Problem 179
Question
If \(A=\left[\begin{array}{cc}2 & -1 \\ 0 & 1\end{array}\right]\) and \(B=\left[\begin{array}{cc}1 & 0 \\ -1 & -1\end{array}\right]\), show that \((A+B)^{2}=A^{2}+A B+B A+B^{2} \neq A^{2}+2 A B+B^{2}\)
Step-by-Step Solution
Verified Answer
Adding the two given matrices A and B and then squaring the result gives a different matrix compared to squaring the matrices A and B individually, adding the results to the matrix resulting from multiplying A and B twice in different order. So, \( (A+B)^2 \ne A^2+2AB+B^2 \), proving that matrix multiplication is not commutative.
1Step 1: Compute A+B
Add the matrices A and B together element by element. Apply matrix addition, which is simply the addition of corresponding elements: \( C_{ij} = A_{ij} + B_{ij} \) where \( A_{ij} \) and \( B_{ij} \) are elements of matrices A and B respectively. This results in a new matrix C where \( C = A + B \)
2Step 2: Compute (A+B)^2
To compute \( (A+B)^2 \), you multiply matrix \( (A+B) \) by itself. Matrix multiplication requires sum of the products of corresponding elements from the rows of the first matrix and the columns of the second. The result is a new matrix.
3Step 3: Compute A^2, AB, BA and B^2
Compute these four elements. For A^2 and B^2, multiply each matrix by itself. For AB, multiply matrix A by matrix B. For BA, multiply matrix B by matrix A. It's crucial to remember that in matrix multiplication, the order of multiplication matters.
4Step 4: Compare Results
We then compare the matrix we got in step 2 \( (A+B)^2 \) with the sum of the matrices we got in step 3. This is how we demonstrate the inequality \( (A+B)^2 \ne A^2+2AB+B^2 \) proving that matrix multiplication is not commutative.
Key Concepts
Matrix AdditionMatrix MultiplicationNon-Commutative PropertyMatrix Square
Matrix Addition
Matrix addition is a process similar to how we add numbers, only here we add the corresponding elements of two matrices. For instance, if we have two matrices,
The result is a new matrix:
- \( A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \)
- \( B = \begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix} \)
The result is a new matrix:
- \( C = A + B = \begin{bmatrix} a_{11} + b_{11} & a_{12} + b_{12} \ a_{21} + b_{21} & a_{22} + b_{22} \end{bmatrix} \)
Matrix Multiplication
Matrix multiplication is a bit more involved than addition. It involves a row-by-column operation. To multiply two matrices,
The result is the matrix \( C \), where each element \( c_{ij} \) is calculated as follows:
- lets say \( A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \)
- and \( B = \begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix} \)
The result is the matrix \( C \), where each element \( c_{ij} \) is calculated as follows:
- \( c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} \)
Non-Commutative Property
In matrix algebra, the non-commutative property is crucial to understand. Unlike normal arithmetic—where, for example, multiplication is commutative (i.e., \( a \times b = b \times a \))—matrix multiplication does not share this property.
If we have two matrices A and B, the product \( AB \) is typically different from \( BA \). This signifies that the order in which you multiply matrices affects the result.
If we have two matrices A and B, the product \( AB \) is typically different from \( BA \). This signifies that the order in which you multiply matrices affects the result.
- In simpler terms, changing the order will lead often lead to different outcomes—\( AB eq BA \)
Matrix Square
Squaring a matrix entails multiplying the matrix by itself. If you have a matrix \( A \), computing \( A^2 \) means performing the operation \( A \times A \).
This involves matrix multiplication, and you follow the same process where each element \( c_{ij} \) is based on the multiplication of corresponding row and column elements.
This involves matrix multiplication, and you follow the same process where each element \( c_{ij} \) is based on the multiplication of corresponding row and column elements.
- For example, if \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), then \( A^2 = A \times A = \begin{bmatrix} a^2 + bc & ab + bd \ ac + cd & bc + d^2 \end{bmatrix} \)
Other exercises in this chapter
Problem 176
Given \(A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\) and \(B=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\), find \(A B\).
View solution Problem 177
Given \(A=\left[\begin{array}{ll}1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{l}0 \\ 1\end{array}\right]\), find \(A B\) and \(B A\).
View solution Problem 180
If \(A=\left[\begin{array}{cc}-1 & 2 \\ 2 & 3\end{array}\right], B=\left[\begin{array}{ll}3 & 0 \\ 1 & 1\end{array}\right]\). Verify that \((A+B)^{2}=A^{2}+A B+
View solution Problem 181
If \(A=\left[\begin{array}{cc}0 & 1 \\ 1 & 1\end{array}\right], B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\), show that \((A+B)(A-B) \neq A^{2}-
View solution