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In the text we state that \((A+B)^{2} \neq\) \(A^{2}+2 A B+B^{2} .\) We investigate that claim here. (a) Let \(A=\left[\begin{array}{cc}5 & 3 \\ -3 & -2\end{array}\right]\) and let \(B=\) $$ \left[\begin{array}{cc} -5 & -5 \\ -2 & 1 \end{array}\right] . \text { Compute } A+B $$ (b) Find \((A+B)^{2}\) by using your answer from (a). (c) Compute \(A^{2}+2 A B+B^{2}\). (d) Are the results from (a) and (b) the same? (e) Carefully expand the expression \((A+B)^{2}=(A+B)(A+B)\) and show why this is not equal to \(A^{2}+2 A B+B^{2}\)

Short Answer

Expert verified
No, \((A+B)^2 \neq A^2 + 2AB + B^2\).

Step by step solution

01

Compute A + B

Start by adding matrices \(A\) and \(B\). The sum of two matrices is obtained by adding their corresponding entries. For matrices \(A\) and \(B\): \[A = \begin{bmatrix} 5 & 3 \ -3 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} -5 & -5 \ -2 & 1 \end{bmatrix}\]The sum \(A + B\) is:\[A + B = \begin{bmatrix} 5 + (-5) & 3 + (-5) \ -3 + (-2) & -2 + 1 \end{bmatrix} = \begin{bmatrix} 0 & -2 \ -5 & -1 \end{bmatrix}\].
02

Compute (A + B)^2

To find \((A+B)^2\), we need to multiply the result from (a) by itself:\[(A+B)^2 = \begin{bmatrix} 0 & -2 \ -5 & -1 \end{bmatrix} \times \begin{bmatrix} 0 & -2 \ -5 & -1 \end{bmatrix} = \begin{bmatrix} (0)(0) + (-2)(-5) & (0)(-2) + (-2)(-1) \ (-5)(0) + (-1)(-5) & (-5)(-2) + (-1)(-1) \end{bmatrix}\]Simplify the expressions:\[(A+B)^2 = \begin{bmatrix} 10 & 2 \ 5 & 11 \end{bmatrix}\]
03

Compute A^2 + 2AB + B^2

First, calculate \(A^2\), \(B^2\), and \(2AB\).\(A^2\):\(A^2 = A \times A = \begin{bmatrix} 5 & 3 \ -3 & -2 \end{bmatrix} \times \begin{bmatrix} 5 & 3 \ -3 & -2 \end{bmatrix} = \begin{bmatrix} 34 & 21 \ -21 & -12 \end{bmatrix}\)\(B^2\):\(B^2 = B \times B = \begin{bmatrix} -5 & -5 \ -2 & 1 \end{bmatrix} \times \begin{bmatrix} -5 & -5 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} 35 & 0 \ 0 & 9 \end{bmatrix}\)\(2AB\):\(AB = A \times B = \begin{bmatrix} 5 & 3 \ -3 & -2 \end{bmatrix} \times \begin{bmatrix} -5 & -5 \ -2 & 1 \end{bmatrix} = \begin{bmatrix} -31 & -22 \ 19 & 17 \end{bmatrix}\)\(2AB = 2 \times AB = \begin{bmatrix} -62 & -44 \ 38 & 34 \end{bmatrix}\)Combine these results:\[A^2 + 2AB + B^2 = \begin{bmatrix} 34 & 21 \ -21 & -12 \end{bmatrix} + \begin{bmatrix} -62 & -44 \ 38 & 34 \end{bmatrix} + \begin{bmatrix} 35 & 0 \ 0 & 9 \end{bmatrix} = \begin{bmatrix} 7 & -23 \ 17 & 31 \end{bmatrix}\]
04

Compare Results from Step 2 and 3

Now compare the matrices obtained in steps 2 and 3:\[(A+B)^2 = \begin{bmatrix} 10 & 2 \ 5 & 11 \end{bmatrix}\]\[A^2 + 2AB + B^2 = \begin{bmatrix} 7 & -23 \ 17 & 31 \end{bmatrix}\]The matrices are not the same, thus supporting the claim that \((A+B)^2 eq A^2 + 2AB + B^2\).
05

Expand (A + B)^2

Let's expand the expression \((A+B)^2 = (A+B)(A+B)\):\[(A + B)^2 = A^2 + AB + BA + B^2\]This equation includes the matrices \(AB\) and \(BA\), which are not equal unless under special circumstances. The term \(2AB\) in the expression \(A^2 + 2AB + B^2\) implies that matrix multiplication is commutative, which it is not. Thus, \(AB + BA eq 2AB\), further validating that \((A + B)^2 eq A^2 + 2AB + B^2\).

