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.

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.

cut 138 centimeters off

step-by-step explanation:

250-112 = 138

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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