Monthly Archives: March 2011

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 →

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 →

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 →

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 →

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 →

Exercise 1.6.4. (a) Given AB = AC, show that B = C if A is invertible. (b) Given find B and C such that AB = AC but . Answer: (a) If A is invertible then (b) If B is … Continue reading →

Exercise 1.6.3. Given AB = C, express in terms of B and C. Similar, given PA = LU, express in terms of P, L, and U. Answer: Assume that both B and C are invertible (see below). We then have … Continue reading →

Exercise 1.6.2. (a) Find the inverses of the following matrices: (b) Why does for any permutation matrix P? Answer: (a) The first permutation matrix exchanges the first and third rows of the identity matrix I, while leaving the second row … Continue reading →

Exercise 1.6.1. Find the inverses of the following matrices: Answer: We can use the standard formula for the inverse of a 2 by 2 matrix A: First, we have We then have Finally we have NOTE: This continues a series … Continue reading →