Given the matrices: \[ A=\left(\begin{array}{ccc} 2 & 0 & 1 \\ 3 & 0 & 0 \\ 5 & 1 & 1 \end{array}\right) ; \quad B=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 0 \end{array}\right) \] Calculate: (1) \( A \times B ; \) (2) \( B \times A \) and show shat \( A B \neq B A \)

To determine whether \( A \times B = B \times A \), let's compute both products step by step. Given the matrices: \[ A = \begin{pmatrix} 2 & 0 & 1 \\ 3 & 0 & 0 \\ 5 & 1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 0 \end{pmatrix} \] ### 1. Calculating \( A \times B \) The product \( AB \) is found by multiplying rows of \( A \) with columns of \( B \): \[ AB_{i,j} = \sum_{k=1}^{3} A_{i,k} \times B_{k,j} \] **First Row of \( A \) and All Columns of \( B \):** \[ \begin{aligned} AB_{1,1} &= (2 \times 1) + (0 \times 1) + (1 \times 1) = 2 + 0 + 1 = 3 \\ AB_{1,2} &= (2 \times 0) + (0 \times 2) + (1 \times 1) = 0 + 0 + 1 = 1 \\ AB_{1,3} &= (2 \times 1) + (0 \times 1) + (1 \times 0) = 2 + 0 + 0 = 2 \\ \end{aligned} \] **Second Row of \( A \) and All Columns of \( B \):** \[ \begin{aligned} AB_{2,1} &= (3 \times 1) + (0 \times 1) + (0 \times 1) = 3 + 0 + 0 = 3 \\ AB_{2,2} &= (3 \times 0) + (0 \times 2) + (0 \times 1) = 0 + 0 + 0 = 0 \\ AB_{2,3} &= (3 \times 1) + (0 \times 1) + (0 \times 0) = 3 + 0 + 0 = 3 \\ \end{aligned} \] **Third Row of \( A \) and All Columns of \( B \):** \[ \begin{aligned} AB_{3,1} &= (5 \times 1) + (1 \times 1) + (1 \times 1) = 5 + 1 + 1 = 7 \\ AB_{3,2} &= (5 \times 0) + (1 \times 2) + (1 \times 1) = 0 + 2 + 1 = 3 \\ AB_{3,3} &= (5 \times 1) + (1 \times 1) + (1 \times 0) = 5 + 1 + 0 = 6 \\ \end{aligned} \] **Resulting Matrix \( AB \):** \[ AB = \begin{pmatrix} 3 & 1 & 2 \\ 3 & 0 & 3 \\ 7 & 3 & 6 \end{pmatrix} \] ### 2. Calculating \( B \times A \) Similarly, the product \( BA \) is found by multiplying rows of \( B \) with columns of \( A \): \[ BA_{i,j} = \sum_{k=1}^{3} B_{i,k} \times A_{k,j} \] **First Row of \( B \) and All Columns of \( A \):** \[ \begin{aligned} BA_{1,1} &= (1 \times 2) + (0 \times 3) + (1 \times 5) = 2 + 0 + 5 = 7 \\ BA_{1,2} &= (1 \times 0) + (0 \times 0) + (1 \times 1) = 0 + 0 + 1 = 1 \\ BA_{1,3} &= (1 \times 1) + (0 \times 0) + (1 \times 1) = 1 + 0 + 1 = 2 \\ \end{aligned} \] **Second Row of \( B \) and All Columns of \( A \):** \[ \begin{aligned} BA_{2,1} &= (1 \times 2) + (2 \times 3) + (1 \times 5) = 2 + 6 + 5 = 13 \\ BA_{2,2} &= (1 \times 0) + (2 \times 0) + (1 \times 1) = 0 + 0 + 1 = 1 \\ BA_{2,3} &= (1 \times 1) + (2 \times 0) + (1 \times 1) = 1 + 0 + 1 = 2 \\ \end{aligned} \] **Third Row of \( B \) and All Columns of \( A \):** \[ \begin{aligned} BA_{3,1} &= (1 \times 2) + (1 \times 3) + (0 \times 5) = 2 + 3 + 0 = 5 \\ BA_{3,2} &= (1 \times 0) + (1 \times 0) + (0 \times 1) = 0 + 0 + 0 = 0 \\ BA_{3,3} &= (1 \times 1) + (1 \times 0) + (0 \times 1) = 1 + 0 + 0 = 1 \\ \end{aligned} \] **Resulting Matrix \( BA \):** \[ BA = \begin{pmatrix} 7 & 1 & 2 \\ 13 & 1 & 2 \\ 5 & 0 & 1 \end{pmatrix} \] ### 3. Comparing \( AB \) and \( BA \) \[ AB = \begin{pmatrix} 3 & 1 & 2 \\ 3 & 0 & 3 \\ 7 & 3 & 6 \end{pmatrix}, \quad BA = \begin{pmatrix} 7 & 1 & 2 \\ 13 & 1 & 2 \\ 5 & 0 & 1 \end{pmatrix} \] Clearly, \( AB \) is **not equal** to \( BA \). For instance: - \( AB_{1,1} = 3 \neq 7 = BA_{1,1} \) - \( AB_{2,1} = 3 \neq 13 = BA_{2,1} \) - \( AB_{3,1} = 7 \neq 5 = BA_{3,1} \) This demonstrates that matrix multiplication is **not commutative**, i.e., \( A \times B \neq B \times A \).