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Question
Using elementary row operations, find the inverse of the matrix A = `((3, 3,4),(2,-3,4),(0,-1,1))` and hence solve the following system of equations : 3x - 3y + 4z = 21, 2x -3y + 4z = 20, -y + z = 5.
Solution
Here A = `((3,-3,4),(2,-3,4),(0,-1,1))`
Using elementary row operations, `"A" = "I.A"`
⇒ `((3,-3,4),(2,-3,4),(0,-1,1)) = ((1,0,0),(0,1,0),(0,-1,1))"A" ..."By" "R"_1->"R"_1 -"R"_2`
⇒ `((1,0,0),(2,-3,4),(0,-1,1)) = ((1,-1,0),(0,1,0),(0,0,1))"A" ..."By" "R"_2->"R"_2 -2"R"_1`
⇒ `((1,0,0),(0,-3,4),(0,-1,1)) = ((1,-1,0),(-2,3,0),(0,0,1))"A" ..."By" "R"_2->"R"_2 -4"R"_3`
⇒ `((1,0,0),(0,1,0),(0,-1,1)) = ((1,-1,0),(-2,3,-4),(0,0,1))"A" ..."By" "R"_3->"R"_3 +"R"_2`
⇒ `((1,0,0),(0,1,0),(0,0,1)) = ((1,-1,0),(-2,3,-4),(-2,3,-3))"A"`
Since we know that `"I" = "A"^-1 "A" "so", "A"^-1 = ((1,-1,0),(-2,3,-4),(-2,3,-3))`.
Now consider the equations 3x - 3y + 4z = 21, 2x -3y + 4z = 20, -y + z = 5.
The matrix form of these equations is, `((3,-3,4),(2,-3,4),(0,-1,1)) (("x"),("y"),("z")) = ((21),(20),(5))`
Where A = `((3,-3,4),(2,-3,4),(0,-1,1)), "X" = (("x"),("y"),("z")) and "B" = ((21),(20),(5))`
So, `"AX" = "B"`
⇒ `"X" = "A"^-1 "B"`
⇒ `"X" = ((1,-1,0),(-2,3,-4),(-2,3,-3))((21),(20),(5))`
⇒ `(("x"),("y"),("z")) = ((1),(-2),(3))`
Therefore, x = 1, y = -2, Z = 3.
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