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\(A, B,\) and \(C\) denote \(n \times n\) matrices. Does \((A+B)(A-B)=A^{2}-B^{2} ?\)

Short Answer

Expert verified
No, it's not true unless matrices commute (i.e., \(AB = BA\)).

Step by step solution

01

Understand the Expression

We need to determine if the equation \((A + B)(A - B) = A^2 - B^2\) holds true for \(n \times n\) matrices. This resembles the difference of squares formula from arithmetic: \((x + y)(x - y) = x^2 - y^2\).
02

Expand the Left Side

Let's expand the left side of the equation. The expression \((A + B)(A - B)\) can be expanded using distributive properties of matrices as follows: \((A + B)(A - B) = A(A - B) + B(A - B) = AA - AB + BA - BB\).
03

Expand the Right Side

The right side of the equation is \(A^2 - B^2\), which represents the straightforward subtraction of \(B^2\) from \(A^2\). This means: \(A^2 - B^2 = AA - BB\).
04

Compare Both Sides

After expanding both sides, compare the expressions: The left side: \(AA - AB + BA - BB\). The right side: \(AA - BB\). For these to be equal, the terms \(-AB\) and \(+BA\) must cancel out, which occurs if \(AB = BA\), i.e., if \(A\) and \(B\) commute. However, matrices do not necessarily commute.
05

Conclusion

Since \(-AB + BA\) does not always equal zero (due to non-commutative property of matrices in general), the expression \((A + B)(A - B) = A^2 - B^2\) is not true for all \(n \times n\) matrices. It only holds when \(A\) and \(B\) commute, i.e., \(AB = BA\).

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Key Concepts

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

Matrix Multiplication
Matrix multiplication involves combining two matrices to form a new matrix. It follows specific rules that are slightly different from other types of multiplication.
To multiply two matrices, say matrix \( A \) with dimensions \( m \times n \) and matrix \( B \) with dimensions \( n \times p \), the resulting matrix, often named \( C \), will have dimensions \( m \times p \).
Here's how it works:
  • The element in the \( i \)-th row and \( j \)-th column of matrix \( C \) is calculated by taking the dot product of the \( i \)-th row of \( A \) and the \( j \)-th column of \( B \).
  • This means multiplying each element of the row with the corresponding element of the column and summing up all these products.
Matrix multiplication is not commutative; this means \( AB eq BA \) in most cases. Therefore, even though **matrix multiplication** follows associative and distributive properties, you need to be careful with the order of multiplication.
Commutative Property
The commutative property is a fundamental principle in algebra. It suggests that the order in which you multiply two elements doesn't change the result. **In arithmetic**, for example, \( 2 \times 3 = 3 \times 2 \).
However, in matrix algebra, the commutative property doesn’t generally hold. That means for most matrices \( A \) and \( B \), \( AB \) is not equal to \( BA \). This non-commutativity can have important implications:
  • When performing calculations that involve matrices, the order in which matrices are multiplied can change the answer or even make it impossible to carry out the calculations.
  • Specific conditions when matrices commute include situations involving diagonal matrices, identity matrices, or when one matrix is a scalar multiple of another.
Understanding when matrices do or do not commute is crucial in solving problems involving matrices.
Difference of Squares Formula
The difference of squares formula is a handy algebraic shortcut. It states that the product of the sum and difference of two numbers equals the difference of their squares:
For numbers, we express it as \[ (x + y)(x - y) = x^2 - y^2 \].
Matrix multiplication attempts to mimic this pattern, but due to the non-commutative nature of matrices, it doesn't always pan out the same way.
Consider matrices \( A \) and \( B \): if you try to use \[ (A + B)(A - B) \] it leads to \[ AA - AB + BA - BB \].
The expression doesn't neatly cancel out to \[ A^2 - B^2 \] unless \( AB = BA \).
This formula highlights why understanding matrix properties is essential, as assumptions based on regular arithmetic don't always apply to matrices.

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Most popular questions from this chapter

Show that adding a multiple of the first row of a matrix to the second row leaves the determinant unchanged; that is, $$ \left|\begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{2}+\lambda a_{1} & b_{2}+\lambda b_{1} & c_{2}+\lambda c_{1} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=\left|\begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right| $$ [In fact, adding a multiple of any row (column) of a matrix to another row (column) leaves the determinant unchanged.]

Interpret these results geometrically in terms of the parallelogram formed by \(x\) and \(y\). Verify the Cauchy-Schwarz inequality and the triangle inequality for the vectors. $$\mathbf{x}=(1,0,0,1), \mathbf{y}=(-1,0,0,1)$$

(This exercise assumes a knowledge of integration of continuous functions of one variable.) Note that the proof of the Cauchy-Schwarz inequality (Theorem 4) depends only on the properties of the inner product listed in Theorem \(1 .\) Use this observation to establish the following inequality for continuous functions \(f, g:[0,1] \rightarrow \mathbb{R}\): $$\left|\int_{0}^{1} f(x) g(x) d x\right| \leq \sqrt{\int_{0}^{1}[f(x)]^{2} d x} \sqrt{\int_{0}^{1}[g(x)]^{2} d x}$$. Do this by (a) verifying that the space of continuous functions from [0,1] to \(\mathbb{R}\) forms a vector space; that is, we may think of functions \(f, g\) abstractly as "vectors" that can be added to each other and multiplied by scalars. (b) introducing the inner product of functions $$f \cdot g=\int_{0}^{1} f(x) g(x) d x$$ and verifying that it satisfies conditions (i) to (iv) of Theorem 3.

Compute \(\|\mathbf{u}\|,\|\mathbf{v}\|,\) and \(\mathbf{u} \cdot \mathbf{v}\) for the given vectors in \(\mathbb{R}^{3}\). $$\mathbf{u}=5 \mathbf{i}-\mathbf{j}+2 \mathbf{k}, \mathbf{v}=\mathbf{i}+\mathbf{j}-\mathbf{k}$$

Find an equation for the plane that passes through. (a) \((0,0,0),(2,0,-1),\) and (0,4,-3). (b) \((1,2,0),(0,1,-2),\) and (4,0,1). (c) \((2,-1,3),(0,0,5),\) and (5,7,-1).

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