# Solutions Manual for Elementary Linear Algebra 8th Edition by Larson IBSN 9781305658004

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Solutions Manual for Elementary Linear Algebra 8th Edition by Larson IBSN 9781305658004 Download at: https://goo.gl/fPQXHc People also search: elementary linear algebra 8th edition pdf download elementary linear algebra 8th edition solutions elementary linear algebra larson pdf larson linear algebra 8th edition pdf elementary linear algebra 8th edition solutions pdf elementary linear algebra cengage elementary linear algebra 5th edition pdf elementary linear algebra 4th edition pdf free
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Solutions Manual for Elementary Linear Algebra 8th Edition by Larson IBSN 9781305658004 Full clear download (no formatting errors) at: http://downloadlink.org/p/solutions-manual-for-elementary-linear-algebra-8th-edition-by-larson-ibsn-9781305658004/  C H A P T E R 2   Matrices   Section 2.1 Operations with M a trices   ..................................................................... 30   Section 2.2 Properties of Matrix Operations........................................................... 36   Section 2.3 The Inverse of a Matrix   ........................................................................ 41   Section 2.4 Elementary Matrices............................................................................. 46   Section 2.5 Markov Chains ..................................................................................... 53   Section 2.6 More Applications of Matrix Operations ............................................ 62   Review Exercises .......................................................................................................... 66   Project Solutions ........................................................................................................... 75    4 4 2 5 2 2 7 2 5 2 2 3   2 5 2 4 0 2 8   C H A P T E R 2   Matrices   Section 2.1 Operations with Matrices   2.  x = 13,  y = 12  4.  x + 2 = 2  x + 6 2    y = 18 − 4 =  x y = 9   2    x = − 8    x = − 4    y + 2 = 11  y = 9 6 − 1  1 4 6 + 1  − 1 + 4 7 3  6. (a)  A +  B = 2 4 + − 1 5 =   2 + ( − 1 )   4 + 5 = 1 9  − 3 5 1 10  − 3 + 1 5 + 10  − 2 15 6 − 1  1 4 6 −   1  − 1 −   4 5  − 5 (b)  A −    B = 2 4 −   − 1 5 =   2 −   ( − 1 )   4 −   5 =   3 − 1  − 3 5 1 10  − 3 −   1 5 −   10  − 4  − 5 6 − 1   2 ( 6 ) 2 ( − 1 )   12  − 2(c) 2  A = 2 2 =   2 ( 2 )   2 ( 4 ) = 8  − 3 5   2 ( − 3 )   2 ( 5 )   − 6 10 12 − 2  1 4 12 −   1  − 2 −   4 11  − 6 (d) 2  A −    B = 4 8 −   − 1 5 =   4 −   ( − 1 )   8 −   5 = 5 3  − 6 10 1 10  − 6 −   1 10 −   10  − 7 0 1 4 6  − 1  1 4   3 −   1   4 7 2 2 (e)   B + 1  A = − 1 5 + 1  2 4 = − 1 5  + 1 2  = 0 7 2 2 1 10  − 3 5 1 10   −   3 5 −   1 25 2 2 2 2 3 2 − 1  0 2 1 3 + 0 2 + 2  − 1 + 1 3 4 0 8. (a)  A +  B = 4 + 5 4 = 2 + 5 4 + 4 5 + = 8 7   0 1 2 2 1 0 0 + 2 1 + 1 2 + 0 2 2 2   3 2 − 1  0 2 1 3 −   0 2 −   2  − 1 −   1 3 0  − 2(b)  A −    B = 4 −   5 4 = 2 −   5 4 −   4 5 −   = − 3 0 0 1 2 2 1 0 0 −   2 1 −   1 2 −   0  − 2 0 2   3 2  − 1   2 ( 3 ) 2 ( 2 )  2 ( − 1 )   6 4  − 2 (c) 2  A =   22 4 5  =   2 ( 2 )   2 ( 4 )   2 ( 5 )   = 4 8 100 1 2   2 ( 0 )   2 ( 1 )  2 ( 2 )   0 2 4 3 2 − 1  0 2 1 6 4  − 2  0 2 1 6 2  − 3(d) 2  A −    B = 2 4 −   5 4 = 8 1 −   5 4 = − 1 4 0 1 2 2 1 0 0 2 4 2 1 0  − 2 1 4   0 2 1 3 2  − 1  0 2 1   3 1  −   1  3 3 1 2 2 2 2 (e)   B + 1  A =   5 4 2  + 1 2 4 5 =   5 4 2  + 1 2  5 = 6 6 9 2 2 2 2 2 1 0 0 1 2 2 1 0   0 1  1 2 3 1    2 2    − 1 − 2 2 3 2   Section 2.1 Ope r atio ns w ith Matrices 31   10. (a)  A +  B is not possible.  