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 left-hand side of the reduced row echelon augmented matrix is the identity matrix, the right-hand 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 left-hand side of the reduced row echelon augmented matrix is the identity matrix, the right-hand 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 one-to-one 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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