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प्रश्न
If A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)] "and B" = [(1, 2, 3),(1, 1, 5),(2, 4, 7)]`, then find a matrix X such that XA = B.
उत्तर १
A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` and B = `[(1, 2, 3),(1, 1, 5),(2, 4, 7)]`
XA = B
Post multiplying by A–1, we get
XAA–1 = BA–1
∴ X = BA–1 ...(i)
|A| = `|(1, 0, 1),(0, 2, 3),(1, 2, 1)|`
= 1(2 – 6) – 0 + 1(0 – 2)
= – 4 – 2
= – 6 ≠ 0
∴ A–1 exists.
A11 = (– 1)1+1 M11 = `1 |(2, 3),(2, 1)|` = 1(2 – 6) = – 4
A12 = (– 1)1+2 M12 = `-1 |(0, 3),(1, 1)|` = –1(0 – 3) = 3
A13 = (– 1)1+3 M13 = `1 |(0, 2),(1, 2)|` = 1(0 – 2) = – 2
A21 = (– 1)2+1 M21 = `-1 |(0, 1),(2, 1)|` = –1(0 – 2) = 2
A22 = (– 1)2+2 M22 = `1 |(1, 1),(1, 1)|` = 1(1 – 1) = 0
A23 = (– 1)2+3 M23 = `-1 |(1, 0),(1, 2)|` = –1(2 – 0) = – 2
A31 = (– 1)3+1 M31 = `1 |(0, 1),(2, 3)|` = 1(0 – 2) = – 2
A32 = (– 1)3+2 M32 = `-1 |(1, 1),(0, 3)|` = –1(3 – 0) = – 3
A33 = (– 1)3+3 M33 = `1 |(1, 0),(0, 2)|` = 1(2 – 0) = 2
∴ The matrix of the co-factors is
[Aij]3×3 = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)]`
= `[(-4, 3, -2),(2, 0, -2),(-2, -3, 2)]`
Now, adj A = `["A"_"ij"]_(3 xx 3)^"T"`
= `[(-4, 2, -2),(3, 0, -3),(-2, -2, 2)]`
∴ A–1 = `1/|"A"| ("adj A")`
= `-1/6[(-4, 2, -2),(3, 0, -3),(-2, -2, 2)]`
X = BA–1 ...[From (i)]
∴ X = `[(1, 2, 3),(1, 1, 5),(2, 4, 7)] {(1/6) [(-4, 2, -2),(3, 0, -3),(-2, -2, 2)]}`
= `(1)/(6) [(1, 2, 3),(1, 1, 5),(2, 4, 7)] [(4, -2, 2),(-3, 0, 3),(2, 2, -2)]`
= `(1)/(6) [(4 - 6 + 6, -2 + 0 + 6, 2 + 6 - 6),(4 - 3 + 10, -2 + 0 + 10, 2 + 3 - 10),(8 - 12 + 14, -4 + 0 + 14, 4 + 12 - 14)]`
∴ X = `(1)/(6)[(4, 4, 2),(11, 8, -5),(10, 10, 2)]`
उत्तर २
Consider XA = B
∴ `X[(1, 0, 1),(0, 2, 3),(1, 2, 1)] = [(1, 2, 3),(1, 1, 5),(2, 4, 7)]`
By C3 – C1, we get
`[(1, 0, 0),(0, 2, 3),(1, 2, 0)] = [(1, 2, 2),(1, 1, 4),(2, 4, 5)]`
By `(1/2)C_2`, we get
`[(1, 0, 0),(0, 1, 3),(1, 1, 0)] = [(1, 1, 2),(1, 1/2, 4),(2, 2, 5)]`
By C3 – 3C2, we get
`[(1, 0, 0),(0, 1, 0),(1, 1, -3)] = [(1, 1, -1),(1, 1/2, 5/2),(2, 2, -1)]`
By `(-1/3)C_3`, we get
`[(1, 0, 0),(0, 1, 0),(1, 1, 1)] = [(1, 1, 1/3),(1, 1/2, -5/6),(2, 2, 1/3)]`
By C1 – C3 and C2 – C3, we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] = [(2/3, 2/3, 1/3),(11/6, 4/3, -5/6),(5/3, 5/3, 1/3)]`
∴ X = `1/6[(4, 4, 2),(11, 8, -5),(10, 10, 2)]`
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