If \( A=\begin{bmatrix}2 & -1 & 3\\-3 & 4 & 5\end{bmatrix} \) and \( B=\begin{bmatrix}2 & 1\\3 & 5\\2 & 3\end{bmatrix} \) and \( BA=(b_{ij}) \), then \( b_{22}+b_{31} \) is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 12
B is 3 × 2 and A is 2 × 3, so BA is 3 × 3. The entry \( b_{ij} \) is row i of B multiplied into column j of A; only the two required entries need to be computed.
\( \displaystyle b_{22}=(3)(-1)+(5)(4) \)
\( \displaystyle =-3+20=17 \)
\( \displaystyle b_{31}=(2)(2)+(3)(-3) \)
\( \displaystyle =4-9=-5 \)
\( \displaystyle b_{22}+b_{31}=17-5=12 \)
Why the other options are incorrect
Option (2) 22 — Incorrect. 22 comes from taking \( b_{31}=+5 \); the product (3)(−3) = −9 makes \( b_{31} \) negative.
Option (3) 17 — Incorrect. 17 is \( b_{22} \) alone; \( b_{31}=-5 \) still has to be added.
Option (4) 21 — Incorrect. The sum is 17 + (−5) = 12, not 21.