Let the system of linear equations x + y + k z = 2 2 x + 3 y - z = 1 3 x + 4 y + 2 z = k have infinitely many solutions. Then the system ( k + 1 ) x + ( 2 k - 1 ) y = 7 ( 2 k + 1 ) x + ( k + 5 ) y = 10 has [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

Q.
Let the system of linear equations

$x + y + k z = 2$

$2 x + 3 y - z = 1$

$3 x + 4 y + 2 z = k$

have infinitely many solutions. Then the system

$(k + 1) x + (2 k - 1) y = 7$

$(2 k + 1) x + (k + 5) y = 10$

has [2023]

1 infinitely many solutions

2 unique solution satisfying $x - y = 1$

3 no solution

4 unique solution satisfying $x + y = 1$

Ans.
(4)

We have system of linear equations,

$x + y + k z = 2,$

$2 x + 3 y - z = 1,$

$3 x + 4 y + 2 z = k$

System has infinitely many solutions, then $| A | = 0 and (adj A) B = 0$

Here $A = [\begin{matrix} 1 & 1 & k \\ 2 & 3 & - 1 \\ 3 & 4 & 2 \end{matrix}]$

$\Rightarrow | A | = 0 \Rightarrow 1 (6 + 4) - 1 (4 + 3) + k (8 - 9) = 0$

$\Rightarrow 3 - k = 0, k = 3$

For $k = 3$ , $(adj A) B = 0$

Now, new system will be $4 x + 5 y = 7, 7 x + 8 y = 10$

After solving these two equations, we get $x = - 2, y = 3$

$\Rightarrow x - y = - 2 - (3) = - 5 and x + y = - 2 + 3 = 1$

Solution

Q.
Let the system of linear equations

$x + y + k z = 2$

$2 x + 3 y - z = 1$

$3 x + 4 y + 2 z = k$

have infinitely many solutions. Then the system

$(k + 1) x + (2 k - 1) y = 7$

$(2 k + 1) x + (k + 5) y = 10$

has [2023]

1 infinitely many solutions

2 unique solution satisfying $x - y = 1$

3 no solution

4 unique solution satisfying $x + y = 1$

Want Us to Email you About Special Offers & Updates?

Solution

Q. Let the system of linear equations x+y+kz=2 2x+3y-z=1 3x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k-1)y=7 (2k+1)x+(k+5)y=10 has [2023]

1 infinitely many solutions

2 unique solution satisfying x-y=1

3 no solution

4 unique solution satisfying x+y=1

Want Us to Email you About Special Offers & Updates?

Q.
Let the system of linear equations

$x + y + k z = 2$

$2 x + 3 y - z = 1$

$3 x + 4 y + 2 z = k$

have infinitely many solutions. Then the system

$(k + 1) x + (2 k - 1) y = 7$

$(2 k + 1) x + (k + 5) y = 10$

has [2023]

2 unique solution satisfying $x - y = 1$

4 unique solution satisfying $x + y = 1$