For the system of linear equations x + y + z = 6 , x + y + 7 z = 3 , x + 2 y + 3 z = 14 , which of the following is NOT true? [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
For the system of linear equations $x + y + z = 6, α x + β y + 7 z = 3, x + 2 y + 3 z = 14,$ which of the following is NOT true [2023]

1 If $α = β$ and $α \neq 7$ , then the system has a unique solution

2 There is a unique point $(α, β)$ on the line $x + 2 y + 18 = 0$ for which the system has infinitely many solutions

3 For every point $(α, β) \neq (7, 7)$ on the line $x - 2 y + 7 = 0$ , the system has infinitely many solutions

4 If $α = β = 7$ , then the system has no solution

Ans.
(3)

We have, $x + y + z = 6, α x + β y + 7 z = 3, x + 2 y + 3 z = 14$

Now, let
$A = [\begin{matrix} 1 & 1 & 1 \\ α & β & 7 \\ 1 & 2 & 3 \end{matrix}], B = [\begin{matrix} 6 \\ 3 \\ 14 \end{matrix}]$

$\Rightarrow$ $| A |$ $= 1 (3 β - 14) - 1 (3 α - 7) + 1 (2 α - β) = 2 β - α - 7$

If $α = β$ and $α \neq 7$ , then $| A | \neq 0$ , so the system has unique solutions.

Hence (1) is true.

If $α = β = 7$ , then $| A | = 0$ and the system of equations is:

$x + y + z = 6 ...(i)$

$7 x + 7 y + 7 z = 3 ...(ii)$

$x + 2 y + 3 z = 14 ...(iii)$

From equations (i) and (ii), two rows in matrix $A$ will get identical; therefore, the system has no solutions.

Hence (1) is true.

For every point $(α, β) \neq (7, 7)$ on the line $x - 2 y + 7 = 0$ ,

$| A | \neq 0$ . Hence, the system has unique solutions. Therefore, (3) is false.

$(α, β) \equiv (3, 3) does not satisfy x + 2 y + 18 = 0, then$

$Δ_{1} =$ $| \begin{matrix} 6 & 1 & 1 \\ 3 & β & 7 \\ 14 & 2 & 3 \end{matrix} |$ $= 6 (3 β - 14) - 1 (9 - 98) + 1 (6 - 14 β)$

$= 4 β + 11$

$Δ_{2} =$ $| \begin{matrix} 1 & 6 & 1 \\ α & 3 & 7 \\ 1 & 14 & 3 \end{matrix} |$ $= 1 (9 - 98) - 6 (3 α - 7) + 1 (14 α - 3)$

$= - 4 α - 50$

$Δ_{3} =$ $| \begin{matrix} 1 & 1 & 6 \\ α & β & 3 \\ 1 & 2 & 14 \end{matrix} |$ $= 1 (14 β - 6) - 1 (14 α - 3) + 6 (2 α - β)$

$= 8 β - 2 α - 3$

For infinite solution, $| A | = 0$ and $Δ_{1} = Δ_{2} = Δ_{3} = 0$ , then

$2 β - α = 7, β = \frac{- 11}{4}, α = \frac{- 25}{2}, and 8 β - 2 α = 3$

Hence, there is a unique $(α, β)$ on the line $x + 2 y + 18 = 0$ for which the system has infinitely many solutions.

Hence (2) is true.

Want Us to Email you About Special Offers & Updates?