{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Midterm1practice_sol - Math 415 — Midterm 1 practice...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 415 — Midterm 1 practice solutions Problem 1: We can start reducing the matrix using Gaussian elimination. We get to 1 0 1 1 0 1- 2- 1 0 0- 3- 3 0 0- 3- 3 and because two rows are equal, we know the matrix is singular and the determinant is 0. Problem 2: We start by writing the augmented matrix 1 1 1 1 0 0 1 2 2 0 1 0 2 1 2 0 0 1 We then reduce the left side to the identity matrix using Gauss-Jordan re- duction, and we do the same operations on the right side to get 1 0 0 2- 1 0 1 0 2- 1 0 0 1- 3 1 1 The right side is the inverse we needed. You may want to verify this by multiplying the original matrix with the inverse we found. Problem 3: Let’s call A the original matrix. We start by reducing A to reduced row echelon form using Gauss-Jordan reduction. We obtain U = 1 0 0 6 / 5 0 1 0- 3 / 5 0 0 1 1 / 5 0 0 0 There are three leading coefficients in the first three columns of U . The basis....
View Full Document

{[ snackBarMessage ]}

Page1 / 3

Midterm1practice_sol - Math 415 — Midterm 1 practice...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online