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

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प्रश्न

Examine the consistency of the system of equations.

3x − y − 2z = 2

2y − z = −1

3x − 5y = 3

योग
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उत्तर

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

|A| `= |(3,-1,-2),(0,2,-1),(3,-5,0)|`

= 3[2 × 0 + 5 × (−1)] + 1(0 + 3) − 2(0 − 6)

= −15 + 3 + 12

= 0

Cofactors of the elements of |A|:

A11 = `|(2,-1),(-5,0)|`

= 0 − 5

= −5

A12 = `-|(0,-1),(3,0)|`

= −(0 + 3)

= −3

A13 = `|(0,2),(3,-5)|`

= 0 − 6

= −6

A21 = `-|(-1,-2),(-5,0)|`

= −(0 − 10)

= 10

A22 = `|(3,-2),(3,0)|`

= 0 + 6

= 6

A23 = `-|(3,-1),(3,-5)|`

= −(−15 + 3)

= 12

A31 = `|(-1,-2),(2,-1)|`

= 1 + 4

= 5

A32 = `-|(3,-2),(0,-1)|`

= −(−3 + 0)

= 3

A33 = `|(3,-1),(0,2)|`

= 6 + 0

= 6

∴ |A| = `[(-5,-3,-6),(10,6,12),(5,3,6)]`

∴ (adj A) = `[(-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)] ≠ 0`

∴ |A| = 0 and (adj A)B ≠ 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 Part 1 and 2 [English] Class 12
अध्याय 4 Determinants
Exercise 4.6 | Q 6 | पृष्ठ १३६

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