Consider the system of linear equations
$-$x + y + 2z = 0
3x $-$ ay + 5z = 1
2x $-$ 2y $-$ az = 7
Let S1 be the set of all a$\in$R for which the system is inconsistent and S2 be the set of all a$\in$R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then
Solution
$$\Delta = \left| {\matrix{
{ - 1} & 1 & 2 \cr
3 & { - a} & 5 \cr
2 & { - 2} & { - a} \cr
} } \right|$$<br><br>$= - 1({a^2} + 10) - 1( - 3a - 10) + 2( - 6 + 2a)$<br><br>$= - {a^2} - 10 + 3a + 10 - 12 + 4a$<br><br>$\Delta = - {a^2} + 7a - 12$<br><br>$\Delta = - [{a^2} - 7a + 12]$<br><br>$\Delta = - [(a - 3)(a - 4)]$<br><br>$${\Delta _1} = \left| {\matrix{
0 & 1 & 2 \cr
1 & { - a} & 5 \cr
7 & { - 2} & { - a} \cr
} } \right|$$<br><br><br>$= a + 35 - 4 + 14a$<br><br>= $15a + 31$<br><br>Now, ${\Delta _1} = 15a + 31$<br><br>For inconsistent $\Delta$ = 0 $\therefore$ a = 3, a = 4 and for a = 3 and 4, $\Delta$<sub>1</sub> $\ne$ 0<br><br>n(S<sub>1</sub>) = 2<br><br>For infinite solution : $\Delta$ = 0 and $\Delta$<sub>1</sub> = $\Delta$<sub>2</sub> = $\Delta$<sub>3</sub> = 0<br><br>Not possible<br><br>$\therefore$ n(S<sub>2</sub>) = 0
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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