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Unformatted text preview: 1.5. Inverses of Matrices* There are actually two things we need to do, to make B the X = formula usable. A • Find a matrix I such that IM = M for all matrices M (where the product is deﬁned); • Given a matrix A, ﬁnd a matrix C so that CA = I. Then our idea will go through . . . * And one last part of Section 1.4. AX = B C (AX ) = CB (CA)X = CB IX = CB X = CB So which matrix will make IM = M for all matrices 1 −1 2 M ? (In particular, what if M = 2 −2 8 ?) 1 −2 0 So which matrix will make IM = M for all matrices 1 −1 2 M ? (In particular, what if M = 2 −2 8 ?) 1 −2 0 What are the dimensions of I ? I ? ×? · M 3×3 = M 3×3 So which matrix will make IM = M for all matrices 1 −1 2 M ? (In particular, what if M = 2 −2 8 ?) 1 −2 0 What are the dimensions of I ? I ? ×3 · = M 3×3 = M 3×3 So which matrix will make IM = M for all matrices 1 −1 2 M ? (In particular, what if M = 2 −2 8 ?) 1 −2 0 What are the dimensions of I ? I 3×3 · M 3×3 = M 3×3 So I must be a 3 × 3 matrix. How should we ﬁll in the entries? So which matrix will make IM = M for all matrices 1 −1 2 M ? (In particular, what if M = 2 −2 8 ?) 1 −2 0 111 No, 1 1 1 doesn’t work . . . 111 1 1 1 1 1 1 1 1 −1 2 4 1 · 2 −2 8 = 4 1 1 −2 0 4 −5 −5 −5 10 10 10 So which matrix will make IM = M for all matrices 1 −1 2 M ? (In particular, what if M = 2 −2 8 ?) 1 −2 0 However, 1 0 0 0 1 0 0 1 −1 2 1 −1 2 0 · 2 −2 8 = 2 −2 8 1 1 −2 0 1 −2 0 (In fact, this is the only such matrix that works.) In general, the m × m matrix 1 0 ··· 0 1 ··· . . . .. Im = . . . . 0 0 ··· 0 0 ··· 0 0 . . . 1 0 0 0 . . . 0 1 has the property that Im A = A, whenever A is m × n. The matrix Im is called the identity matrix. This solves one of our problems. Now, let’s suppose we have a matrix A, and we want to ﬁnd a matrix C such that CA = I . For technical reasons, we will also require that AC = CA = I ; if we can ﬁnd this matrix C is called the inverse of A and is denoted A−1 . Note that the matrix A has to be square; that is, the number of rows and columns must be the same. The 1 × 1 case is not interesting, so we’ll move on to the 2 × 2 case. Derivation For Illustrative Purposes Only Suppose A = ab cd (where a, b, c and d are xy . Then CA = I means zw numbers), and C = xy ab 10 · = zw cd 01 Derivation For Illustrative Purposes Only Suppose A = ab cd (where a, b, c and d are xy . Then CA = I means zw bx + dy 10 = bz + dw 01 numbers), and C = ax + cy az + cw Derivation For Illustrative Purposes Only Suppose A = ab cd (where a, b, c and d are numbers), and C = xy . Then CA = I means zw ax + cy =1 bx + dy =0 az + cw = 0 bz + dw = 1 So we have to solve a system of linear equations. Derivation For Illustrative Purposes Only Solving this system yields the formulas d x= ad − bc c z=− ad − bc ab cd
−1 b y=− ad − bc a w= ad − bc So 1 d −b = ad − bc −c a . Derivation For Illustrative Purposes Only Solving this system yields the formulas d x= ad − bc c z=− ad − bc
−1 b y=− ad − bc a w= ad − bc 1 = ad − bc So ab cd d c b 8 . E a Solving this system yields the formulas d x= ad − bc c z=− ad − bc
