Consider the following matrix a, and let b = a−1. find b31, b32, and b33. (i. e., find the entries in the third row of a−1.
a = 1 1 -2
-9 -1 0
-2 -1 2

The result of operation A - 1 is:
 0     0    -3
-10  -2    -1
-3    -2     1

which is equal to matrix B.

Looking at the third row we have:
b31 = -3
b32 = -2
b33 = 1
which are the entries in the third row of A - 1

Idon’t know what the answer is i wish i could

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt