Explanation Given Information: The provided matrices are, I = [ 1 0 0 0 1 0 0 0 1 ] , A = [ .1 0 .1 .2 .1 .1 .1 .2 0 ] , X = [ x y z ] and D = [ 300 100 100 ] . Calculation: The 3 × 3 identity matrix is provided as, I = [ 1 0 0 0 1 0 0 0 1 ] Consider the provided matrices, A = [ .1 0 .1 .2 .1 .1 .1 .2 0 ] , X = [ x y z ] and D = [ 300 100 100 ] Compute the matrix ( I − A ) as, ( I − A ) = [ 1 0 0 0 1 0 0 0 1 ] − [ .1 0 .1 .2 .1 .1 .1 .2 0 ] = [ 1 − .1 0 − 0 0 − .1 0 − .2 1 − .1 0 − .1 0 − .1 0 − .2 1 − 0 ] = [ .9 0 − .1 − .2 .9 − .1 − .1 − .2 1 ] The equivalent matrix for inverse of ( I − A ) is [ .9 0 − .1 − .2 .9 − .1 − .1 − .2 1 1 0 0 0 1 0 0 0 1 ] . Compute the equivalent matrix for inverse of ( I − A ) as follows: Multiply row 1 by 1 0.9 to get a leading 1 : [ .9 0 − .1 − .2 .9 − .1 − .1 − .2 1 1 0 0 0 1 0 0 0 1 ] → ( 1 0.9 ) R 1 [ 1 0 − .11 − .2 .9 − .1 − .1 − .2 1 1.11 0 0 0 1 0 0 0 1 ] Multiply row 1 by 0.2 and add the result to row 2 and multiply row 1 by 0.1 and add the result to row 3 . [ .9 0 − .1 − .2 .9 − .1 − .1 − .2 1 1 0 0 0 1 0 0 0 1 ] → R 3 + ( 0.1 ) R 1 R 2 + ( 0.2 ) R 1 [ 1 0 − .11 0 .9 − .12 0 − .2 0.99 1.11 0 0 0.22 1 0 0.11 0 1 ] Multiply row 2 by 1 0.9 to get a leading 1 : [ 1 0 − .11 0 .9 − .12 0 − .2 0.99 1.11 0 0 0.22 1 0 0.11 0 1 ] → ( 1 0.9 ) R 2 [ 1 0 −

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ISBN: 9780134464053

Author: HAIR

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Chapter 2.6, Problem 2CYU

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To calculate: The solution of matrix equation (I−A)X=D, where I=[100010001], A=[.10.1.2.1.1.1.20], X=[xyz] and D=[300100100].

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Chapter 2.6, Problem 1CYU

Chapter 2.6, Problem 1E

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Chapter 2 Solutions

EBK FINITE MATHEMATICS & ITS APPLICATIO

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The solution of matrix equation ( I − A ) X = D , where I = [ 1 0 0 0 1 0 0 0 1 ] , A = [ .1 0 .1 .2 .1 .1 .1 .2 0 ] , X = [ x y z ] and D = [ 300 100 100 ] .

Solution Summary: The author calculates the solution of matrix equation (I-A)X=D.

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To calculate: The solution of matrix equation (I−A)X=D, where I=[100010001], A=[.10.1.2.1.1.1.20], X=[xyz] and D=[300100100].

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Chapter 2 Solutions