{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

mat22a-practiceexam1-sol

mat22a-practiceexam1-sol - Math 22A UC Davis Winter 2011...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 22A UC Davis, Winter 2011 Prof. Dan Romik Solutions to practice question set 1 1. Find the general form of the solution of the linear system x 3 - 2 x 4 = - 2 x 1 + 4 x 2 + x 3 + 7 x 4 = 1 2 x 1 + 8 x 2 + 3 x 3 + 12 x 4 = 0 x 1 + 4 x 2 - x 3 + 11 x 4 = 5 Solution. First, write the system as an augmented matrix 0 0 1 - 2 - 2 1 4 1 7 1 2 8 3 12 0 1 4 - 1 11 5 Next, perform the Gaussian elimination procedure to bring the aug- mented matrix to reduced row echelon form (RREF) by applying a sequence of elementary row operations: 0 0 1 - 2 - 2 1 4 1 7 1 2 8 3 12 0 1 4 - 1 11 5 swap R 1 ,R 2 ------→ 1 4 1 7 1 0 0 1 - 2 - 2 2 8 3 12 0 1 4 - 1 11 5 R 3 - 2 R 1 R 4 - R 1 ------→ 1 4 1 7 1 0 0 1 - 2 - 2 0 0 1 - 2 - 2 0 0 - 2 4 4 R 1 - R 2 R 3 - R 2 R 4 +2 R 2 ------→ 1 4 0 9 3 0 0 1 - 2 - 2 0 0 0 0 0 0 0 0 0 0 Translating this back into a system of equations, we have reached the equivalent system x 1 + 4 x 2 + 9 x 4 = 3 x 3 - 2 x 4 = - 2 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Here we have two independent (non-pivot) variables x 2 and x 4 , so the solution can be written as x 2 = s, x 4 = t, x 1 = 3 - 4 s - 9 t, x 3 = - 2 + 2 t, where s, t are parameters taking arbitrary real values. In vector nota-
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}