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Matrices and Determinants question

2018 · Shift 2 · Q21
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  2. /JEE Advanced
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  4. /Matrices and Determinants
  5. /2018 · Shift 2 · Q21

Matrices and Determinants question

2018 · Shift 2 · Q21

JEE AdvancedMathematicsMatrices and DeterminantsMultiple correct+4 / −1
Let S be the set of all column matrices [b1b2b3]\left[ {\begin{matrix} {{b_1}} \\ {{b_2}} \\ {{b_3}} \\ \end{matrix} } \right]​b1​b2​b3​​​ such that b1,b2,b3∈R{b_1},{b_2},{b_3} \in Rb1​,b2​,b3​∈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}​−x+2y+5z=b1​2x−4y+3z=b2​x−2y+2z=b3​​ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each [b1b2b3]\left[ {\begin{matrix} {{b_1}} \\ {{b_2}} \\ {{b_3}} \\ \end{matrix} } \right]​b1​b2​b3​​​ ∈\in∈ S?
  1. A
    x+2y+3z=b1x + 2y + 3z = {b_1}x+2y+3z=b1​,  4y+5z=b2\,4y + 5z = {b_2}4y+5z=b2​ and x+2y+6z=b3x + 2y + 6z = {b_3}x+2y+6z=b3​
  2. B
    x+y+3z=b1x + y + 3z = {b_1}x+y+3z=b1​, 5x+2y+6z=b25x + 2y + 6z = {b_2}5x+2y+6z=b2​ and −2x−y−3z=b3- 2x - y - 3z = {b_3}−2x−y−3z=b3​
  3. C
    −x+2y−5z=b1- x + 2y - 5z = {b_1}−x+2y−5z=b1​,  2x−4y+10z=b2\,2x - 4y + 10z = {b_2}2x−4y+10z=b2​ and x−2y+5z=b3x - 2y + 5z = {b_3}x−2y+5z=b3​
  4. D
    x+2y+5z=b1x + 2y + 5z = {b_1}x+2y+5z=b1​, 2x+3z=b22x + 3z = {b_2}2x+3z=b2​ and x+4y−5z=b3x + 4y - 5z = {b_3}x+4y−5z=b3​
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