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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 393
Question 393 of 472
Medium
MCQ
Consider the following system of equations : x + 2y - 3z = a 2x + 6y - 11z = b x - 2y + 7z = c, where a, b and c are real constants. Then the system of equations :
A
has no solution for all a, b and c
B
has a unique solution when 5a = 2b + c
C
has infinite number of solutions when 5a = 2b + c
D
has a unique solution for all a, b and c
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Official solution
\[D = \left| {
(
1
2
−
3
2
6
−
11
1
−
2
7
)
\begin{pmatrix} 1 & 2 & { - 3} \\ 2 & 6 & { - 11} \\ 1 & { - 2} & 7 \\ \end{pmatrix}
1
2
1
2
6
−
2
−
3
−
11
7
} \right|\] = 20 - 2(25) -3(-10) = 20 - 50 + 30 = 0 \[{D_1} = \left| {
(
a
2
−
3
b
6
−
11
c
−
2
7
)
\begin{pmatrix} a & 2 & { - 3} \\ b & 6 & { - 11} \\ c & { - 2} & 7 \\ \end{pmatrix}
a
b
c
2
6
−
2
−
3
−
11
7
} \right|\] = 20a - 2(7b + 11c) -3(-2b - 6c) = 20a - 14b - 22c + 6b +18c = 20a - 8b - 4c = 4(5a - 2b - c) \[{D_2} = \left| {
(
1
a
−
3
2
b
−
11
1
c
7
)
\begin{pmatrix} 1 & a & { - 3} \\ 2 & b & { - 11} \\ 1 & c & 7 \\ \end{pmatrix}
1
2
1
a
b
c
−
3
−
11
7
} \right|\] = 7b + 11c - a(25) -3(2c - b) = 7b + 11c - 25a - 6c + 3b = -25a + 10b + 5c = -5(5a - 2b - c) \[{D_3} = \left| {
(
1
2
a
2
6
b
1
−
2
c
)
\begin{pmatrix} 1 & 2 & a \\ 2 & 6 & b \\ 1 & { - 2} & c \\ \end{pmatrix}
1
2
1
2
6
−
2
a
b
c
} \right|\] = 6c + 2b - 2(2c - b) - 10a = -10a + 4b + 2c = -2(5a - 2b - c) for infinite solution
D
=
D
1
=
D
2
=
D
3
=
0
D = {D_1} = {D_2} = {D_3} = 0
D
=
D
1
=
D
2
=
D
3
=
0
\[ \Rightarrow \] 5a = 2b + c
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