Find the value of k for which the system of equations x+y+z=2, 2x+y+3z=1, 3x-2y+z=k is solvable and then solve it
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Step-by-step explanation:
x+2y+3z=1
2x+y+3z=2
5x+5y+9z=4
we have,
determinant,
∣A∣=
∣
∣
∣
∣
∣
∣
∣
∣
1
2
5
2
1
5
3
3
9
∣
∣
∣
∣
∣
∣
∣
∣
∣A∣=1(9−15)−2(18−15)+3(10−5)
∣A∣=−6−6+15
∣A∣=3
∣A∣
=0
∣A∣= determinant of coefficient matrix
=0
Therefore there exists a unique solution ( only one solution)
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