# Monthly Archives: March 2011

## Linear Algebra and Its Applications, Exercise 1.6.9

Exercise 1.6.9. Given the singular matrix show that A has no inverse. If it did have an inverse then multiplying the third row of by the columns of A should give the third row of I. Explain why this is … Continue reading

## Linear Algebra and Its Applications, Exercise 1.6.8

Exercise 1.6.8. The matrix has no inverse. Demonstrate this by trying to solve the following: Answer: Multiplying the first row of the first marix by the first column of the second matrix gives However multiplying the second row of the … Continue reading

## Linear Algebra and Its Applications, Exercise 1.6.7

Exercise 1.6.7. Find three 2 by 2 matrices A such that and A is neither I nor -I. Answer: We first note that the transpose of I is its own inverse: Note that this also follows from the result of … Continue reading

## Linear Algebra and Its Applications, Exercise 1.6.6

Exercise 1.6.6. Invert the following matrices using the Gauss-Jordan method: Answer: For the first matrix Gauss-Jordan elimination proceeds as follows: We first subtract 1 times the first row from the second row: This completes the process of forward elimination. We … Continue reading

## Linear Algebra and Its Applications, Exercise 1.6.5

Exercise 1.6.5. For a matrix A assume that is invertible and has inverse B. Prove that A is also invertible, with inverse AB. Answer: We have We then have We also have So AB is both a left and right … Continue reading