A and  B have different sizes.  14. Simplifying the right side of the equation produces(b)  A −    B is not possible.  A and  B have different sizes.  w x   − 4 + 2  y  3 + 2 w   = . 3 6    y x  2 + 2  z    − 1 + 2  x (c) 2  A = 2 2 = 4 By setting corresponding entries equal to each other, you obtain four equations.(d) 2    A −    B is not possible.  A and  B have different  w = − 4 + 2  y   − 2  y + w = − 4   (e) sizes.  B + 1  A is not possible.  A and  B have different  x = 3 + 2 w  y = 2 + 2  z     x −   2 w = 3   y −   2  z = 2sizes.  x = − 1 + 2  x    x = 1   12. (a) c 23   = 5 a 23 + 2 b 23 = 5 ( 2 ) + 2 ( 11 ) = 32 The solution to this linear system is:  x = 1, 1   y = 2   ,(b) c 32   = 5 a 32 + 2 b 32   = 5 ( 1 ) + 2 ( 4 ) = 13   z = −   4   , and w = − 1.  2 −   2 4 1   2 ( 4 ) + ( −   2 )( 2 ) 2 ( 1 ) + ( −   2 )( −   2 )   4 6 16. (a)   AB = = = − 1 4  2 4 1 2  −   2   −   2   − 1 ( 4 ) + 4 ( 2 )  4 ( 2 ) + 1 ( − 1 )   − 1 ( 1 ) + 4 ( −   2 )  4 ( −   2 ) + 1 ( 4 )  4 7  − 9   −   4(b)   BA = = = 2 −   2  − 1 4   2 ( 2 ) + ( −   2 )( − 1 )   2 ( −   2 ) + ( −   2 )( 4 )  6 − 12 1 − 1 7  1 1 2   1 ( 1 )   +   ( − 1 )( 2 )   +   7 ( 1 )   1 ( 1 ) + ( − 1 )( 1 ) + 7 ( − 3 )   1 ( 2 ) + ( − 1 )( 1 ) + 7 ( 2 )   6 − 21 15  18. (a)  AB = 2  − 1 8  2 1 1  = 2 ( 1 ) + ( − 1 )( 2 ) + 8 ( 1 )  2 ( 1 ) + ( − 1 )( 1 ) + 8 ( − 3 )   2 ( 2 ) + ( − 1 )( 1 ) + 8 ( 2 )   = 8  − 23 193 1  − 1 1  − 3 2   3 ( 1 )   +   1 ( 2 )   +   ( − 1 )( 1 )   3 ( 1 ) + 1 ( 1 ) + ( − 1 )( − 3 )   3 ( 2 ) + 1 ( 1 ) + ( − 1 )( 2 )   4 7 5 1 1 2 1  − 1 7   1 ( 1 ) + 1 ( 2 ) + 2 ( 3 )  1 ( − 1 ) + 1 ( − 1 ) 1 + 2 ( 1 )  1 ( 7 ) + 1 ( 8 ) + 2 ( − 1 )   9 0 13 (b)  BA =   2 1 1 2 − 1 8 =   2 ( 1 ) + 1 ( 2 ) + 1 ( 3 ) 2 ( − 1 ) + 1 ( − 1 ) + 1 ( 1 ) 2 ( 7 ) + 1 ( 8 ) + 1 ( − 1 ) = 7  − 2 21 1 − 3 2  3 1  − 1   1 ( 1 ) + ( − 3 )( 2 ) + 2 ( 3 )  1 ( − 1 ) + ( − 3 )( − 1 ) + 2 ( 1 )  1 ( 7 ) + ( − 3 )( 8 ) + 2 ( − 1 )   1 4  − 19 3 2 1 1 2   3 ( 1 )   +   2 ( 2 )   +   1 ( 1 )   3 ( 2 ) + 2 ( − 1 ) + 1 ( −   2 )  8 2 20. (a)  AB = 3 0 4 2 − 1 =   3 ( 1 )   +   0 ( 2 )   +   4 ( 1 )   − 3 ( 2 ) + 0 ( − 1 ) + 4 ( −   2 ) = 1 − 14  −   −   4 −   2  −   4 1  −   2   4 ( 1 ) + ( −   2 )( 2 ) + ( −   4 )( 1 )  4 ( 2 ) + ( −   2 )( − 1 ) + ( −   4 )( −   2 )   −   4 18 (b)  BA is not defined because  B is 3 × 2 and  A is 3 × 3.  − 1   − 1 ( 2 )   − 1 ( 1 )   − 1 ( 3 )   − 1 ( 2 )   −   2  − 1 − 3  −   2  22. (a)   AB = 2 [ 2 1 3 2 ] = 2 ( 2 )   2 ( 1 )  2 ( 3 )  2 ( 2 )   =   4 2 6 4 −   2   −   2 ( 2 ) −   2 ( 1 )  −   2 ( 3 )  −   2 ( 2 )   −   4 −   2 −   6  −   4 1 − 1   1 ( 2 )   1 ( 1 )  1 ( 3 )  1 ( 2 )  2 1 3 2 2(b)   BA = [ 2 1 3 2 ]   = 2 ( − 1 ) + 1 ( 2 ) + 3 ( −   2 ) + 2 ( 1 )   = [ −   4 ]   −   1   24. (a)  AB is not defined because  A is 2 × 2 and  B is 3 × 2.   2 1   2 ( 2 ) + 1 ( 5 )  2 ( − 3 ) + 1 ( 2 ) 9 −   4 2  −   3 (b)   BA = 1 3  = 1 ( 2 ) + 3 ( 5 )  1 ( − 3 ) + 3 ( 2 )  = 17 3 5 2 2 − 1   2 ( 2 ) + ( − 1 )( 5 )   2 ( − 3 ) + ( − 1 )( 2 )   − 1  − 8

May 6, 2018

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May 6, 2018
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