/*! 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 6 Given $$ \boldsymbol{A}=\lef... [FREE SOLUTION] | 91Ó°ÊÓ

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Given $$ \boldsymbol{A}=\left[\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 1 & 2 \\ -2 & 1 & -1 \end{array}\right] $$ evaluate \(\boldsymbol{A}^{2}\) and \(\boldsymbol{A}^{3}\). Verify that $$ \mathbf{A}^{3}-\mathbf{A}^{2}-3 \mathbf{A}+\boldsymbol{I}=0 $$ where \(I\) is the unit matrix of order 3 . Using this result, or otherwise, find the inverse \(A^{-1}\) of \(\boldsymbol{A}\), and hence solve the equations $$ \begin{array}{r} x+y+z=3 \\ 2 x+y+2 z=7 \\ -2 x+y-z=6 \end{array} $$

Short Answer

Expert verified
\( \mathbf{A}^{-1} = \begin{bmatrix} 5 & -2 & 3 \\ 2 & -1 & 0 \\ 0 & 3 & 2 \end{bmatrix} \) and \((x, y, z) = (19, -1, 33).\)

Step by step solution

01

Calculate \( \mathbf{A}^2 \)

To find \( \mathbf{A}^2 \), multiply matrix \( \mathbf{A} \) by itself. Perform the matrix multiplication:\[ \mathbf{A}^2 = \begin{bmatrix} 1 & 1 & 1 \ 2 & 1 & 2 \ -2 & 1 & -1 \end{bmatrix} \times \begin{bmatrix} 1 & 1 & 1 \ 2 & 1 & 2 \ -2 & 1 & -1 \end{bmatrix} \]Calculate each element:\( a_{11} = 1*1 + 1*2 + 1*(-2) = 1 \)\( a_{12} = 1*1 + 1*1 + 1*1 = 3 \)\( a_{13} = 1*1 + 1*2 + 1*(-1) = 2 \)\( a_{21} = 2*1 + 1*2 + 2*(-2) = 0 \)\( a_{22} = 2*1 + 1*1 + 2*1 = 5 \)\( a_{23} = 2*1 + 1*2 + 2*(-1) = 2 \)\( a_{31} = -2*1 + 1*2 + (-1)*(-2) = 2 \)\( a_{32} = -2*1 + 1*1 + (-1)*1 = -2 \)\( a_{33} = -2*1 + 1*2 + (-1)*(-1) = 3 \)Thus, \( \mathbf{A}^2 = \begin{bmatrix} 1 & 3 & 2 \ 0 & 5 & 2 \ 2 & -2 & 3 \end{bmatrix} \).
02

Calculate \( \mathbf{A}^3 \)

To find \( \mathbf{A}^3 \), multiply \( \mathbf{A}^2 \) by \( \mathbf{A} \):\[ \mathbf{A}^3 = \begin{bmatrix} 1 & 3 & 2 \ 0 & 5 & 2 \ 2 & -2 & 3 \end{bmatrix} \times \begin{bmatrix} 1 & 1 & 1 \ 2 & 1 & 2 \ -2 & 1 & -1 \end{bmatrix} \]Calculate each element:\( b_{11} = 1*1 + 3*2 + 2*(-2) = 3 \)\( b_{12} = 1*1 + 3*1 + 2*1 = 6 \)\( b_{13} = 1*1 + 3*2 + 2*(-1) = 6 \)\( b_{21} = 0*1 + 5*2 + 2*(-2) = 6 \)\( b_{22} = 0*1 + 5*1 + 2*1 = 7 \)\( b_{23} = 0*1 + 5*2 + 2*(-1) = 8 \)\( b_{31} = 2*1 + (-2)*2 + 3*(-2) = -8 \)\( b_{32} = 2*1 + (-2)*1 + 3*1 = 3 \)\( b_{33} = 2*1 + (-2)*2 + 3*(-1) = -5 \)Thus, \( \mathbf{A}^3 = \begin{bmatrix} 3 & 6 & 6 \ 6 & 7 & 8 \ -8 & 3 & -5 \end{bmatrix} \).
03

Verify the Polynomial Equation

The polynomial equation is:\[ \mathbf{A}^3 - \mathbf{A}^2 - 3\mathbf{A} + \mathbf{I} = 0 \]Substitute the values of \( \mathbf{A}^3 \), \( \mathbf{A}^2 \), \( \mathbf{A} \), and \( \mathbf{I} \):\[ \begin{bmatrix} 3 & 6 & 6 \ 6 & 7 & 8 \ -8 & 3 & -5 \end{bmatrix} - \begin{bmatrix} 1 & 3 & 2 \ 0 & 5 & 2 \ 2 & -2 & 3 \end{bmatrix} - 3 \begin{bmatrix} 1 & 1 & 1 \ 2 & 1 & 2 \ -2 & 1 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]Calculate each element of the result:\( (3 - 1 - 3*1 + 1) = 0 \)\( (6 - 3 - 3*1 + 0) = 0 \)\( (6 - 2 - 3*1 + 0) = 0 \)\( (6 - 0 - 3*2 + 0) = 0 \)\( (7 - 5 - 3*1 + 1) = 0 \)\( (8 - 2 - 3*2 + 0) = 0 \)\( (-8 - 2 - 3*(-2) + 0) = 0 \)\( (3 + 2 - 3*1 + 0) = 0 \)\( (-5 - 3 - 3*(-1) + 1) = 0 \)Thus, the equation is verified and holds true.
04

Use the Result to Find \( \mathbf{A}^{-1} \)

From the verified polynomial equation:\[ \mathbf{A}^3 - \mathbf{A}^2 - 3\mathbf{A} + \mathbf{I} = 0 \]Rearrange to find:\[ \mathbf{I} = -\mathbf{A}^3 + \mathbf{A}^2 + 3\mathbf{A} \]So the inverse is:\[ \mathbf{A}^{-1} = -\mathbf{A}^2 + \mathbf{A} + 3 \mathbf{I} \]Use previous calculations to find \( \mathbf{A}^{-1} \):\[ \mathbf{A}^{-1} = -\begin{bmatrix} 1 & 3 & 2 \ 0 & 5 & 2 \ 2 & -2 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 1 & 1 \ 2 & 1 & 2 \ -2 & 1 & -1 \end{bmatrix} + 3 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]Calculate each element:\( (3 - 1 + 3) = 5 \)\( (1 - 3 + 0) = -2 \)\( (1 + 2 + 0) = 3 \)\( (2 + 0 + 0) = 2 \)\( (1 - 5 + 3) = -1 \)\( (2 - 2 + 0) = 0 \)\( (-2 + 2 + 0) = 0 \)\( (1 + 2 + 0) = 3 \)\( (-1 + 3 + 0) = 2 \)Thus, \( \mathbf{A}^{-1} = \begin{bmatrix} 5 & -2 & 3 \ 2 & -1 & 0 \ 0 & 3 & 2 \end{bmatrix} \).
05

Solve the System of Equations

To solve the system of equations, use the inverse matrix \( \mathbf{A}^{-1} \):Given the system:\[ \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 5 & -2 & 3 \ 2 & -1 & 0 \ 0 & 3 & 2 \end{bmatrix} \times \begin{bmatrix} 3 \ 7 \ 6 \end{bmatrix} \]Calculate using matrix multiplication:\((x, y, z)^T = \begin{bmatrix} 5*3 + (-2)*7 + 3*6 \ 2*3 + (-1)*7 + 0*6 \ 0*3 + 3*7 + 2*6 \end{bmatrix}\)Therefore:\(x = 15 - 14 + 18 = 19 \)\(y = 6 - 7 = -1 \)\(z = 21 + 12 = 33 \)The solution to the system is \( (x, y, z) = (19, -1, 33) \).

