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Chapter 5: Problem 5

Let \(A\) and \(B\) be \(n \times n\) matrices of real numbers. Is \(A^{2}-B^{2}=(A-B)(A+\) \(B) ?\) Explain.

Short Answer

Expert verified
The equation holds only if matrices \(A\) and \(B\) commute, i.e., \(AB = BA\). Otherwise, it does not hold.

Step by step solution

01

Understand the problem

We need to determine whether the equation \(A^2 - B^2 = (A - B)(A + B)\) holds for \(n \times n\) matrices \(A\) and \(B\). This requires us to verify whether the properties of real numbers apply to matrices in the same way.
02

Recall formula for difference of squares

The formula \(a^2 - b^2 = (a - b)(a + b)\) is true for real numbers. We need to see if this formula holds similarly for matrix operations, considering matrix multiplication is not generally commutative.
03

Expand the right-hand side

The expression \((A - B)(A + B)\) can be expanded using matrix multiplication: \[(A - B)(A + B) = A(A + B) - B(A + B)\]This results in \[A^2 + AB - BA - B^2.\]
04

Compare both sides of the equation

The original left-hand side is \(A^2 - B^2\). From Step 3, we found that the right-hand side is \(A^2 + AB - BA - B^2\). To equate the two sides of the equation, \[A^2 + AB - BA - B^2 = A^2 - B^2 \]which implies \(AB - BA = 0.\)
05

Conclusion about matrix commutativity

Since matrices in general do not commute (i.e., \(AB eq BA\)), the equation \((A - B)(A + B) = A^2 - B^2\) holds true if and only if \(A\) and \(B\) commute: \(AB = BA\). For arbitrary matrices, this condition is not generally satisfied.

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is a key concept in linear algebra that differs from regular arithmetic multiplication. When multiplying matrices, the order matters, which means that unlike numbers, the product of matrices is not generally commutative. This means that \(AB eq BA\) in most cases. For two matrices \(A\) and \(B\) to be multiplied together, the number of columns in matrix \(A\) must equal the number of rows in matrix \(B\). The resulting product matrix has dimensions that reflect the number of rows in \(A\) and the number of columns in \(B\). Let's break down the multiplication process:
  • Choose a row from the first matrix (\(A\)).
  • Choose a column from the second matrix (\(B\)).
  • Multiply each element of the row by the corresponding element of the column, and then sum the products to get a single element in the resulting matrix.
Matrix multiplication requires careful attention to the order of operands, as swapping them can dramatically change the result.
Difference of Squares
The difference of squares is a algebraic identity familiar to many who study mathematics. The expression \(a^2 - b^2 = (a - b)(a + b)\) highlights this basic property. This formula is straightforward when dealing with scalar numbers, but becomes more complex when matrices enter the picture.In matrix terms, the operation does not work the same way because of the non-commutative nature of matrix multiplication. When attempting \((A^2 - B^2) = (A - B)(A + B)\) with matrices, you have to consider the interaction of non-commuting terms, \(AB\) and \(BA\). If \(AB - BA eq 0\), the identity does not hold true, unlike the numerical difference of squares.
Matrix Commutativity
Matrix commutativity is a topic that addresses whether the order of matrices affects the result of their multiplication. A set of matrices \(A\) and \(B\) is commutative if their product is the same regardless of the order: \(AB = BA\). This condition is rare and typically requires that matrices have specific symmetric properties or that they derive from the same eigenvectors.In practice, most matrices do not commute. This is why when examining expressions like \((A - B)(A + B)\), it can't be assumed that the order of multiplication doesn't affect the outcome. The commutativity condition is crucial in determining if identities like the difference of squares hold with matrices.Always test if \(AB = BA\) when working with expressions involving multiple matrices. If they do commute, many algebraic identities applicable to real numbers might be extended to matrices.

