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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 $100+\frac{2 \alpha+11 \beta}{\gamma}=$
Options:
$\left(\begin{array}{ccc}7 & 5 & \alpha \\ \beta & 2 & 11 \\ 3 & \gamma & 1\end{array}\right)\left(\begin{array}{l}1 \\ 3 \\ 2\end{array}\right)=\left(\begin{array}{c}\alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma\end{array}\right)$ then $100+\frac{2 \alpha+11 \beta}{\gamma}=$
Solution:
1446 Upvotes
Verified Answer
The correct answer is:
27
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,
$$
\begin{aligned}
& \alpha=-13, \beta=9, \gamma=-1, \text { then, } 100+\frac{2 \alpha+11 \beta}{\gamma} \\
& \Rightarrow 100-73=27
\end{aligned}
$$
$\left[\begin{array}{ccc}7 & 5 & \lambda \\ \beta & 2 & 11 \\ 3 & \gamma & 1\end{array}\right]\left[\begin{array}{l}1 \\ 3 \\ 2\end{array}\right]=\left[\begin{array}{c}\alpha+\beta \\ -2 \alpha+\beta-2 \gamma \\ \alpha+2 \beta+3 \gamma\end{array}\right]$
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,
$$
\begin{aligned}
& \alpha=-13, \beta=9, \gamma=-1, \text { then, } 100+\frac{2 \alpha+11 \beta}{\gamma} \\
& \Rightarrow 100-73=27
\end{aligned}
$$
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