2.3 Matrix Inverses. Numerical equivalent How would we solve for x in: ax = b ? –a -1 a x = a -1 b –x=a -1 b since a -1 a = 1 and 1x = x We use the same.

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Presentation transcript:

2.3 Matrix Inverses

Numerical equivalent How would we solve for x in: ax = b ? –a -1 a x = a -1 b –x=a -1 b since a -1 a = 1 and 1x = x We use the same principle with matrices: – AX = B what mtx, when mult x X, just leaves X? – the identity matrix (verify) – so we need to find an mtx to multiply x A to get I, and we will call it A -1

Definition If A is a square matrix, a matrix B is called the inverse of A iff: –AB = I and BA = I –We say B = A -1 so AA -1 = I = A -1 A A matrix, A, that has an inverse is called an invertible matrix.

Examples Show that B is the inverse of A: Show that C has no inverse:

Theorem If B and C are both inverses of A, then B = C (i.e. the inverse of a matrix is unique) Proof: – CA = AB = I (since both are inverses) – also, B = IB = (CA)B = C(AB) = CI = C

Helpful Example Given A= and ad - bc ≠ 0, show that: Now we have a simple way of finding the inverse of a 2 x 2 matrix.

Solving Systems using inverses If AX = B, and we want to solve for X (assuming A is invertible: –A -1 AX=A -1 B –IX = X = A -1 B Note that the solution X will be unique.

To think about How could we have no solution or infinite solutions then? –Matrix is not invertible A is not a square matrix A begins as a square matrix, but one row is a linear combination of the other rows (so row-ech. Form will have a row of 0’s)

Example We showed that B = A -1 : Find the solution to the following system using inverses:

General Method of Inversion If we want to find the inverse of square matrix A, we need AA -1 =I.(We show for 2 x 2)

(continued) We could solve by solving for individual entries: first column second column

(continued) Notice that to solve each, we will go through the same row operations, so why not do it all together?

Generalize to (n x n) matrix Given that A is an (n x n) matrix, we can find its inverse by: – augmenting A with I n – use elementary row op’s to write lhs in reduced row echelon form If get I n on lhs, invertible If don’t, then must have row of 0’s, so no unique sol’n, so not invertible – matrix that remains on rhs is A -1

Theorem 3 If A is an n x n matrix, either A can be reduced to I by elementary row operations or it cannot. In the first case, the algorithm will produce A -1 ; in the second case, A -1 does not exist.

Example Given A, find A -1 :

Properties of Inverses Given that all matrices are square n x n’s: 1I is invertible and I -1 = I 2If A is invertible, so is A -1 and (A -1 ) -1 =A 3If A and B are invertible, so is AB, and (AB) -1 =B -1 A -1. 4If A 1, A 2, …,A k are all invertible, so is their product A 1 A 2 …A k, and (A 1 A 2 …A k ) -1 =A k -1 …A 2 -1 A If A is invertible, so is A k for k≥1, and (A k ) -1 =(A -1 ) k. 6If A is invertible and a≠0, then aA is also and

Properties of Inverses 7If A is invertible, so is its transpose A T, and (A T ) -1 = (A -1 ) T. Proof of Property 7: –A T (A -1 ) T = (A -1 A) T = I T = I –(A -1 ) T A T = (AA -1 ) T = I T = I –So (A -1 ) T is the inverse of A T

Proof of Property 3 Property 3: If A and B are invertible, so is AB, and (AB) -1 = B -1 A -1 Proof: –B -1 A -1 (AB) = B -1 (A -1 A)B = B -1 IB = B -1 B = I –(AB)(B -1 A -1 ) = A(BB -1 )A -1 = AIA -1 = AA -1 = I –So (B -1 A -1 ) is the inverse of AB

Example Find A when Hint: Take inverse of both sides.

Another Property If A is invertible, –and if AB = AC, then B = C. –and if BA = CA, then B = C. Proof: Given AB = AC, then A -1 AB = A -1 AC, so IB = IC, so B = C (other part is similar with right mult.)

One more useful property If A is invertible, then the only matrix X such that AX = 0 is X = 0. Proof: Given AX = 0, left multiplication gives A -1 AX = A -1 0 so X = 0

Proving that a matrix is not invertible Given a matrix A, if we can find a non-zero matrix X such that AX = 0, then this would imply that X = 0 (if A is invertible). But we know that X ≠ 0. Therefore A is not invertible. Example: Show that A is not invertible: Hint: augment w/ the 0 matrix.