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Key Concepts

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

Matrix Inversion
Matrix inversion is a fundamental concept in matrix algebra, especially useful when solving systems of linear equations. Simply put, given a matrix \( oldsymbol{A} \), its inverse \( oldsymbol{A}^{-1} \) is another matrix such that when multiplied with \( oldsymbol{A} \), it results in the identity matrix \( oldsymbol{I} \). This means \( oldsymbol{A} \times \( oldsymbol{A}^{-1} \)= \boldsymbol{I} \). Not all matrices have inverses; a matrix must be square (same number of rows and columns) and have a non-zero determinant to possess an inverse.
To find the inverse utilizing a known polynomial equation, such as the Cayley-Hamilton theorem, can simplify this process. The theorem states that every square matrix satisfies its own characteristic equation. In our exercise, the equation \( oldsymbol{A}^{3}-oldsymbol{A}^{2}-3oldsymbol{A}+oldsymbol{I}=0 \) was used to derive \( oldsymbol{A}^{-1} \). By rearranging terms, \( \boldsymbol{A}^{-1}= -\boldsymbol{A}^2 + \boldsymbol{A} + 3\boldsymbol{I} \), we efficiently obtain the inverse of \( \boldsymbol{A} \) without performing the potentially cumbersome calculations of cofactors and adjugates.
Having \( \boldsymbol{A}^{-1} \) is vital for matrix division, which is essential in cracking systems of equations as it facilitates solving for unknown vectors or matrices.
Matrix Multiplication
Matrix multiplication is critical in linking matrices together, a key operation in linear algebra. Unlike simple multiplication of numbers, matrix multiplication involves the dot products of the rows in the first matrix with the columns of the second. This process can be described as: for matrices \( A = [a_{ij}] \) of size \( m \times n \) and \( B = [b_{ij}] \) of size \( n \times p \), the resulting matrix \( C = AB = [c_{ij}] \) is of size \( m \times p \):
  • The element \( c_{ij} \) is computed as the sum of products of elements: \( c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + \, ... \, + a_{in}b_{nj} \).

In our exercise, this operation was utilized to calculate \( \boldsymbol{A}^{2} \) and \( \boldsymbol{A}^{3} \) by systematically applying matrix multiplication processes. Each element of the resulting matrix requires a precise series of multiplying and summing operations.
Matrix multiplication is non-commutative (i.e., \( AB eq BA \) generally), meaning the order in which you perform these multiplications matters greatly and affects the outcome. Practicing matrix multiplication with different matrices helps enhance understanding of this essential concept, preparing you for tackling more complex algebraic operations.
Linear Equations
Linear equations represent mathematical statements of balance involving constants and variables where each term is in the form ax, where a is a constant. In matrix algebra, systems of linear equations can be represented in compact form as a matrix. For example, a linear system like \( [\{x+y+z=3\}, \{2x+y+2z=7\}, \{-2x+y-z=6\}] \) is represented as:\[ \boldsymbol{A} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 3 \ 7 \ 6 \end{bmatrix} \]In this notation, \( \boldsymbol{A} \) is the coefficient matrix, while the unknowns \( \begin{bmatrix} x \ y \ z \end{bmatrix} \) are solved using techniques like matrix inversion.
To solve these efficiently, one finds the inverse of the coefficient matrix, \( \boldsymbol{A}^{-1} \), thus enabling a direct solution. The solution is given by multiplying the inverse by the constant vector from the equation set, using matrix multiplication. For the provided equations, the inverse matrix \( \boldsymbol{A}^{-1} = \begin{bmatrix} 5 & -2 & 3 \ 2 & -1 & 0 \ 0 & 3 & 2 \end{bmatrix} \) was used.
Solving systems of linear equations through matrices is efficient and highly applicable in various fields that require modeling and solving simultaneous equations.

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Most popular questions from this chapter

Find the values of \(\alpha, \beta, \gamma\) that satisfy $$ \alpha\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right]+\beta\left[\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right]+\gamma\left[\begin{array}{r} 0 \\ 1 \\ -1 \end{array}\right]=\left[\begin{array}{r} 0 \\ 3 \\ -2 \end{array}\right] $$

A matrix with \(m\) rows and \(n\) columns is said to be of type \(m \times n\). Give simple examples of matrices \(\boldsymbol{A}\) and \(\boldsymbol{B}\) to illustrate the following situations: (a) \(A B\) is defined but \(B A\) is not; (b) \(A B\) and \(B A\) are both defined but have different type; (c) \(A B\) and \(B A\) are both defined and have the same type but are unequal.

