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MathematicsMultiple correct+4 / -2Question 39 of 54

JEE Advanced 2018 · Paper 2, Mathematics question 3

Let SS be the set of all column matrices [b1b2b3]\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} such that b1,b2,b3∈Rb_1, b_2, b_3 \in \mathbb{R} and the system of equations (in real variables)

−x+2y+5z=b12x−4y+3z=b2x−2y+2z=b3\begin{aligned} -x + 2y + 5z &= b_1 \\ 2x - 4y + 3z &= b_2 \\ x - 2y + 2z &= b_3 \end{aligned}

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each [b1b2b3]∈S\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \in S?

Choose one or more options
Answer and worked solution

Official answer: A, D

Describe SS. Look for a row dependency a r1+b r2+c r3=0a\,r_1+b\,r_2+c\,r_3=0 in the original system:
−a+2b+c=0,5a+3b+2c=0  ⟹  b=7a, c=−13a.-a+2b+c=0,\quad5a+3b+2c=0\implies b=7a,\ c=-13a.
So the system is solvable exactly when b1+7b2−13b3=0b_1+7b_2-13b_3=0. SS is this plane.

  • (A) det⁡(123045126)=14+10−12=12≠0\det\begin{pmatrix}1&2&3\\0&4&5\\1&2&6\end{pmatrix}=14+10-12=12\ne0. Always solvable. True.
  • (B) The determinant is 0, and the rows satisfy r1+r2+3r3=0r_1+r_2+3r_3=0, so it needs b1+b2+3b3=0b_1+b_2+3b_3=0. This is a different plane from SS. False.
  • (C) All rows are multiples of (1,−2,5)(1,-2,5), which needs b2=−2b1b_2=-2b_1 and b3=−b1b_3=-b_1. That is a line, not all of SS. False.
  • (D) det⁡(12520314−5)=−12+26+40=54≠0\det\begin{pmatrix}1&2&5\\2&0&3\\1&4&-5\end{pmatrix}=-12+26+40=54\ne0. Always solvable. True.

(A), (D)

Topics:Systems of linear equationsDeterminants and properties
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