inverse of the coefficient matrix

[Tiger-T]

Solve each system of equations by using the inverse of the coefficient matrix. (The inverses for the first four problems were found in Exercises 19, 20, 23, and 24 above.)
2x + 4z = -8
3x + y + 5z = -2
-x + y – 2z = 4
A = [2 0 4] x = [x] B = [-8]
[3 1 5] [y] [2]
[-1 1 -2] [z] [4]
[2 0 4?1 0 0]
[3 1 5?0 1 0]
[-1 1 -2?0 0 1]
1/2R1-----R1
3R3 + R2-----R2
3R3 + R2-----R3
[1 0 2?1/2 0 0]
[0 4 -1?0 1 3]
[0 4 -1?0 1 3]

1/2R2----R2
-1R2 + R3------R3
[1 0 2?1/2 0 0 ]
[0 1 0?0 ¼ 3/4]
[0 0 -1?0 0 0 ]

2R3+ R1-----R1
-1R3-----R3
[1 0 0?1/2 0 0 ]
[0 1 0?0 ¼ ¾ ]
[0 0 1?0 0 0 ]


X = [1/2 0 0 ] -8
[0 ¼ 3/4] B = 2
[0 0 0 ] 4

I got this far; the book does not explain how to get the last three numbers arranged like B. And I have no clue how else to explain it. Sorry!

 

[soroban]

Hello, Tiger-T!

Solve by using the inverse of the coefficient matrix.

. . \(\displaystyle \begin{array}{ccccc}2x \quad + 4z &=& \text{-}8 & [1] \\ 3x + y + 5z &=& \text{-}2 & [2] \\ \text{-}x + y - 2z &=& 4 & [3] \end{array}\)

\(\displaystyle \text{Move equation [1] to the bottom: }\;\begin{array}{ccc} 3x + y + 5z &=& \text{-}2 \\ \text{-}x + y - 2z &=& 4 \\ 2x \quad + 4z &=& \text{-}8 \end{array}\)


\(\displaystyle \text{We have: }\;\left[\begin{array}{ccc|ccc} 3&1&5 & 1 & 0 & 0 \\ \text{-}1 & 1 & \text{-}2 & 0 & 1 & 0 \\ \text{-}2 & 0 & 4 & 0 & 0 & 1 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1 - R_3 \\ \\ \\ \end{array} \left[\begin{array}{ccc|ccc} 1&1&1 & 1 & 0 & \text{-}1 \\ \text{-}1 & 1 & \text{-}2 & 0 & 1 & 0 \\ 2 & 0 & 4 & 0 & 0 & 1 \end{array}\right]\)


\(\displaystyle \begin{array}{c}\\ R_2 + R_1 \\ R_3 - 2R_1 \end{array} \left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & \text{-}1 \\ 0 & 2 & \text{-}1 & 1 & 1 & \text{-}1 \\ 0 & \text{-}2 & 2 & \text{-}2 & 0 & 3 \end{array}\right]\)


. . . \(\displaystyle \begin{array}{c} \\ \frac{1}{2}R_2 \\ \\ \end{array} \left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & \text{-}1 \\ 0 & 1 & \text{-}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \text{-}\frac{1}{2} \\ 0 & \text{-}2 & 2 & \text{-}2 & 0 & 3 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1-R_2 \\ \\ R_3+2R_2 \end{array} \left[\begin{array}{ccc|ccc} 1 & 0 & \frac{3}{2} & \frac{1}{2} & \text{-}\frac{1}{2} & \text{-}\frac{1}{2} \\ 0 & 1 & \text{-}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \text{-}\frac{1}{2} \\ 0 & 0 & 1 & \text{-}1 & 1 & 2 \end{array}\right]\)


\(\displaystyle \begin{array}{c}R_1-\frac{3}{2}R_3 \\ R_2 + \frac{1}{2}R_3 \\ \\ \end{array} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 2 & \text{-}2 & \text{-}\frac{7}{2} \\ 0 & 1 & 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 1 & \text{-}1 & 1 & 2 \end{array}\right]\)


\(\displaystyle \text{Therefore: }\;A^{-1} \;=\;\begin{bmatrix} 2 & \text{-}2 & \text{-}\frac{7}{2} \\ 0 & 1 & \frac{1}{2} \\ \text{-}1 & 1 & 2 \end{bmatrix}\)


I assume you can complete the problem now . . .