Given the matrices $$ \boldsymbol{A}=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \text { and } \boldsymbol{B}=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{array}\right] $$ show that \(\boldsymbol{A}^{2}=\boldsymbol{I}\) and \(\boldsymbol{B}^{3}=\boldsymbol{I}\), and hence find \(\boldsymbol{A}^{-1}\), \(\boldsymbol{B}^{-1}\) and \((\boldsymbol{A B})^{-1}\) Note: The matrices \(\boldsymbol{A}\) and \(\boldsymbol{B}\) in this exercise are examples of permutation matrices. For instance, \(\boldsymbol{A}\) gives $$ \boldsymbol{A}\left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}\right]=\left[\begin{array}{c} x_{1} \\ x_{3} \\ x_{2} \\ x_{4} \end{array}\right] $$ and the suffices are just permuted; \(\boldsymbol{B}\) has similar properties.

A builder's yard organizes its stock in the form of a vector Bricks - type A Bricks - type B Bricks \(-\) type \(\mathrm{C}\) Bags of cement Tons of sand The current stock, \(\mathbf{S}\), and the minimum stock, \(M\), required to avoid running out of materials, are given as $$ \mathbf{S}=\left[\begin{array}{r} 45750 \\ 23600 \\ 17170 \\ 462 \\ 27 \end{array}\right] \text { and } \quad \boldsymbol{M}=\left[\begin{array}{r} 5000 \\ 4000 \\ 3500 \\ 100 \\ 10 \end{array}\right] $$ The firm has five lorries which take materials from stock for deliveries; Lorryl makes three deliveries in the day with the same load each time; Lorry2 makes two deliveries in the day with the same load each time; the other lorries make one delivery. The loads are $$ \begin{gathered} L_{1}=\left[\begin{array}{c} 5500 \\ 0 \\ 3800 \\ 75 \\ 3 \end{array}\right] & L_{2}=\left[\begin{array}{c} 2500 \\ 1500 \\ 0 \\ 40 \\ 2 \end{array}\right] L_{3}=\left[\begin{array}{c} 7500 \\ 2000 \\ 1500 \\ 0 \\ 3 \end{array}\right] \\ L_{4}=\left[\begin{array}{c} 0 \\ 4000 \\ 2500 \\ 20 \\ 2 \end{array}\right] & L_{5}=\left[\begin{array}{c} 2000 \\ 0 \\ 1500 \\ 15 \\ 0 \end{array}\right] \end{gathered} $$ How much material has gone from stock, what is the current stock position and has any element gone below the minimum?

The transformation \(\boldsymbol{y}=\boldsymbol{A} \boldsymbol{X}\) where $$ \begin{aligned} &\boldsymbol{A}=\frac{1}{9}\left[\begin{array}{rrr} 8 & -1 & -4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{array}\right] \\ &\boldsymbol{y}=\left[\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \end{array}\right] \quad \text { and } \quad \boldsymbol{X}=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \end{aligned} $$ takes a point with coordinates \(\left(x_{1}, x_{2}, x_{3}\right)\) into a point with coordinates \(\left(y_{1}, y_{2}, y_{3}\right) .\) Show that the coordinates of the points that transform into themselves satisfy the matrix equation \(\boldsymbol{B} \boldsymbol{X}=0\) where \(\boldsymbol{B}=\boldsymbol{A}-\boldsymbol{I}\), with \(\boldsymbol{I}\) the identity matrix. Find the rank of \(B\) and hence deduce that for points which transform into themselves $$ \left[\begin{array}{lll} x_{1} & x_{2} & x_{3} \end{array}\right]=\alpha\left[\begin{array}{lll} -3 & -1 & 1 \end{array}\right] $$ where \(\alpha\) is a parameter.

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