Answer :
Certainly! Let's step through the solution to show that the product of matrices \(A\) and \(B\) is equal to the product of matrices \(B\) and \(A\).
Given matrices:
[tex]\[ A = \begin{pmatrix} k & -1 \\ 1 & k \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 2 & 2 \\ -2 & 2 \end{pmatrix} \][/tex]
First, we will calculate the product \(AB\).
### Calculation of \(A \cdot B\)
[tex]\[ AB = \begin{pmatrix} k & -1 \\ 1 & k \end{pmatrix} \begin{pmatrix} 2 & 2 \\ -2 & 2 \end{pmatrix} \][/tex]
We perform matrix multiplication as follows:
[tex]\[ (AB)_{11} = k \cdot 2 + (-1) \cdot (-2) = 2k + 2 \][/tex]
[tex]\[ (AB)_{12} = k \cdot 2 + (-1) \cdot 2 = 2k - 2 \][/tex]
[tex]\[ (AB)_{21} = 1 \cdot 2 + k \cdot (-2) = 2 - 2k \][/tex]
[tex]\[ (AB)_{22} = 1 \cdot 2 + k \cdot 2 = 2k + 2 \][/tex]
Thus, the matrix \(AB\) is:
[tex]\[ AB = \begin{pmatrix} 2k + 2 & 2k - 2 \\ 2 - 2k & 2k + 2 \end{pmatrix} \][/tex]
### Calculation of \(B \cdot A\)
[tex]\[ BA = \begin{pmatrix} 2 & 2 \\ -2 & 2 \end{pmatrix} \begin{pmatrix} k & -1 \\ 1 & k \end{pmatrix} \][/tex]
We again perform matrix multiplication:
[tex]\[ (BA)_{11} = 2 \cdot k + 2 \cdot 1 = 2k + 2 \][/tex]
[tex]\[ (BA)_{12} = 2 \cdot (-1) + 2 \cdot k = 2k - 2 \][/tex]
[tex]\[ (BA)_{21} = -2 \cdot k + 2 \cdot 1 = 2 - 2k \][/tex]
[tex]\[ (BA)_{22} = -2 \cdot (-1) + 2 \cdot k = 2k + 2 \][/tex]
Thus, the matrix \(BA\) is:
[tex]\[ BA = \begin{pmatrix} 2k + 2 & 2k - 2 \\ 2 - 2k & 2k + 2 \end{pmatrix} \][/tex]
### Conclusion
Since both products yield the same matrix:
[tex]\[ AB = BA = \begin{pmatrix} 2k + 2 & 2k - 2 \\ 2 - 2k & 2k + 2 \end{pmatrix} \][/tex]
Thus, we have shown that [tex]\(A \cdot B = B \cdot A\)[/tex].
Given matrices:
[tex]\[ A = \begin{pmatrix} k & -1 \\ 1 & k \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 2 & 2 \\ -2 & 2 \end{pmatrix} \][/tex]
First, we will calculate the product \(AB\).
### Calculation of \(A \cdot B\)
[tex]\[ AB = \begin{pmatrix} k & -1 \\ 1 & k \end{pmatrix} \begin{pmatrix} 2 & 2 \\ -2 & 2 \end{pmatrix} \][/tex]
We perform matrix multiplication as follows:
[tex]\[ (AB)_{11} = k \cdot 2 + (-1) \cdot (-2) = 2k + 2 \][/tex]
[tex]\[ (AB)_{12} = k \cdot 2 + (-1) \cdot 2 = 2k - 2 \][/tex]
[tex]\[ (AB)_{21} = 1 \cdot 2 + k \cdot (-2) = 2 - 2k \][/tex]
[tex]\[ (AB)_{22} = 1 \cdot 2 + k \cdot 2 = 2k + 2 \][/tex]
Thus, the matrix \(AB\) is:
[tex]\[ AB = \begin{pmatrix} 2k + 2 & 2k - 2 \\ 2 - 2k & 2k + 2 \end{pmatrix} \][/tex]
### Calculation of \(B \cdot A\)
[tex]\[ BA = \begin{pmatrix} 2 & 2 \\ -2 & 2 \end{pmatrix} \begin{pmatrix} k & -1 \\ 1 & k \end{pmatrix} \][/tex]
We again perform matrix multiplication:
[tex]\[ (BA)_{11} = 2 \cdot k + 2 \cdot 1 = 2k + 2 \][/tex]
[tex]\[ (BA)_{12} = 2 \cdot (-1) + 2 \cdot k = 2k - 2 \][/tex]
[tex]\[ (BA)_{21} = -2 \cdot k + 2 \cdot 1 = 2 - 2k \][/tex]
[tex]\[ (BA)_{22} = -2 \cdot (-1) + 2 \cdot k = 2k + 2 \][/tex]
Thus, the matrix \(BA\) is:
[tex]\[ BA = \begin{pmatrix} 2k + 2 & 2k - 2 \\ 2 - 2k & 2k + 2 \end{pmatrix} \][/tex]
### Conclusion
Since both products yield the same matrix:
[tex]\[ AB = BA = \begin{pmatrix} 2k + 2 & 2k - 2 \\ 2 - 2k & 2k + 2 \end{pmatrix} \][/tex]
Thus, we have shown that [tex]\(A \cdot B = B \cdot A\)[/tex].
