Question

Express the following equations in matrix form and solve them by the method of reduction:

x − y + z = 1, 2x − y = 1, 3x + 3y − 4z = 2

Solution:

The given equations can be written in the matrix form as:

`[(1,−1,1),(2,−1,0),(3,3,−4)] [(x),(y),(z)] = [(1),(1),(2)]`

By R₂ → R₂ − 2R₁,

`[(1,−1,1),(square,square,square),(3,3,−4)] [(x),(y),(z)] = [(1),(−1),(2)]`

By R₃ → R₃ − 3R₁

`[(1,−1,1),(0,1,−2),(square,square,square)] [(x),(y),(z)] = [(1),(−1),(−1)]`

By R₃ → R₃ − 6R₂

`[(1,−1,1),(0,1,−2),(0,0,5)] [(x),(y),(z)] = [(1),(−1),(square)]`

We write equations as

x − y + z = 1   ...(I)

y − 2z = − 1    ...(II)

5z = 5              ...(III)

Solving equations (I), (II) and (III)

We get x = `square, y = square, z = square`

Answer 1

The given equations can be written in the matrix form as:

`[(1,−1,1),(2,−1,0),(3,3,−4)] [(x),(y),(z)] = [(1),(1),(2)]`

By R₂ → R₂ − 2R₁,

\[\begin{bmatrix} 1 & -1 & 1 \\ \boxed{0} & \boxed{1} & \boxed{-2} \\ 3 & 3 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}\]

By R₃ → R₃ − 3R₁

\[\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ \boxed{0} & \boxed{6} & \boxed{-7} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}\]

By R₃ → R₃ − 6R₂

\[\begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} 1 \\ -1 \\ \boxed{5} \end{bmatrix}\]

We write equations as

x − y + z = 1   ...(I)

y − 2z = − 1    ...(II)

5z = 5              ...(III)

Solving equations (I), (II) and (III)

We get x = \[\boxed{1}\], y =  \[\boxed{1}\], z = \[\boxed{1}\]