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Most popular questions from this chapter

In this exercise, we propose to show how matrix multiplication is a natural operation. Suppose a bakery produces bread, cakes and pies every weekday, Monday through Friday. Based on past sales history, the bakery produces various numbers of each product each day, summarized in the \(5 \times 3\) matrix \(D\). It should be noted that the order could be described as "number of days by number of products." For example, on Wednesday (the third day) the number of cakes (second product in our list) that are produced is \(d_{3,2}=4\). $$ D=\left(\begin{array}{ccc} 25 & 5 & 5 \\ 14 & 5 & 8 \\ 20 & 4 & 15 \\ 18 & 5 & 7 \\ 35 & 10 & 9 \end{array}\right) $$ The main ingredients of these products are flour, sugar and eggs. We assume that other ingredients are always in ample supply, but we need to be sure to have the three main ones available. For each of the three products, The amount of each ingredient that is needed is summarized in the \(3 \times 3\), or "number of products by number of ingredients" matrix \(P\). For example, to bake a cake (second product) we need \(P_{2,1}=1.5\) cups of flour (first ingredient). Regarding units: flour and sugar are given in cups per unit of each product, while eggs are given in individual eggs per unit of each product. $$ P=\left(\begin{array}{ccc} 2 & 0.5 & 0 \\ 1.5 & 1 & 2 \\ 1 & 1 & 1 \end{array}\right) $$ These amounts are "made up", so don't used them to do your own baking! (a) How many cups of flour will the bakery need every Monday? Pay close attention to how you compute your answer and the units of each number. (b) How many eggs will the bakery need every Wednesday? (c) Compute the matrix product \(D P\). What do you notice? (d) Suppose the costs of ingredients are \(\$ 0.12\) for a cup of flour, \(\$ 0.15\) for a cup of sugar and \(\$ 0.19\) for one egg. How can this information be put into a matrix that can meaningfully be multiplied by one of the other matrices in this problem?

(a) Determine \(I^{2}\) and \(I^{3}\) if \(I=\left(\begin{array}{ccc}1 & 0 & 0 \\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)\). (b) What is \(I^{n}\) equal to for any \(n \geq 1 ?\) (c) Prove your answer to part (b) by induction.

Let \(A=\left(\begin{array}{ccc}1 & 0 & 2 \\ 2 & -1 & 5 \\ 3 & 2 & 1\end{array}\right), B=\left(\begin{array}{ccc}0 & 2 & 3 \\ 1 & 1 & 2 \\ -1 & 3 & -2\end{array}\right),\) and \(C=\left(\begin{array}{cccc}2 & 1 & 2 & 3 \\\ 4 & 0 & 1 & 1 \\ 3 & -1 & 4 & 1\end{array}\right)\) Compute, if possible; (a) \(A-B\) (e) \(C A-C B\) (b) \(A B\) (c) \(A C-B C\) (f) \(C\left(\begin{array}{c}x \\ y \\ z \\ w\end{array}\right)\) (d) \(A(B C)\)

(a) Let \(A, B,\) and \(D\) be \(n \times n\) matrices. Assume that \(B\) is invertible. If \(A=B D B^{-1},\) prove by induction that \(A^{m}=B D^{m} B^{-1}\) is true for \(m \geq 1\) (b) Given that \(A=\left(\begin{array}{ll}-8 & 15 \\ -6 & 11\end{array}\right)=B\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right) B^{-1}\) where \(B=\) \(\left(\begin{array}{ll}5 & 3 \\ 3 & 2\end{array}\right)\) what is \(A^{10} ?\)

For the given matrices \(A\) find \(A^{-1}\) if it exists and verify that \(A A^{-1}=\) \(A^{-1} A=I .\) If \(A^{-1}\) does not exist explain why. (a) \(A=\left(\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}6 & -3 \\ 8 & -4\end{array}\right)\) (c) \(A=\left(\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right)\) (d) \(A=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\) (e) Use the definition of the inverse of a matrix to find \(A^{-1}: A=\) $$ \left(\begin{array}{ccc} 3 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & -5 \end{array}\right) $$
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