 matrix involves using the formula:
  • For matrix \(A\): \[ \text{det}(A) = a(ei \text{ - } fh) - b(di \text{ - } fg) + c(dh \text{ - } eg) \]
  • Where \(a, b, c, d, e, f, g, h, i\) represent the elements of the matrix.
Understanding determinants can help determine solvability in systems of linear equations and provide insights into transformations represented by matrices.
Matrix Inverse
The inverse of a matrix, denoted as \( A^{-1} \), plays a significant role in solving systems of linear equations, among other applications. A matrix must be square and have a non-zero determinant to possess an inverse. When a matrix is multiplied by its inverse, it results in the identity matrix:
  • \( A \times A^{-1} = I \)
To find an inverse for a 3x3 matrix, the method involves:
  • Calculating the determinant.
  • Finding the matrix of minors, then the cofactor matrix.
  • Taking the transpose of the cofactor matrix.
  • Finally, dividing by the determinant.
Knowing how to compute an inverse is extremely useful in matrix algebra, particularly for solving equations.
Identity Matrix
An identity matrix is a pivotal concept as it acts as the multiplicative identity in matrix multiplication. For a 3x3 identity matrix, each element along the diagonal is 1, and all other elements are 0:
  • \[ I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]
The identity matrix fulfills the property:
  • For any matrix \( A \), \( A \times I = A \text{ and } I \times A = A \)
This matrix is an important tool in matrix operations because it does not change any matrix it multiplies with, thus maintaining all original properties of matrix \( A \).
Matrix Transpose
A matrix transpose, denoted \( A^T \), is obtained by flipping a matrix over its diagonal. This operation switches the matrix's row and column indices. For example, if matrix \( A \) is:
  • \[ A = \begin{bmatrix} 1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} \]
Then, its transpose \( A^T \) is:
  • \[ A^T = \begin{bmatrix} 1 & 2 & 1 \ -1 & -1 & 0 \ 1 & 0 & 0 \end{bmatrix} \]
The transpose is used in various contexts, such as simplifying matrix equations or computing the inverse more efficiently. It maintains the original matrix's properties, such as determinant and rank, but presents these in a different orientation.

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

\(\underline{A}=\left|\begin{array}{ccc}-1 & 2-3 i & 3+4 i \\ 2+3 i & 5 & 1+i \\\ 3-4 i & 1-i & 4\end{array}\right|\) then det \(\mathrm{A}\) is \(\ldots .\) (a) purely real (b) purely imaginary (c) complex number (d) 0

Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) be such that \(\mathrm{b}(\mathrm{a}+\mathrm{c}) \neq 0\) If \(\left|\begin{array}{ccc}\mathrm{a} & \mathrm{a}+1 & \mathrm{a}-1 \\\ -\mathrm{b} & \mathrm{b}+1 & \mathrm{~b}-1 \\ \mathrm{c} & \mathrm{c}-1 & \mathrm{c}+1\end{array}\right|+\left|\begin{array}{lll}\mathrm{a}+1 & \mathrm{~b}+1 & \mathrm{c}-1 \\ \mathrm{a}-1 & \mathrm{~b}-1 & \mathrm{c}+1 \\\ (-1)^{\mathrm{n}+2} \mathrm{a} & (-1)^{\mathrm{n}+1} \mathrm{~b} & (-1)^{\mathrm{n}} \mathrm{c}\end{array}\right|=0\) then \(\mathrm{n}\) is (a) Zero (b) any even integer (c) any odd integer (d) any integer

Read the following paragraph carefully and answer the following questions. $$ \mathrm{A}=\left|\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right| $$ if, \(\mathrm{U}_{1}, \mathrm{U}_{2}\) and \(\mathrm{U}_{3}\) are column matrices satisfying $$ \mathrm{AU}_{1}=\left|\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right|, \mathrm{AU}_{2}=\left|\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right|, \mathrm{AU}_{3}=\left|\begin{array}{l} 2 \\ 3 \\ 1 \end{array}\right| $$ and \(\mathrm{U}\) is \(3 \times 3\) matrix whose columns are \(\mathrm{U}_{1}, \mathrm{U}_{2}, \mathrm{U}_{3}\), then The value of \(|\mathrm{U}|\) is \(\ldots \ldots\) (a) 3 (b) \(-3\) (c) \((3 / 2)\) (d) 2

If \(\left|\begin{array}{ccc}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 2 x-3 \\\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=24 x+B\) then \(B=\ldots\) (a) \(-12\) (b) 12 (c) 24 (d) \(-8\)

If the equations \(x-2 y+3 z=0,-2 x+3 y+2 z=0\) \(-8 \mathrm{x}+\lambda \mathrm{y}=0\) have non-trivial solutions then \(\lambda=\) (a) 18 (b) 13 (c) \(-10\) (d) 4
See all solutions

Recommended explanations on English Textbooks

Creative Story

Read Explanation

Graphology

Read Explanation

Semiotics

Read Explanation

Free Response Essay

Read Explanation

Cues and Conventions

Read Explanation

Essay Plan

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.