Which of the following are correct in respect of the system of equation x + y + z = 8, x – y + 2z = 6 and 3x – y + 5z = k?

1. They have no solution, if k = 15

2. They have infinitely many solutions, if k = 20

3. They have unique solution, if k = 25

Select the correct answer using the code given below:This question was previously asked in

NDA (Held On: 18 Sept 2016) Maths Previous Year paper

Option 1 : 1 and 2 only

General Principles of Business Management - 1

3787

15 Questions
15 Marks
15 Mins

__Concept__

Let the system of equations be,

a_{1}x + b_{1}y + c_{1}z = d_{1}

a_{2}x + b_{2}y + c_{2}z = d_{2}

a_{3}x + b_{3}y + c_{3}z = d_{3}

\({\rm{\;}} ⇒ \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A^{-1} B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}{\rm{\;B}}\)

⇒ If det (A) ≠ 0, system is consistent having **unique solution**.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with **infinitely many solutions**.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is **inconsistent (no solution)**

__Calculation:__

Given system of equation x + y + z = 8, x – y + 2z = 6 and 3x – y + 5z = k

\(⇒ \left[ {\begin{array}{*{20}{c}} 1&1&1\\ 1&{ - 1}&2\\ 3&{ - 1}&5 \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{k}} \end{array}} \right]\)

⇒ AX = B

Determinant of A = |A| = 1 (-5 + 2) – 1 (5 – 6) + 1 (-1 + 3) = -3 + 1 + 2 = 0

So we can say that equations have either infinite solution or no solution.

A unique solution is not possible.

∴ Statement 3 is wrong.

We have adj A = \(\left[ {\begin{array}{*{20}{c}} -3&-6&3\\ 1&{2}&2\\ 2&{ 4}&-2 \end{array}} \right]\)

If K = 15,

B = \(\left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{15}} \end{array}} \right]\)

Now (adj A). B will be

\(\left[ {\begin{array}{*{20}{c}} -3&-6&3\\ 1&{2}&2\\ 2&{ 4}&-2 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{15}} \end{array}} \right] \ne 0 \)

⇒ no solution

If K = 20,

B = \(\left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{20}} \end{array}} \right]\)

Now (adj A). B will be

\(\left[ {\begin{array}{*{20}{c}} -3&-6&3\\ 1&{2}&2\\ 2&{ 4}&-2 \end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}} 8\\ 6\\ {\rm{20}} \end{array}} \right] = 0 \)

⇒ infinitely many solutiions

Hence Option 1 is correct.