If the system of linear equations
$$ \begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned} $$
has infinitely many solutions, then the value of $22 \beta-9 \alpha$ is :
Solution
<p>$$\begin{aligned}
&\begin{aligned}
& 3 x+y+\beta z=3 \\
& 2 x+\alpha y-z=-3 \\
& x+2 y+z=4
\end{aligned}\\
&\text { has infinite solution }\\
&\begin{aligned}
& \Rightarrow \Delta=0, \Delta_1=\Delta_2=\Delta_3 \\
& \Delta=0 \Rightarrow\left|\begin{array}{ccc}
3 & 1 & \beta \\
2 & \alpha & -1 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Delta_2=0 \Rightarrow\left|\begin{array}{ccc}
3 & 3 & \beta \\
2 & -3 & -1 \\
1 & 4 & 1
\end{array}\right|=0 \\
& \Rightarrow 3(-3+4)-3(2+1)+\beta(8+3)=0 \\
& \Rightarrow 3-9+11 \beta=0 \\
& \Rightarrow \quad \beta=\frac{6}{11} \\
& \Delta_3=0 \Rightarrow\left|\begin{array}{ccc}
3 & 1 & 3 \\
2 & \alpha & -3 \\
1 & 2 & 4
\end{array}\right|=0 \\
& \Rightarrow \quad 3(4 \alpha+6)-1(8+3)+3(4-\alpha)=0 \\
& \quad 12 \alpha+18-11+12-3 \alpha=0 \\
& \quad 9 \alpha=-19 \\
& \quad \alpha=\frac{-19}{9}
\end{aligned}
\end{aligned}$$</p>
<p>$\therefore \quad 22 \beta-9 \alpha=31$</p>
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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