8 - Math 1313 Section 3.2 Example 5 Solve the system of linear equations using the Gauss-Jordan elimination method 2 x 4 y 6z = 38 x 2 y 3z = 7 3 x 4 y

8 - Math 1313 Section 3.2 Example 5 Solve the system of...

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Math 1313 Section 3.2 Example 5:Solve the system of linear equations using the Gauss-Jordan elimination method.
Math 1313 Section 3.2 Popper 1: Is the following matrix in row reduced form? °120800190000± Example 6:Solve the system of linear equations using the Gauss-Jordan elimination method.
Math 1313 Section 3.2 Popper 2: Given the following matrix, state the solution. ° 1 0 0 −3 0 1 0 0 0 0 1 2 ± a. (-3,2) b. (-3, 0, 2) c. (3, 0, -2) d. None of the above Infinite Number of Solutions Example 7:The following augmented matrix in row-reduced form is equivalent to the augmented matrix of a certain system of linear equations. Use this result to solve the system of equations. --000025103101Example 8:Solve the system of linear equations using the Gauss-Jordan elimination method. 2
Math 1313 Section 3.2 A System of Equations That Has No Solution In using the Gauss-Jordan elimination method the following equivalent matrix was obtained (note this matrix is not in row-reduced form, let’s see why): - - - 1 0 0 0 1 4 4 0 1 1 1 1 Look at the last row. It reads: 0x + 0y + 0z = -1, in other words, 0 = -1!!! This is never true. So the system is inconsistent and has no solution. Popper 3 is C Systems with No Solution If there is a row in the augmented matrix containing all zeros to the left of the vertical line and a nonzero entry to the right of the line, then the system of equations has no solution. Example 9: Solve the system of linear equations using the Gauss-Jordan elimination method. 32323=--=+=yxyxyx 2
Math 1313 Section 3.2 Popper 4: Given the following matrix state the solution. °131−4±a.x = 3, y = 4 b.x = 3, y = -4, z = 0 c.Infinitely many d.x =3, y = -4 Example 10:Solve the system of linear equations using the Gauss-Jordan elimination method. −² + 3³ − 4´ = 124² − 12³ + 16´ = −36 Example 11: Solve the system of linear equations using the Gauss-Jordan elimination method. 2² − 3³ = 13² + ³ = −1² − 4³ = 14

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