Q120 / 170 · Manipal (MET) 2018 · Mathematics
If $\left[\begin{array}{rrr}a & 2 & 3 \\ b & 5 & -1\end{array}\right] \times\left[\begin{array}{rr}1 & 2 \\ 3 & 4 \\ -1 & 1\end{array}\right]=\left[\begin{array}{rr}4 & 13 \\ 12 & 11\end{array}\right],(a, b)$ is
- A$(1,-2)$
- B$(-1,-4)$
- C$(1,3)$
- Correct answerD$(1,-4)$
Solution
$\begin{aligned} & {\left[\begin{array}{rrr}a & 2 & 3 \\ b & 5 & -1\end{array}\right]\left[\begin{array}{rr}1 & 2 \\ 3 & 4 \\ -1 & 1\end{array}\right]} \\ & =\left[\begin{array}{ll}a+6-3 & 2 a+8+3 \\ b+15+1 & 2 b+20-1\end{array}\right]\end{aligned}$
$\Rightarrow\left[\begin{array}{cc} a+3 & 2 a+11 \\ b+16 & 2 b+19 \end{array}\right]=\left[\begin{array}{cc} 4 & 13 \\ 12 & 11 \end{array}\right]$ (given) $\begin{aligned} & \Rightarrow \quad a+3=4 \Rightarrow a=1 \\ & \text { and } \quad b+16=12 \Rightarrow b=12-16=-4\end{aligned}$ $\therefore \quad(a, b)=(1,-4)$
$\Rightarrow\left[\begin{array}{cc} a+3 & 2 a+11 \\ b+16 & 2 b+19 \end{array}\right]=\left[\begin{array}{cc} 4 & 13 \\ 12 & 11 \end{array}\right]$ (given) $\begin{aligned} & \Rightarrow \quad a+3=4 \Rightarrow a=1 \\ & \text { and } \quad b+16=12 \Rightarrow b=12-16=-4\end{aligned}$ $\therefore \quad(a, b)=(1,-4)$
Hence, the correct answer is option: 'D'.