Let α, β, γ be real numbers. If ( 7 5 α β 2 11 3 γ 1 ) ( 1 3 2 )= ( α+β -2 α+β-2 γ α+2 β+3 γ ) then 100+(2 α+11 β)/(γ)=

Correct answer: (A) 27. We have [ 7 5 λ β 2 11 3 γ 1 ] [ 1 3 2 ]= [ α+β -2 α+β-2 γ α+2 β+3 γ ] On solving we get, α=-13, β=9, γ=-1, then, 100+(2 α+11 β)/(γ) ⇒ 100-73=27 On solving we get, α=-13, β=9, γ=-1, then, 100+(2 α+11 β)/(γ) ⇒ 100-73=27

Q46 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

Let $\alpha, \beta, \gamma$ be real numbers. If $\left(\begin{array}{ccc}7 & 5 & \alpha \\ \beta & 2 & 11 \\ 3 & \gamma & 1\end{array}\right)\left(\begin{array}{l}1 \\ 3 \\ 2\end{array}\right)=\left(\begin{array}{c}\alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma\end{array}\right)$ then $100+\frac{2 \alpha+11 \beta}{\gamma}=$

Correct answer
A
27
B
-25
C
225
D
-227

Solution

We have $\left[\begin{array}{ccc}7 & 5 & \lambda \\ \beta & 2 & 11 \\ 3 & \gamma & 1\end{array}\right]\left[\begin{array}{l}1 \\ 3 \\ 2\end{array}\right]=\left[\begin{array}{c}\alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma\end{array}\right]$ On solving we get, $$ \begin{aligned} & \alpha=-13, \beta=9, \gamma=-1, \text { then, } 100+\frac{2 \alpha+11 \beta}{\gamma} \\ & \Rightarrow 100-73=27 \end{aligned} $$ On solving we get, $$ \begin{aligned} & \alpha=-13, \beta=9, \gamma=-1, \text { then, } 100+\frac{2 \alpha+11 \beta}{\gamma} \\ & \Rightarrow 100-73=27 \end{aligned} $$

Hence, the correct answer is option: 'A'.

Solution

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