For the system of linear equations
$x+y+z=6$
$\alpha x+\beta y+7 z=3$
$x+2 y+3 z=14$
which of the following is NOT true ?
Solution
$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & 7 \\ 1 & 2 & 3\end{array}\right|$
<br/><br/>$$
\begin{aligned}
& =1(3 \beta-14)-1(3 \alpha-7)+1(2 \alpha-\beta) \\\\
& =3 \beta-14+7-3 \alpha+2 \alpha-\beta \\\\
& =2 \beta-\alpha-7
\end{aligned}
$$
<br/><br/>So, for $\alpha=\beta \neq 7, \Delta \neq 0$ so unique solution.
<br/><br/>$\alpha=\beta=7$, equation (i) and (ii) represent 2 parallel planes so no solution.
<br/><br/>If $\alpha-2 \beta+7=0$, but $(\alpha, \beta) \neq(7,7)$, then no solution.
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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