Compute AB and BA
Compute AB and BA
Question:
Given
\[
A=\begin{bmatrix}3 & 2 \\ -1 & 0 \\ -1 & 1\end{bmatrix}, \quad
B=\begin{bmatrix}4 & 5 & 6 \\ 0 & 1 & 2\end{bmatrix}
\]
compute \(AB\) and \(BA\), wherever possible.
Solution:
Step 1: Check order
\(A\) is \(3 \times 2\), \(B\) is \(2 \times 3\)
- \(AB\): Possible → \(3×3\)
- \(BA\): Possible → \(2×2\)
Step 2: Compute \(AB\)
\[
AB =
\begin{bmatrix}
3 & 2 \\
-1 & 0 \\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
4 & 5 & 6 \\
0 & 1 & 2
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
3(4)+2(0) & 3(5)+2(1) & 3(6)+2(2) \\
-1(4)+0(0) & -1(5)+0(1) & -1(6)+0(2) \\
-1(4)+1(0) & -1(5)+1(1) & -1(6)+1(2)
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
12 & 17 & 22 \\
-4 & -5 & -6 \\
-4 & -4 & -4
\end{bmatrix}
\]
Step 3: Compute \(BA\)
\[
BA =
\begin{bmatrix}
4 & 5 & 6 \\
0 & 1 & 2
\end{bmatrix}
\begin{bmatrix}
3 & 2 \\
-1 & 0 \\
-1 & 1
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
4(3)+5(-1)+6(-1) & 4(2)+5(0)+6(1) \\
0(3)+1(-1)+2(-1) & 0(2)+1(0)+2(1)
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
12-5-6 & 8+0+6 \\
-1-2 & 2
\end{bmatrix}
=
\begin{bmatrix}
1 & 14 \\
-3 & 2
\end{bmatrix}
\]
Final Answer:
\[
AB =
\begin{bmatrix}
12 & 17 & 22 \\
-4 & -5 & -6 \\
-4 & -4 & -4
\end{bmatrix}
\]
\[
BA =
\begin{bmatrix}
1 & 14 \\
-3 & 2
\end{bmatrix}
\]
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