Explanation Procedure used: 1. Form an augmented matrix [ A | I ] where a matrix A is placed along with a same dimension identity matrix I separated by a line. 2. Apply row operations on [ A | I ] in order to obtain another augmented matrix [ I | B ] where I , is again an identity matrix and B is an inverse of A. 3. If the reduction process yields a row of zeros in the left half of the augmented matrix, then the inverse of A does not exist. Calculation: Take A = [ − 1 1 0 − 2 3 − 2 2 − 2 1 ] and X = [ x y z ] and transform [ − 1 1 0 − 2 3 − 2 2 − 2 1 ] [ x y z ] = [ 3 5 8 ] as follows. A − 1 A X = A − 1 [ 3 5 8 ] ( Multiply with A − 1 on RHS of both sides ) I X = A − 1 [ 3 5 8 ] ( ∵ A − 1 A = I ) X = A − 1 [ 3 5 8 ] Obtain the augmented matrix as follows. [ − 1 1 0 1 0 0 − 2 3 − 2 0 1 0 2 − 2 1 0 0 1 ] Apply the row operation − 2 R 1 + R 2 → R 2 and 2 R 1 + R 3 → R 3 and reduce the matrix. [ − 1 1 0 1 0 0 0 1 − 2 − 2 1 0 0 0 1 2 0 1 ] Apply the row operation R 2 + 2 R 3 → R 2 and reduce the matrix

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Chapter 7.4, Problem 17E

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To find: The matrix [xyz] such that [−110−23−22−21][xyz]=[358].

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Chapter 7.4, Problem 16E

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The matrix [ x y z ] such that [ − 1 1 0 − 2 3 − 2 2 − 2 1 ] [ x y z ] = [ 3 5 8 ] .

Solution Summary: The author explains how to form an augmented matrix left[A|Iright], where a matrix A is placed along with an identity matrix and inverse of A. If the reduction process yields

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To find: The matrix [xyz] such that [−110−23−22−21][xyz]=[358].

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