Answer to Question #234080 in Linear Algebra for iman munaim

Question

The sum of three numbers is 20. If we multiply the first number by 2 and add the second

number and subtract the third number, then we get 23. If we multiply the first number by 3

and add second and third number to it, then we get 46. Let x be the first number, y be the

second number and z

be the third number.

(a) Obtain a system of linear equations to represent the given information.

(b) Write down the system in (a) as a matrix equation.

(c) Use inverse matrix to solve for x , y and z .

Answer 1

(a) Obtain a system of linear equations to represent the given information.

Construct a system containing three linear equations.

$$x+y+z=20$$

$$2x+y-z=23$$

$$3x+y+z=46$$

 (b) Write down the system in (a) as a matrix equation.

$$AX=B$$

$$A=\begin{pmatrix} 1 & 1 & 1\\ 2 & 1 & -1\\ 3 & 1 & 1\\ \end{pmatrix}, X=\begin{pmatrix} x\\ y \\ z \end{pmatrix}, B=\begin{pmatrix} 20\\ 23\\ 46 \end{pmatrix}$$

$$\begin{pmatrix} 1 & 1 & 1\\ 2 & 1 & -1\\ 3 & 1 & 1\\ \end{pmatrix}\begin{pmatrix} x\\ y \\ z \end{pmatrix}=\begin{pmatrix} 20\\ 23\\ 46 \end{pmatrix}$$

(c) Use inverse matrix to solve for $x,y$ and $z$

$$AX=B$$

$$A^{-1}AX=A^{-1}B$$

$$X=A^{-1}B$$

$$\det⁡A=\begin{vmatrix} 1 & 1 & 1\\ 2 & 1 & -1\\ 3 & 1 & 1\\ \end{vmatrix}=1\begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}-1\begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix}+1\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix}$$

$$=1+1-(2+3)+2-3=-4\not=0=>A^{-1}\ exists$$

Find the cofactor matrix:

$$C_{11}=(-1)^{(1+1)}\begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}=2$$

$$C_{12}=(-1)^{(1+2)}\begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix}=-5$$

$$C_{13}=(-1)^{(1+3)}\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix}=-1$$

$$C_{21}=(-1)^{(2+1)}\begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix}=0$$

$$C_{22}=(-1)^{(2+2)}\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix}=-2$$

$$C_{23}=(-1)^{(2+3)}\begin{vmatrix} 1 & 1 \\ 3 & 1 \end{vmatrix}=2$$

$$C_{31}=(-1)^{(3+1)}\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix}=-2$$

 

$$C_{32}=(-1)^{(3+2)}\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix}=3$$

$$C_{33}=(-1)^{(3+3)}\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}=-1$$

The cofactor matrix is

$$\begin{pmatrix} 2 & -5 & -1\\ 0 & -2 & 2\\ -2 & 3 & -1\\ \end{pmatrix}$$

The adjugate matrix is

$$\begin{pmatrix} 2 & 0 & -2\\ -5 & -2 & 3\\ -1 & 2 & -1\\ \end{pmatrix}$$

$$A^{-1}=\dfrac{1}{-4}\begin{pmatrix} 2 & 0 & -2\\ -5 & -2 & 3\\ -1 & 2 & -1\\ \end{pmatrix}$$

$$=\begin{pmatrix} -1/2 & 0 &1/2\\ 5/4 & 1/2 & -3/4\\ 1/4 & -1/2 & 1/4\\ \end{pmatrix}$$

$$\begin{pmatrix} x\\ y \\ z \end{pmatrix}=\begin{pmatrix} -1/2 & 0 &1/2\\ 5/4 & 1/2 & -3/4\\ 1/4 & -1/2 & 1/4\\ \end{pmatrix}\begin{pmatrix} 20\\ 23\\ 46 \end{pmatrix}$$

$$=\begin{pmatrix} -10+0+23 \\ 25+23/2-69/2 \\ 5-23/2+23/2\\ \end{pmatrix}=\begin{pmatrix} 13\\ 2\\ 5\\ \end{pmatrix}$$

Then $x=13, y=2, z=5.$

 

$(13, 2, 5).$