Consider the system of equations x − 2 y + 3 z = − 1 , x − 3 y + 4 z = 1 and − x + y − 2 z = k Statement 1 The system of equations has no solution for k  = 3 . Statement 2 The determinant 1 − 1 1

Question

Consider the system of equations x − 2 y + 3 z = − 1 , x − 3 y + 4 z = 1 and − x + y − 2 z = k Statement 1 The system of equations has no solution for k  = 3 . Statement 2 The determinant ​ 1 − 1 1 ​ 3 − 2 4 ​ − 1 k 1 ​ ​  = 0 , for k  = 3 .

TestHub · Accepted Answer

The given system of equations can be expressed as When k  = 3 , the given system of equations has no solution. ⇒ Statement 1 is true.Clearly, Statement 2 is also true as it is rearrangement of rows and columns of Hence, option (a) is correct.

Mathematics - Determinant Question with Solution | TestHub

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