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Examine the consistency of the system of equations. 3x − y − 2z = 2 2y − z = −1 3x − 5y = 3 - Mathematics

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प्रश्न

Examine the consistency of the system of equations.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

योग

उत्तर

Let, A `= [(3,-1,-2),(0,2,-1),(3,-5,0)] , X = [(x),(y),(z)], B = [(2),(-1),(3)]`

`abs A = [(3,-1,-2),(0,2,-1),(3,-5,0)] = 3 [ 2 xx 0 + 5 xx (-1)] + 1 (0 + 3) - 2(0 - 6)`

`= -15 + 3 + 12 = 0`

Cofactors of the elements of `abs A`

`A_11 = abs ((2,-1),(-5,0)) = 0 - 5 = -5`

`A_12 = - abs ((0,-1),(3,0)) = -3`

`A_13 = abs ((0,2),(3,-5)) = - 6`

`A_21 = - abs ((-1,-2),(-5,0)) = 10`

`A_22 = abs ((3,-2),(3,0)) = 6`

`A_23 = - abs ((3,-1),(3,-5)) = -(- 15 + 3) = 12`

`A_31 = abs ((-1,-2),(2,-1)) = 1 + 4 = 5`

`A_32 = - abs ((3,-2),(0,-1)) = 3`

`A_33 = abs ((3,-1),(0,2)) = 6`

The cofactor matrix of `therefore abs A` is C = `[(-5,-3,-6),(10,6,12),(5,3,6)]`

`therefore adj (A) = C = [(-5,10,5),(-3,6,3),(-6,12,6)]`

(adj A) B = ` [(-5,10,5),(-3,6,3),(-6,12,6)] [(2),(-1),(3)] = [(-10-10 + 15),(-6 - 6 + 9),(-12 - 12 + 18)] = [(-5),(-3),(-6)] ne 0`

`therefore abs A = 0  "and"  (adj A) B ne 0`

Hence, The system of equations is inconsistent.

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अध्याय 4: Determinants - Exercise 4.6 [पृष्ठ १३६]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 4 Determinants
Exercise 4.6 | Q 6 | पृष्ठ १३६

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