Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
6. Solving Systems Using Matrices
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Exercise 31 Page 180

In the matrix form of the system, the coefficients of the variables form the columns on the left-hand side of the bar and the constants form the column on the right-hand side of the bar.

See solution.

Practice makes perfect

Our town made an investment into two different funds last year. Now we will find how much money was invested in each found. Let's first define the unknowns. In this situation, the unknowns are the money invested in two separate funds. Money in 4 % -interest:& x Money in 6 % -interest:& y Now we can make an organized table to write the equation that represents the situations.

Verbal Expression Algebraic Expression
Total amount of money is $25 000. x+ y=25 000
Total earning from the investment is $1300. 0.4 x+ 0.6 y=1300

We have two equations that we can use to form a system. x+ y&=25 000 0.4x+0.6y&=1300 From here we can use a matrix to solve the system. To do so, we need to consider how the elements of the system relate to the elements of a matrix.

  • The equals signs in the system are represented with a vertical bar in the matrix.
  • The coefficients of the variables form the columns on the left-hand side of the bar.
  • The constants form the column on the right-hand side of the bar.
Below we demonstrate this in a generalized form. a_(11)x+ a_(12)y=c_1 a_(21)x+ a_(22)y=c_2 ⇕ [ cc|c a_(11) & a_(12) & c_1 a_(21) & a_(22) & c_2 ] When each equation in the system is written in the same order we can consider the coefficients of the variables and the constants. Then we can place them in a matrix. 1x+ 1y&=25 000 0.04x+ 0.06y&=1300 ⇕ [ cc|c 1 & 1 & 25 000 0.04 & 0.06 & 1300 ] Finally we will solve the matrix. We will use row operations to obtain a matrix as follows. [ cc|c 1 & 0 & a 0 & 1 & b ] This final matrix represents the solution of the system of equations, where x= a, y= b. Let's solve the matrix!
[ cc|c1 & 1 & 25 000 0.04 & 0.06 & 1300 ]
[ cc|c1 & 1 & 25 000 4 & 6 & 130 000 ]
â–Ľ
Evaluate
[ cc|c4 & 4 & 100 000 4 & 6 & 130 000 ]
[ cc|c4 & 4 & 100 000 4- 4 & 6- 4 & 130 000- 100 000 ]
[ cc|c4 & 4 & 100 000 0 & 2 & 30 000 ]
[ cc|c4 & 4 & 100 000 0 & 1 & 15 000 ]
[ cc|c1 & 1 & 25 000 0 & 1 & 15 000 ]
[ cc|c1- 0 & 1- 1 & 25 000- 15 000 0 & 1 & 15 000 ]
[ cc|c1 & 0 & 10 000 0 & 1 & 15 000 ]
We obtained the matrix that is in the expected form. [ cc|c 1 & 0 & 10 000 0 & 1 & 15 000 ] The amount of money that is invested at 4 %-interest is $ 10 000 and amount of money that is invested at 6 %-interest is $ 15 000.