Problem 83 - College Algebra

Question

If \(A B=-B A,\) then \(A\) and \(B\) are said to be anticommutative. Are \(A=\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\) anticommutative?

Step-by-Step Solution

Answer

Yes, matrices A and B are anticommutative because the product AB equals the negation of the product BA.

Step 1: Multiply A and B

The first operation is to multiply matrices \(A\) and \(B\). This will provide the matrix \(AB\). Use standard matrix multiplication techniques. Each element in the product matrix is the dot product of its corresponding row in the first matrix and column in the second matrix.\r\n \( A \cdot B = \left[ \begin{array}{cc} 0 & -1 \ 1 & 0 \end{array} \right] \cdot \left[ \begin{array}{cc} 1 & 0 \ 0 & -1 \end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \ 1 & 0 \end{array} \right] = AB \)

Step 2: Multiply B and A

Next, do the multiplication in the opposite order to get \(BA\). Note that in general, the result will be different due to the non-commutative property of matrix multiplication: \( BA = \left[ \begin{array}{cc} 1 & 0 \ 0 & -1 \end{array} \right] \cdot \left[ \begin{array}{cc} 0 & -1 \ 1 & 0 \end{array} \right] = \left[ \begin{array}{cc} 0 & -1 \ -1 & 0 \end{array} \right] = BA. \)

Step 3: Check if AB equals -BA

Now we negate \(BA\) and compare with \(AB\) to see if \( A \) and \( B \) are anticommutative. Negating \( BA \) we have \( -BA = \left[ \begin{array}{cc} 0 & 1 \ 1 & 0 \end{array} \right] \). Comparing with the result of step 1 (i.e. \( AB \)), we see that they are equal, thus \( A \) and \( B \) are indeed anticommutative.