The system of equations kx + y + z = 1, x + ky + z = k and x + y + zk = k2 has no solution if k is equal to :
Solution
$$D = \left| {\matrix{
k & 1 & 1 \cr
1 & k & 1 \cr
1 & 1 & k \cr
} } \right| = 0$$<br><br>$\Rightarrow k({k^2} - 1) - (k - 1) + (1 - k) = 0$<br><br>$\Rightarrow (k - 1)({k^2} + k - 1 - 1) = 0$<br><br>$\Rightarrow (k - 1)({k^2} + k - 2) = 0$<br><br>$\Rightarrow (k - 1)(k - 1)(k + 2) = 0$<br><br>$\Rightarrow k = 1,k = -2$<br><br>for k = 1 equation identical so k = $-$2 for no solution.
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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