Question 447 of 472MediumMCQ
If the system of linear equations
\[x−2y+z=−42x+αy+3z=53x−y+βz=3\]
has infinitely many solutions, then \[12 \alpha+13 \beta\] is equal to Solutions · 1
Official solution
\[\begin{aligned}
& D=\left|\[\begin{array}{|ccccccc|}1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta\end{array}\end{aligned}\]\right| \\
\& =1(\alpha \beta+3)+2(2 \beta-9)+1(-2-3 \alpha) \\
\& =\alpha \beta+3+4 \beta-18-2-3 \alpha
\]
For infinite solutions \[\mathrm{D}=0, \mathrm{D}_1=0, \mathrm{D}_2=0\] and
\[D3=0D=0\]
\[\alpha \beta-3 \alpha+4 \beta=17\] ..... (1)
\[\begin{aligned}
& \mathrm{D}_1=\left|\[\begin{array}{|ccccccc|}-4 & -2 & 1 \\ 5 & \alpha & 3 \\ 3 & -1 & \beta\end{array}\end{aligned}\]\right|=0 \\
\& \mathrm{D}\_2=\left|\[123−45313β\]\right|=0 \\
\& \Rightarrow 1(5 \beta-9)+4(2 \beta-9)+1(6-15)=0 \\
\& 13 \beta-9-36-9=0
\]
\[13 \beta=54, \beta=\frac{54}{13}\] put in (1)
\[\frac{54}{13} \alpha-3 \alpha+4\left(\frac{54}{13}\right)=17\]
\[54 \alpha-39 \alpha+216=221\]
\[15 \alpha=5 \quad \alpha=\frac{1}{3}\]
Now, \[12 \alpha+13 \beta=12 \cdot \frac{1}{3}+13 \cdot \frac{54}{13}\]
=4+54=58 Answer the question to unlock