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प्रश्न
Solve the following equations by method of inversion.
x + y + z = 1, x – y + z = 2 and x + y – z = 3
उत्तर
Matrix form of the given system of equations is
`[(1,1,1),(1, -1, 1),(1, 1, -1)][(x),(y),(z)] = [(1),(2),(3)]`
This is of the form AX = B,
where A = `[(1,1,1),(1, -1, 1),(1, 1, -1)], "X" = [(x),(y),(z)] "and B"= [(1),(2),(3)]`
Consider AA–1 = I
∴ `[(1,1,1),(1, -1, 1),(1, 1, -1)],"A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Appying R2 → R2 – R1 and R3 → R3 – R1, we get
`[(1, 1, 1),(0, -2, 0),(0, 0, -2)]"A"^-1 = [(1, 0, 0),(-1, 1, 0),(-1, 0, 1)]`
Appying R2 → `(-1/2)` R2, we get
`[(1, 1, 1),(0, 1, 0),(0, 0, -2)]"A"^-1 = [(1, 0, 0),(1/2, -1/2, 0),(-1, 0, 1)]`
Appying R1 → R1 – R2, we get
`[(1, 0, 1),(0, 1, 0),(0, 0, -2)]"A"^-1 = [(1/2, 1/2, 0),(1/2, -1/2, 0),(-1, 0, 1)]`
Appying R3 → `(-1/2)` R3, we get
`[(1, 0, 1),(0, 1, 0),(0, 0, 1)]"A"^-1 = [(1/2, 1/2, 0),(-1/2, -1/2, 0),(-1/2, 0, -1/2)]`
Appying R1 → R1 – R3, we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]"A"^-1 = [(0, 1/2, 1/2),(1/2, -1/2, 0),(1/2, 0, -1/2)]`
∴ A–1 = `[(0, 1/2, 1/2),(1/2, -1/2, 0),(1/2, 0, -1/2)]`
pre-multiplying AX = B by A–1, we get
A–1(AX) = A–1B
∴ (A–1A)X = A–1B
∴ IX = A–1B
∴ X = A–1B
∴ `[(x),(y),(z)] = [(0, 1/2, 1/2),(1/2, -1/2, 0),(1/2, 0, -1/2)][(1),(2),(3)]`
∴ `[(x),(y),(z)] = [(0 + 1 + 3/2),(1/2 - 1 + 0),(1/2 + 0 - 3/2)] = [(5/2),(-1/2),(-1)]`
∴ By equality of matrices, we get
x = `(5)/(2), y = -(1)/(2) "and z = -1`.
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