Explanation Given: The matrix A = [ 2 2 − 1 0 3 − 1 − 1 − 2 1 ] . Calculation: Consider the provided matrix A = [ 2 2 − 1 0 3 − 1 − 1 − 2 1 ] . Compute matrix in form of: [ A | I ] The augment matrix with identity matrix is: [ A | I ] = [ 2 2 − 1 1 0 0 0 3 − 1 0 1 0 − 1 − 2 1 0 0 1 ] Now, use row operation to reduce the matrix in row echelon form for inverse: R 3 → R 3 + 1 2 × R 1 , R 3 → R 3 + 1 3 × R 2 The resulting matrix is: [ 2 2 − 1 1 0 0 0 3 − 1 0 1 0 0 0 1 6 1 2 1 3 1 ] Again, apply the row operations: R 3 → 6 × R 3 , R 2 → R 2 + 1 × R 3 , R 1 → R 1 + 1 × R 2 , R 2 → 1 3 × R 2 , R 1 → R 1 − 2 × R 2 , R 1 → 1 2 × R 1 The resulting matrix is: [ A | I ] = [ 1 0 0 1 0 1 0 1 0 1 1 2 0 0 1 3 2 6 ] = [ I | B ] Where A − 1 = [ B ] Therefore, the inverse of matrix is: A − 1 = [ 1 0 1 1 1 2 3 2 6 ] Now, check the result for A A − 1 = I 3 And A − 1 A = I 3 Here, A = [ 2 2 − 1 0 3 − 1 − 1

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Chapter 9.4, Problem 23E

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To calculate: The inverse of matrix A=[22−103−1−1−21] and also check the provided result.

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Chapter 9.4, Problem 22E

Chapter 9.4, Problem 24E

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Chapter 9 Solutions

ALGEBRA AND TRIGONOMETRY

Ch. 9.1 - Fill in each blank so that the resulting statement...Ch. 9.1 - Fill in each blank so that the resulting statement...Ch. 9.1 - Prob. 3CVC Ch. 9.1 - Prob. 4CVC Ch. 9.1 - Fill in each blank so that the resulting statement...Ch. 9.1 - Fill in each blank so that the resulting statement...Ch. 9.1 - In Exercises 1-8, write the augmented matrix for...Ch. 9.1 - In Exercises 1-8, write the augmented matrix for...Ch. 9.1 - In Exercises 18, write the augmented matrix for...Ch. 9.1 - In Exercises 1-8, write the augmented matrix for...

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The inverse of matrix A = [ 2 2 − 1 0 3 − 1 − 1 − 2 1 ] and also check the provided result.

Solution Summary: The author calculates the inverse of the matrix A=left[cc2& 2& -1 0& 3&

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To calculate: The inverse of matrix A=[22−103−1−1−21] and also check the provided result.

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Chapter 9 Solutions