Matrices — JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 Question with Solution

Consider the system of equations

x - 2 y + 3 z = - 1 - x + y - 2 z = k x - 3 y + 4 z = 1.

STATEMENT-1 : The system of equations has no solution for

k \neq = 3

.

and

STATEMENT-2 : The determinant

1 - 1 1 3 - 2 4 - 1 k 1 \neq = 0

, for

k \neq = 3

(A)STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1✓ Correct
(B)STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
(C)STATEMENT-1 is True, STATEMENT-2 is False
(D)STATEMENT-1 is False, STATEMENT-2 is True

Answer: (A)

Solution

The coefficient matrix determinant:

D = 1 - 1 1 - 2 1 - 3 3 - 2 4 = 1 (4 - 6) + 2 (- 4 + 2) + 3 (3 - 1) = - 2 - 4 + 6 = 0

Since

D = 0

, the system either has no solution or infinitely many solutions.

For no solution, we need at least one of

D_{x}, D_{y}, D_{z} \neq = 0

D_{x} = - 1 k 1 - 2 1 - 3 3 - 2 4 = - 1 (4 - 6) + 2 (4 k + 2) + 3 (- 3 k - 1)

= 2 + 8 k + 4 - 9 k - 3 = 3 - k

D_{x} = 0

when

k = 3

, and

D_{x} \neq = 0

when

k \neq = 3

.

Therefore, for

k \neq = 3

, the system has no solution. Statement 1 is TRUE.

Statement 2: The determinant given is:

1 - 1 1 3 - 2 4 - 1 k 1 = 1 (- 2 - 4 k) - 3 (- 1 - k) + (- 1) (- 4 + 2)

= - 2 - 4 k + 3 + 3 k + 2 = 3 - k

This equals zero when

k = 3

and is non-zero when

k \neq = 3

. Statement 2 is TRUE.

Statement 2 (showing a related determinant is non-zero for

k \neq = 3

) correctly explains why the system has no solution for

k \neq = 3

.

Answer: (A)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option