Ask Question
17 January, 12:13

Prove that if the product AB of two n * n matrices is invertible, then both A and B are invertible. Even if you know about determinants, do not use them, we did not cover them yet. Hint: use previous 2 problems.

+2
Answers (1)
  1. 17 January, 12:32
    0
    Step-by-step explanation:

    Given two matrices A and B

    The product of the matrix A and B is AB

    And we know that AB is an "n*n" matrix

    We can defined invertible by the rank of the matrix, then, RANK of a matrix is equal to the maximum number of linearly independent columns or rows in the matrix

    Also, the rank of an "n*n" matrix is at most "n", an "n*n" having a rank "n" is invertible.

    Since, AB is an "n*n" matrix,

    AB will be invertible if and only if RANK (AB) = n, and it has a pivots in every rows and columns, so that it ranks will be n,

    NOTE: If matrix A is an "n*n"

    , if matrix B is an "n*n"

    Then, the rank of A cannot be more than "n" and the rank of B cannot be more than "n".

    Therefore, Both A and B have rank "n" and they have a pivots in every row and column, so they are both invertible.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Prove that if the product AB of two n * n matrices is invertible, then both A and B are invertible. Even if you know about determinants, do ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers