Skip to main contentSkip to solution

If A=[2−13−345]A=\begin{bmatrix}2 & -1 & 3 \\ -3 & 4 & 5\end{bmatrix} and B=[213523]B=\begin{bmatrix}2 & 1 \\ 3 & 5 \\ 2 & 3\end{bmatrix} and BA=(bij)BA=(b_{ij}), then b22+b31b_{22}+b_{31} is :

Solution

✅ Correct Option: 1

BABA is a 3×33\times3 matrix; compute only the entries needed.

b22=b_{22} = (row 2 of BB)·(column 2 of AA) =3(−1)+5(4)=17= 3(-1)+5(4) = 17.

b31=b_{31} = (row 3 of BB)·(column 1 of AA) =2(2)+3(−3)=−5= 2(2)+3(-3) = -5.

Sum =17−5=12= 17-5 = 12.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question