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.

Answer 1

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.