Explanation Theorem used: “If an n × n matrix A can be transformed into the n × n identity I by a sequence of elementary row operations, then A is nonsingular. The same sequence of operations that transforms A into the identity I will also transform I into A − 1 ”. A matrix is said to be singular if its determinant is equal to 0 and non-singular if its determinant is not equal to 0. Also, inverse of any matrix exists if and only if it is non singular. Calculation: Calculate the determinant value of the matrix [ 1 2 3 0 1 4 0 0 8 ] . det A = det [ 1 2 3 0 1 4 0 0 8 ] = 1 ⋅ det [ 1 4 0 8 ] − 2 ⋅ det [ 0 4 0 8 ] + 3 ⋅ det [ 0 1 0 0 ] = 1 ⋅ 8 − 2 ⋅ 0 + 3 ⋅ 0 =8 Here, the determinant of given matrix is 8 . So, the given matrix is a nonsingular matrix. Hence, its inverse exists. Now, the augmented matrix corresponding to the given matrix is shown below as [ A | I 3 ] . [ A | I 3 ] = [ 1 2 3 1 0 0 0 1 4 0 1 0 0 0 8 0 0 1 ] Perform row operations on the augmented matrix [ A | I 3 ] to obtain a matrix of the form [ I 3 | B ] , where the matrix B = A − 1 . Apply R 3 → 1 8 ⋅ R 3 in the above matrix. [ 1 2 3 1 0 0 0 1 4 0 1 0 0 0 1 0 0 1 8 ] Apply R 2 → R 2 − 4 ⋅ R 3 in the above matrix

The inverse matrix A − 1 of the matrix A = [ 1 2 3 0 1 4 0 0 8 ] or show that no inverse exists.

Advanced Engineering Mathematics

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 8.6, Problem 24E

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The inverse matrix A−1 of the matrix A=[123014008] or show that no inverse exists.

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Chapter 8.6, Problem 23E

Chapter 8.6, Problem 25E

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The inverse matrix A − 1 of the matrix A = [ 1 2 3 0 1 4 0 0 8 ] or show that no inverse exists.

Solution Summary: The author calculates the inverse matrix A-1 of the matrix, which is non-singular if its determinant is equal to 0.

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The inverse matrix A−1 of the matrix A=[123014008] or show that no inverse exists.

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