Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 384 Question 384 of 472 Medium MCQ
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
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\[\Delta = \left| { ( − 1 1 2 3 − a 5 2 − 2 − a ) \begin{pmatrix} { - 1} & 1 & 2 \\
3 & { - a} & 5 \\
2 & { - 2} & { - a} \\
\end{pmatrix} − 1 3 2 1 − a − 2 2 5 − a } \right|\]
\[ = - 1({a^2} + 10) - 1( - 3a - 10) + 2( - 6 + 2a)\]
= − a 2 − 10 + 3 a + 10 − 12 + 4 a = - {a^2} - 10 + 3a + 10 - 12 + 4a = − a 2 − 10 + 3 a + 10 − 12 + 4 a
\[\Delta = - {a^2} + 7a - 12\]
\[\Delta = - [{a^2} - 7a + 12]\]
\[\Delta = - [(a - 3)(a - 4)]\]
\[{\Delta _1} = \left| { ( 0 1 2 1 − a 5 7 − 2 − a ) \begin{pmatrix} 0 & 1 & 2 \\
1 & { - a} & 5 \\
7 & { - 2} & { - a} \\
\end{pmatrix} 0 1 7 1 − a − 2 2 5 − a } \right|\]
= a + 35 − 4 + 14 a = a + 35 - 4 + 14a = a + 35 − 4 + 14 a
= 15 a + 31 15a + 31 15 a + 31
Now, \[{\Delta _1} = 15a + 31\]
For inconsistent \[\Delta\] = 0 \[\therefore\] a = 3, a = 4 and for a = 3 and 4, \[\Delta\]1 \[\ne\] 0
n(S1) = 2
For infinite solution : \[\Delta\] = 0 and \[\Delta\]1 = \[\Delta\]2 = \[\Delta\]3 = 0
Not possible
\[\therefore\] n(S2) = 0 Answer the question to unlock