Section 2.2: The Inverse of a Matrix


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1 Section 22: The Inverse of a Matrix Recall that a linear equation ax b, where a and b are scalars and a 0, has the unique solution x a 1 b, where a 1 is the reciprocal of a From this result, it is natural to ask the question: Could we solve systems of equations in a similar way? Suppose that A is an n n matrix and b is a vector in R n If the system Ax b has a unique solution, is the solution determined by x A 1 b, where presumably A 1 is another matrix? The answer happens to be yes, but the condition is that A has to be invertible Definition: An n n matrix A is said to be invertible if there is an n n matrix C such that CA I and AC I, where I I n, the n n identity matrix Note that C is an inverse of A and it is uniquely determined To see this, assume that B and C are different inverses of A, and thus Then B AB I and CA AC I B BI B(AC) (BA)C IC C, which shows that an n n invertible matrix has only one inverse Notation: The inverse of A will be denoted by C A 1 Therefore, A 1 A AA 1 I Definition: If a square matrix is not invertible, it is called singular All invertible matrices are called nonsingular A Quick Formula for the Inverse of a 2 2 Matrix a b Theorem: Let A If ad bc 0, then A is invertible and c d A 1 1 d b ad bc c a If ad bc 0, then A is not invertible The expression ad bc is called the determinant of A and is denoted by det(a) We will discuss determinants in more detail in Chapter 3 Example: Let A 2 7 Determine A 1 5 1
2 Solution: Let s first compute the determinant of A: det(a) 2(5) 7(1) 3 Since det(a) 0, A is invertible (ie A has an inverse) Using the formula for the inverse of a 2 2 matrix, we have A 1 1 d b det(a) c a You should verify that A 1 A I and AA 1 I ] Theorem: If A is an invertible n n matrix, then for each b in R n, the equation Ax b has the unique solution x A 1 b Note that this theorem alone tells us that an invertible matrix has a pivot in every row and in every column Clearly, this means that the columns of an n n invertible matrix not only form a linearly independent set, but they span R n as well Example: Solve the system 2 7 x1 1 5 x Solution: Recall that the matrix A is invertible with inverse 1 5 A Therefore, the system has a unique solution Let b x A 1 b ] ] 12 ] 4 2] 12 Then
3 The following theorem is a list of necessary conditions for invertible matrices Theorem: a If A is an invertible matrix, then A 1 is invertible and (A 1 ) 1 A b If A and B are n n invertible matrices, then so is AB, and the inverse of AB is the product of the inverses of A and B in the reverse order That is (AB) 1 B 1 A 1 c If A is an invertible matrix, then so is A T, and the inverse of A T is the transpose of A 1 That is, (A T ) 1 (A 1 ) T Theorem: An n n matrix A is invertible if and only if A is row equivalent to I n, and in this case, any sequence of elementary row operations that reduces A to I n also transforms I n into A 1 This theorem states that a matrix is invertible if and only if its reduced row echelon matrix is the identity matrix A Method for Finding the Inverse of a Square Matrix Let A be an n n matrix and I be the n n identity matrix The following two steps allow us to determine A 1 if it exists (1) Set up the augmented matrix A I ] and reduce it to reduced row echelon form (2) If the reduced row echelon matrix obtained in part (1) has the form I B ], where B is an n n matrix, then the inverse of A exists and B A 1 Otherwise, A is not invertible
4 1 3 Example: Let A Determine A Solution: Set up the augmented matrix A I ] and reduce it to reduced row echelon form: Since the lefthand side of the reduced row echelon augmented matrix is the identity matrix, the righthand side of the matrix must be A 1 Thus, 5 3 A Let s check our result by using the quick formula for the inverse of a 2 2 matrix The determinant of A is det(a) 1(5) 3(2) 1 Since det(a) 0, A 1 exists and is given by A which confirms our result ] ]
5 1 0 2 Example: Let A Determine A 1 Solution: Set up the augmented matrix A I ] and reduce it to reduced row echelon form: Since the lefthand side of the reduced row echelon augmented matrix is the identity matrix, the righthand side of the matrix must be A 1 Thus, A You should verify that AA 1 A 1 A I Remarks: In this section, we assumed that the matrix A has size n n, which means that the systems of equations that we have considered here have n equations and n variables However, as we all know, a system of equations may have more equations than variables or vice versa This begs the following question: Is there a way to solve the equation Ax b using the inverse approach if A is not a square matrix? The answer to this question is yes, but to solve such an equation, we would need what are called left and right inverses of A If A is an m n matrix, then an n m matrix X is called a left inverse of A if XA I, where I is the m m identity matrix If such a matrix X exists, then we say that A is left invertible Right inverses are defined in a similar manner As mentioned before, if A is a square matrix, its inverse (assuming it exists) is unique However, if A is a left invertible m n matrix, then A may have more than one left inverse Surprisingly, this does not mean
6 that Ax b is consistent It is very possible for a system to be inconsistent even if a left inverse exists One known fact is that if A is left invertible, then Ax b can have at most one solution If the system just happens to have a solution, then it must be x Xb, where X is the left inverse of A that yields that solution Note that if A is right invertible, then the system has at least one solution (ie the system is consistent) If A is square and invertible, then the left and right inverses are identical and unique, and thus, Ax b has a unique solution Note that we will not consider left and right inverses in this course However, it is worth mentioning a few things about them to understand why an invertible matrix is assumed to be square Another way to justify this is through the notions of onetoone and onto linear transformations (more concepts that we will not be covering in this course) In Section 23, we will discuss necessary and sufficient conditions for a square matrix to be invertible
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