If \(A=\begin{bmatrix}1 & -2 & 3 \\ -4 & 2 & 4\end{bmatrix}\) and \(B=\begin{bmatrix}1 & -2 \\ 3 & -4 \\ 2 & 4\end{bmatrix}\) then product AB is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}1 & 18 \\ 10 & 16\end{bmatrix}\)
A is of order 2 × 3 and B is of order 3 × 2. The number of columns of A equals the number of rows of B, so AB exists and has order 2 × 2.
Each entry is (row of A) · (column of B):
\((AB)_{11} = 1(1) + (-2)(3) + 3(2) = 1\)
\((AB)_{12} = 1(-2) + (-2)(-4) + 3(4) = 18\)
\((AB)_{21} = (-4)(1) + 2(3) + 4(2) = 10\)
\((AB)_{22} = (-4)(-2) + 2(-4) + 4(4) = 16\)
So
\( AB = \begin{bmatrix}1 & 18 \\ 10 & 16\end{bmatrix} \)
Why the other options are incorrect
Option (1) Not possible — Incorrect. A has 3 columns and B has 3 rows, so the product is defined.
Option (2) \(\begin{bmatrix}1 & -6 & 6 \\ 8 & -8 & 16\end{bmatrix}\) — Incorrect. It is of order 2 × 3, but AB must be of order 2 × 2.
Option (3) \(\begin{bmatrix}9 \\ 19 \\ 15\end{bmatrix}\) — Incorrect. It is of order 3 × 1, but AB must be of order 2 × 2.