−1 b y=− ad − bc a w= ad − bc 1 = ad − bc So ab cd d c b 8 , if ad − bc = 0. E a What happens if ad − bc = 0? Then it turns out that the system of linear equations has no solutions, and A does not have an inverse. More terminology: A matrix which has an inverse is called invertible or nonsingular. A matrix which does not have an inverse is called noninvertible or singular. What happens with larger matrices? Since a system of linear equations arises in a natural way, you might suspect that we’ll be doing GaussJordan Elimination to ﬁnd the inverse in general. You would be right . . . Procedure for Finding the Inverse of a General Matrix A • Set up the augmented matrix [ A  I ], where I is the identity matrix. • Do GaussJordan Elimination on this Matrix. • If the reduced row echelon form looks like [ I  C ], then C = A−1 . Otherwise, the matrix A has no inverse. 1 −1 2 Example. Find the inverse of the matrix 2 −2 8 , 1 −2 0 if it exists. 1 −1 2 Example. Find the inverse of the matrix 2 −2 8 , 1 −2 0 if it exists. We 1 2 1 will do row reduction on the augmented matrix −1 21 0 0 −2 80 1 0 . −2 00 0 1 1 2 1 −1 −2 −2 21 80 00 0 1 0 0 0 1 1 2 1 21 0 0 8 0 1 0 00 0 1 2 1 − − − − → 1 −1 −−−− 0 2 −2 1 0 4 −2 1 −2 0 0 −1 −2 −2 0 1 0 0 0 1 1 2 1 21 0 80 1 00 0 − − − − → 1 −1 −−−− 0 2 −2 1 0 1 −2 −− − − − − − → 1 −1 0 3−1 0 0 −1 −1 −2 −2 0 0 1 2 1 4 −2 0 0 2 1 4 −2 −2 −1 0 1 0 0 1 0 0 0 1 0 0 1 1 2 1 21 0 80 1 00 0 − − − − → 1 −1 −−−− 0 2 −2 1 0 1 −2 −− − − − − − → 1 −1 0 3−1 0 0 −1 −−−− − − − → 1 −1 0 −1 2↔3 0 0 −1 −2 −2 0 0 1 2 1 4 −2 0 0 2 1 4 −2 −2 −1 2 1 −2 −1 4 −2 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 1 0 −−−− −−−→ 1 0 2↔3 0 −1 −1 0 2 1 −2 −1 4 −2 0 0 1 0 1 0 2 1 0 −−−− − − − → 1 −1 0 −1 −2 −1 2↔3 0 0 0 4 −2 1 − − − 1 −1 − −→ 2 1 0 −2 0 1 2 1 0 13 4 0 0 1 −1/2 1/4 0 1 0 0 −1 0 2 1 0 0 −−−− − − − → 1 −1 0 −1 −2 −1 2↔3 0 1 0 0 4 −2 1 0 − − − 1 −1 − −→ 2 1 0 0 −2 0 1 2 1 0 −1 13 4 0 0 1 −1/2 1/4 0 − − − − → 1 −1 −−−− 0 2 −1/2 0 1 −2 3 0 1 0 2 −1/2 −1 2 −2 3 0 0 1 −1/2 1/4 0 −−−− −−−→ 1 0 2↔3 0 − − − 1 −1 − −→ −2 0 1 13 4 0 0 −− − −→1 −−−− 1 −2 3 0 2 −2 3 0 −− − − − − −→ 1 0 1+2 0 −1 −1 0 2 1 −2 −1 4 −2 0 0 1 0 0 1/4 2 1 2 1 1 −1/2 −1 1 0 0 1 0 0 1 0 0 −1 0 −1/2 −1/2 1/4 0 −1 0 −1 −1 0 0 2 0 2 1 −1/2 0 4 0 2 1 −1/2 −1 −1/2 1/4 Since we have the identity matrix on the lefthand side of our ﬁnal matrix 1 0 0 4 −1 −1 0 1 0 2 −1/2 −1 , this means 0 0 1 −1/2 1/4 0 −1 1 −1 2 4 −1 −1 2 −2 8 = 2 −1/2 −1 . 1 −2 0 −1/2 1/4 0 Since we have the identity matrix on the lefthand side of our ﬁnal matrix 1 0 0 4 −1 −1 0 1 0 2 −1/2 −1 , this means 0 0 1 −1/2 1/4 0 −1 1 −1 2 4 −1 −1 2 −2 8 = 2 −1/2 −1 . 1 −2 0 −1/2 1/4 0 Okay; now that we have the inverse of our matrix, what can we do with it? The system of linear equations x − y + 2z = 3 2x − 2y + 8z = 22 x − 2y =2 can be written in the form AX = B , where 1 −1 2 A = 2 −2 8 , 1 −2 0 x X = y , z 3 B = 22 . 2 and The system of linear equations x − y + 2z = 3 2x − 2y + 8z = 22 x − 2y =2 can be written 1 −1 A = 2 −2 1 −2 in the form AX = B , where 2 x 3 8 , X = y , and B = 22 , z 2 0 and the solution to the system is 4 −1 −1 3 −12 X = A− 1 B = 2 −1/2 −1 · 22 = −7 . −1/2 1/4 0 2 4 Some properties of the inverse of a matrix: • (A−1 )−1 = A • You CAN cancel invertible matrices: If AB = AC , and A is invertible, then B = C . • (AB )−1 = B −1 A−1 (not A−1 B −1 ) • (An )−1 = (A−1 )n And one more question: In general, do we have an indicator which will tell us whether a matrix is invertible (like ad − bc for 2 × 2 matrices)? ...
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This note was uploaded on 03/10/2010 for the course MATHEMATIC 242 taught by Professor Heckman during the Spring '10 term at University of Arizona Tucson.
 Spring '10
 heckman
 Matrices

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