Find all real values of $k$ for which the system $x + y + z = 2$ , $2 x + 3 y + k z = 5$ , $3 x + 4 y + (k + 1) z = 7$ has infinitely many solutions.

hard 5 marks Class 12 · Mathematics · Matrices

Answer

k can be any real number (infinitely many solutions for all k)

The correct answer is k can be any real number (infinitely many solutions for all k) — Conclusion: the system possesses infinitely many solutions for all real values of $k$ .

Step-by-step solution

Write the coefficient matrix $A = 123134 1 k k + 1$ and the augmented matrix $[A ∣ B] = 123134 1 k k + 1 ∣ ∣ ∣ 257$ .
Compute the determinant of $A$ : $det A = 1 \cdot (3 (k + 1) - 4 k) - 1 \cdot (2 (k + 1) - 3 k) + 1 \cdot (2 \cdot 4 - 3 \cdot 3) = (3 - k) + (k - 2) - 1 = 0$ . So $det A = 0$ for every $k$ ; the rows of $A$ are linearly dependent.
Observe that the third row equals the sum of the first two rows: $R_{3} = R_{1} + R_{2}$ because $(1 + 2, 1 + 3, 1 + k) = (3, 4, k + 1)$ .
Check the same relation for the RHS: $2 + 5 = 7$ , which matches the third equation’s constant term. Hence the augmented matrix has the same dependency.
Therefore $rank A = rank [A ∣ B] = 2 < 3$ , giving infinitely many solutions for any real $k$ .
Conclusion: the system possesses infinitely many solutions for all real values of $k$ .

Common mistake: Ignoring the consistency condition and assuming that

det A = 0

alone guarantees infinite solutions.

Concept tested: Condition for infinite solutions using rank and determinant

Published 29 August 2026 · How this solution was prepared and checked

Find all real values of k for which the system x+y+z=2, 2x+3y+kz=5, 3x+4y+(k+1)z=7 has infinitely many solutions.

Step-by-step solution

Related questions

Find all real values of $k$ for which the system $x + y + z = 2$ , $2 x + 3 y + k z = 5$ , $3 x + 4 y + (k + 1) z = 7$ has infinitely many solutions.