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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 fundamental operation in matrix algebra. It involves taking two matrices and producing a new matrix by applying specific rules. To multiply matrices, you must have a clear understanding of the dimensions involved.
  • If you have a matrix A with dimensions (m x n) and a matrix B with dimensions (n x p), then the product, AB, will be a matrix of dimensions (m x p).
  • This requires that the number of columns in A matches the number of rows in B.
The multiplication process involves taking each element in row i of the first matrix and multiplying it with the corresponding element in column j of the second matrix, summing the results. For example, when calculating matrix product \[C = A \times B\]The element in the first row and first column of C, denoted \( c_{11} \), is calculated as:\[c_{11} = a_{11}b_{11} + a_{12}b_{21} + \ldots + a_{1n}b_{n1}\]This process is repeated for each element in the resulting matrix.
Matrix Addition
Matrix addition is the simplest of matrix operations and is only performed on matrices of the same dimensions. When adding two matrices, the operation is straightforward:
  • Add corresponding elements from the two matrices to get a new matrix.
  • It is crucial that both matrices involved are of identical dimensions; otherwise, the operation cannot be executed.
  • For example, if matrix A is 2x2 and matrix B is also 2x2, their sum, A + B, will be another 2x2 matrix.
Each element in this resulting matrix is given by:\[(A + B)_{ij} = a_{ij} + b_{ij}\]Where \( a_{ij} \) and \( b_{ij} \) represent elements from matrix A and matrix B respectively. It’s much like adding numbers together but done for each position individually across the two matrices.
Non-Commutative Properties
In matrix algebra, one of the surprising elements for learners is that matrix multiplication is not inherently commutative. This means:
  • For two matrices A and B, generally, \( AB eq BA \).
  • Even if both products are valid (dimensions allow multiplication both ways), the results may not be equal.
This is an essential concept when considering operations like matrix squaring or expanding expressions such as \((A+B)^2\), where the distribution of multiplication through addition does not hold as it would with scalar numbers.The order in which you multiply matrices affects the output. To illustrate:\[AB = \begin{bmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22}\end{bmatrix}\]Contrast this with the calculation of BA, which will almost certainly yield a different matrix unless very specific conditions are met.
Matrix Squaring
Matrix squaring is similar to squaring a number, but within the confines of matrix multiplication rules. Given a square matrix A, the operation A² involves multiplying A by itself.
  • Matrix squaring is only possible if the matrix is square, i.e., the number of rows equals the number of columns.
  • The result A² will have the same dimensions as the original matrix A.
To compute A², apply the matrix multiplication rules to A multiplied by A. It is crucial to note here that since matrix multiplication is not commutative, matrix square should always be expressed explicitly.In expressions such as \((A+B)^2\), which suggests addition before squaring, it is misleading to employ shortcuts like algebraic FOIL (First, Outside, Inside, Last) without considering the non-commutative nature of matrices. Instead, the expansion should be written as:\[(A+B)^2 = A^2 + AB + BA + B^2\]Where the distinction lies in treating \(AB\) and \(BA\) as potentially different calculations, compared to scalar arithmetic.

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Most popular questions from this chapter

Matrices \(A\) and \(B\) are defined. (a) Give the dimensions of \(A\) and \(B\). If the dimensions properly match, give the dimensions of \(A B\) and \(B A\). (b) Find the products \(A B\) and \(B A\), if possible. $$ A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 4 \end{array}\right] B=\left[\begin{array}{cc} 2 & 5 \\ 3 & -1 \end{array}\right] $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \vec{u}=\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right] \quad \vec{v}=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \begin{array}{l} \vec{u}=\left[\begin{array}{cccc} 1 & 2 & 3 & 4 \end{array}\right] \\ \vec{v}=\left[\begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array}\right] \end{array} $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ A=\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right], \vec{b}=\left[\begin{array}{c} 0 \\ -5 \end{array}\right] $$

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cccc} 0 & -3 & -1 & -3 \\ -4 & 2 & -3 & 5 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} 0 \\ 0 \end{array}\right], \vec{u}=\left[\begin{array}{c} 11 \\ 4 \\ -12 \\ 0 \end{array}\right], \\ \vec{v}=\left[\begin{array}{c} 9 \\ -12 \\ 0 \\ 12 \end{array}\right] \end{array} $$

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