Exercise 26 Page 179

To solve the given system of equations using a matrix, we should first write the matrix for the system. 300x - 1y= 130 200x+ 1y= 120 ⇓ [ cc|c300 & -1 & 130 200 & 1 & 120 ] In order to solve the matrix, we will use row operations to obtain a matrix in the following form. [ cc|c1 & 0 & a 0 & 1 & b ] This final matrix represents the solution of the system of equations, where x= a and y= b. Let's solve the matrix!

[ cc|c300 & -1 & 130 200 & 1 & 120 ]

SysEqnAdd

(I): Add (II)

[ cc|c 300+ 200 & -1+ 1 & 130+ 120 200 & 1 & 120 ]

AddTerms

(I):Add terms

[ cc|c 500 & 0 & 250 200 & 1 & 120 ]

MultEqn

(I):LHS * 2/5=RHS* 2/5

[ cc|c 500( 2/5) & 0( 2/5) & 250( 2/5) 200 & 1 & 120 ]

Multiply

(I):Multiply

[ cc|c 200 & 0 & 100 200 & 1 & 120 ]

SysEqnSub

(II):Subtract (I)

[ cc|c 200 & 0 & 100 200- 200 & 1- 0 & 120- 100 ]

SubTerms

(II):Subtract terms

[ cc|c 200 & 0 & 100 0 & 1 & 20 ]

DivEqn

(I):.LHS /200.=.RHS /200.

[ cc|c 1 & 0 & 0.5 0 & 1 & 20 ]

Looking at the right-hand column, we can see that the solution to the system is the unique point (0.5,20). To help visualize this answer, we can also write the matrix that resulted from using row operations in system notation. x+0y=0.5 0x+1y=20 ⇒ x=0.5 y=20