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If A = [42-1x] and such that (A – 2I)(A – 3I) = 0, find the value of x - Mathematics

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प्रश्न

If A = `[(4, 2),(-1, x)]` and such that (A – 2I)(A – 3I) = 0, find the value of x

योग

उत्तर

A = `[(4, 2),(-1, x)]`

A – 2I = `[(4, 2),(-1, x)] - 2[(1, 0),(0, 1)]`

= `[(4, 2),(-1, x)] - [(2, 0),(0, 2)]`

= `[(4 - 2, 2 - 0),(-1 - 0, x - 2)]`

A – 2I = `[(2, 2),(-1, x - 2)]`

A – 3I = `[(4, 2),(-1, x)] - 3[(1, 0),(0, 1)]`

= `[(4, 2),(-1, x)] - 3[(3, 0),(0, 3)]`

= `[(4 - 3, 2 - 0),(-1 - 0, x - 3)]`

A – 3I = `[(1, 2),(-1, x - 3)]`

(A – 2I)(A – 3I) = `[(2, 2),(-1, x  2)] [(1, 2),(-1, x - 3)]`

= `[(2 -2, 4 + 2(x - 3)),(-1 - (x - 2), -2 + (x - 2)(x - 3))]`

= `[(0, 4 + 2x- 6),(-1 - x + 2, -2 + x^2 - 3x - 2x + 6)]`

(A – 2I)(A – 3I) = `[(0, 2x - 2), (-x + 1, x^2 - 5x + 4)]`

Given (A – 2I)(A – 3I) = 0

∴ `[(0, 2x - 2), (-x + 1, x^2 - 5x + 4)] = [(0, 0),(0, 0)]`

Equation the corresponding entries

2x – 2 = 0

⇒ x = 1

x2 – 5x + 4 = 0  ......(1)

Put x = 1 in equation (1)

12 – 5 × 1 + 4 = 5 – 5 = 0

∴ The  reqired value of x is x = 1

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Matrices and Determinants - Exercise 7.1 [पृष्ठ १८]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 7 Matrices and Determinants
Exercise 7.1 | Q 7 | पृष्ठ १८

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