Finding the Inverse of a Matrix Find the inverse of the matrix if it exists.
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Chapter 6 Solutions
College Algebra
- Finding a Value: Find x such that the matrix is equal to its own inverse. A=3x23arrow_forwardVerifying the Inverse of a Matrix Calculate the products AB and BA to verify that B is the inverse of A. A=[131140132]B=[834211101]arrow_forwardWhen Do Matrices Have Inverse? Find the inverse of the matrix. For what value(s) of x, if any, does the matrix have no inverse? [2xx x 2]arrow_forward
- Which of the following operations can we perform for a matrix A of any dimension? i A+A ii 2A iii AAarrow_forwardWhen Do Matrices Have Inverse? Find the inverse of the matrix. For what value(s) of x, if any, does the matrix have no inverse? [ e x e 2x e 2x e 3x]arrow_forwardGuided Proof Prove that the inverse of a symmetric non-singular matrix is symmetric. Getting Started: To prove that the inverse of A is symmetric, you need to show that A-1T=A-1. i Let A be symmetric, nonsingular matrix. ii This means that AT=A and A-1 exists. iii Use the properties of the transpose to show that A-1T is equal to A-1.arrow_forward
- When Do Matrices Have Inverse? Find the inverse of the matrix. For what value(s) of x, if any, does the matrix have no inverse? [1 e x0 e x e 2x0002]arrow_forwardConjecture Consider matrices of the form Aa11000...00a2200...000a330...00000...ann. (a) Write a 22 matrix and a 33 matrix in the form of A. Find the inverse of each. (b) Use the result of part (a) to make a conjecture about the inverses of matrices in the form of A.arrow_forwardGuided Proof Prove that if A is row-equivalent to B, and B is row-equivalent to C, A is row-equivalent to C. Getting Started: to prove that If A is row-equivalent to C, you have to find elementary matrices E1, E2. Ek such that A=EkE2E1C. i Begin by observing that A is row-equivalent to B and B is row-equivalent to C. ii This means that there exist elementary matrices F1F2Fn and G1G2Gm such that A=FnF2F1B and B=GmG2G1C. iii Combine the matrix equations from step ii.arrow_forward
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