MATH136 Lecture Notes - Lecture 25: Invertible Matrix, Identity Matrix, If And Only If
14 May 2018
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Monday, June 26
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Lecture 25 : Invertible matrices. (Refers to 5.1)
Concepts:
1. Define the inverse of a matrix.
2. Find the inverse of a matrix which is 2 by 2
3. Solve a system by using the inverse of a matrix.
4. Recognize that a matrix is invertible if and only if its RREF is identity.
5. Right inverse, left inverse
6. A matrix B is invertible iff Bx = 0 has only trivial solution.
The multiplicative matrix inverse A−1 of A.
25.1 Definition − We say that an n × n matrix A is invertible (or non-singular) if there
exists an n × n matrix B such that both conditions AB = In = BA hold true. We call such a
matrix B, an inverse of A.
Note that only a pair of square matrices of equal dimension can satisfy the condition AB
= BA. So if we refer to matrix A as being invertible then A must be a square matrix
25.1.1 Definition – If A is an n × m matrix for which there exists an m × n matrix B
such that AB = In then we say that B is a right inverse of A and A is a left inverse of
B. If B is both a right inverse and left inverse of A then, by definition of “inverse”,
both A and B are easily seen to be both (square) invertible matrices.
25.1.2 Proposition − Uniqueness of the inverse of a matrix. Let A be an n × n
invertible matrix. Then there exists no more than one inverse B of A.
Proof : Suppose A is invertible. Suppose both B and C are inverses of the matrix A.
Then AB = I = BA and AC = I = CA.
Then C = CI = C(AB) = (CA)B = IB = B.
So C = B.
Then if a matrix A is invertible it will have, at most, one inverse.
25.1.3 Notation − For an invertible matrix A, the unique inverse matrix of A is denoted
by A-1.
If a matrix has dimensions 3 × 3 or larger it is not easy to tell whether it has an inverse
or not simply by looking at its entries. Even if we know a matrix is invertible finding
its unique inverse can still be a considerable amount of work. However, if a matrix is 2
× 2 it is fairly straightforward as the following example shows.
25.1.4 Example − Show that if A is the matrix,
such that ad
−
bc is not zero, A-1 exists and is equal to the matrix :
Solution: Suppose ad
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bc ≠ 0. We compute the product AB:
Then AB = I. Computation BA = I follows similar steps. This means that if ad
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bc ≠ 0
then a two by two matrices has an inverse and is of the form given to the matrix B.
The converse : We claim that if A is invertible then ad
−
bc ≠ 0. Suppose A is
invertible. Suppose both ad – bc = 0. Then
Then
Then for any matrix B, AB will produce a two by two matrix whose second row only
contains zeros and so cannot be the identity matrix contradicting the fact that A is
invertible.
25.1.5 Definition − The number ad − bc obtained in the 2 × 2 matrix A−1 above is
called the determinant of the 2 × 2 matrix A. It is denoted by det A.
- We can always use the derived expression as a way of finding the inverse of a 2 by
2 provided det A ≠ 0.
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Document Summary
Monday, june 26 lecture 25 : invertible matrices. (refers to 5. 1) 25. 1 definition we say that an n n matrix a is invertible (or non-singular) if there exists an n n matrix b such that both conditions ab = in = ba hold true. We call such a matrix b, an inverse of a. Note that only a pair of square matrices of equal dimension can satisfy the condition ab. So if we refer to matrix a as being invertible then a must be a square matrix. 25. 1. 2 proposition uniqueness of the inverse of a matrix. Let a be an n n invertible matrix. Then there exists no more than one inverse b of a. Suppose both b and c are inverses of the matrix a. Then ab = i = ba and ac = i = ca. C = ci = c(ab) = (ca)b = ib = b.
