Q&A - Ask Doubts and Get Answers
Q

Solve system of linear equations, using matrix method. 2x + 3y + 3z = 5, x - 2y + z = -4, 3x - y - 2z = 3.

Q : 13      Solve system of linear equations, using matrix method. 

              \small 2x+3y+3z=5

             \small x-2y+z=-4

             \small 3x-y-2z=3

Answers (1)
Views

The given system of equations

 \small 2x+3y+3z=5

 \small x-2y+z=-4

\small 3x-y-2z=3

can be written in the matrix form of AX =B, where

A = \begin{bmatrix} 2 &3 &3 \\ 1 & -2 &1 \\ 3& -1 &-2 \end{bmatrix}X = \begin{bmatrix} x\\y \\z \end{bmatrix}  and\ B =\begin{bmatrix} 5\\ -4 \\ 3 \end{bmatrix}. 

we have, 

|A| =2(4+1) -3(-2-3)+3(-1+6) = 10+15+15 = 40.

So, A is non-singular, Therefore, its inverse A^{-1} exists.

as we know A^{-1} = \frac{1}{|A|} (adjA)

Now, we will find the cofactors;

A_{11} =(-1)^{1+1}(4+1) = 5      A_{12} =(-1)^{1+2}(-2-3) = 5

A_{13} =(-1)^{1+3}(-1+6) = 5      A_{21} =(-1)^{2+1}(-6+3) = 3

A_{22} =(-1)^{2+2}(-4-9) = -13      A_{23} =(-1)^{2+3}(-2-9) = 11

A_{31} =(-1)^{3+1}(3+6) = 9     A_{32} =(-1)^{3+2}(2-3) = 1

A_{33} =(-1)^{3+3}(-4-3) = -7

(adjA) =\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5&11 & -7 \end{bmatrix}

A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}

So, the solutions can be found by X = A^{-1}B =\frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}\begin{bmatrix} 5\\-4 \\ 3 \end{bmatrix}

\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 25-12+27\\25+52+3 \\ 25-44-21 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40\\80 \\ -40 \end{bmatrix}= \begin{bmatrix} 1\\2 \\ -1 \end{bmatrix}

Hence the solutions of the given system of equations;

x =1,\ y =2,\ and\ \ z=-1.

Exams
Articles
Questions