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2 years ago

The matrix a is 13 by 91. give the smallest possible dimension for nul a.

1 answer:

7 0

Use the rank-nullity theorem. It says that the rank of a matrix $\mathbf A$ , $\mathrm{rank}(\mathbf A)$ , has the following relationship with its nullity $\mathrm{null}(\mathbf A)$ and its number of columns $n$ :

$\mathrm{rank}(\mathbf A)+\mathrm{null}(\mathbf A)=n$

We're given that $\mathbf A$ is $13\times91$ , i.e. has $n=91$ columns. The largest rank that a $m\times n$ matrix can have is $\min\{m,n\}$ ; in this case, that would be 13.

So if we take $\mathbf A$ to be of rank 13, i.e. we maximize its rank, we must simultaneously be minimizing its nullity, so that the smallest possible value for $\mathrm{null}(\mathbf A)$ is given by

$13+\mathrm{null}(\mathbf A)=91\implies\mathrm{null}(\mathbf A)=91-13=78$

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A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Katena32 [7]

Answer:

(a) The fraction of the calls last between 4.50 and 5.30 minutes is 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is 0.1271.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is 0.745.

(e) The time is 5.65 minutes.

Step-by-step explanation:

We are given that the mean length of time per call was 4.5 minutes and the standard deviation was 0.70 minutes.

Let X = the length of the calls, in minutes.

So, X ~ Normal( $\mu=4.5,\sigma^{2} =0.70^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 4.5 minutes

$\sigma$ = standard deviation = 0.7 minutes

(a) The fraction of the calls last between 4.50 and 5.30 minutes is given by = P(4.50 min < X < 5.30 min) = P(X < 5.30 min) - P(X $\leq$ 4.50 min)

P(X < 5.30 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{5.30-4.5}{0.7}$ ) = P(Z < 1.14) = 0.8729

P(X $\leq$ 4.50 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.5-4.5}{0.7}$ ) = P(Z $\leq$ 0) = 0.50

The above probability is calculated by looking at the value of x = 1.14 and x = 0 in the z table which has an area of 0.8729 and 0.50 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.8729 - 0.50 = 0.3729.

(b) The fraction of the calls last more than 5.30 minutes is given by = P(X > 5.30 minutes)

P(X > 5.30 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{5.30-4.5}{0.7}$ ) = P(Z > 1.14) = 1 - P(Z $\leq$ 1.14)

= 1 - 0.8729 = 0.1271

The above probability is calculated by looking at the value of x = 1.14 in the z table which has an area of 0.8729.

(c) The fraction of the calls last between 5.30 and 6.00 minutes is given by = P(5.30 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 5.30 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 5.30 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{5.30-4.5}{0.7}$ ) = P(Z $\leq$ 1.14) = 0.8729

The above probability is calculated by looking at the value of x = 2.14 and x = 1.14 in the z table which has an area of 0.9838 and 0.8729 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.8729 = 0.1109.

(d) The fraction of the calls last between 4.00 and 6.00 minutes is given by = P(4.00 min < X < 6.00 min) = P(X < 6.00 min) - P(X $\leq$ 4.00 min)

P(X < 6.00 min) = P( $\frac{X-\mu}{\sigma}$ < $\frac{6-4.5}{0.7}$ ) = P(Z < 2.14) = 0.9838

P(X $\leq$ 4.00 min) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{4.0-4.5}{0.7}$ ) = P(Z $\leq$ -0.71) = 1 - P(Z < 0.71)

= 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 2.14 and x = 0.71 in the z table which has an area of 0.9838 and 0.7612 respectively.

Therefore, P(4.50 min < X < 5.30 min) = 0.9838 - 0.2388 = 0.745.

(e) We have to find the time that represents the length of the longest (in duration) 5 percent of the calls, that means;

P(X > x) = 0.05 {where x is the required time}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-4.5}{0.7}$ ) = 0.05

P(Z > $\frac{x-4.5}{0.7}$ ) = 0.05

Now, in the z table the critical value of x which represents the top 5% of the area is given as 1.645, that is;

$\frac{x-4.5}{0.7}=1.645$

${x-4.5}{}=1.645 \times 0.7$

x = 4.5 + 1.15 = 5.65 minutes.

SO, the time is 5.65 minutes.

7 0

2 years ago

The water under a draw bridge was 15 feet deep at midnight. As the tide went out, the

Vsevolod [243]

Answer:

80 minutes or 1 h 20 minutes

Step-by-step explanation:

15-13=2

2/0.025=80

So we can conclude that it takes 80 minutes till it reaches 13 feet.

7 0

2 years ago

The profit earned by a sporting goods outlet is modeled in the graph below, where x is the number of years since the outlet open

LenaWriter [7]

The domain would be x ≥ 0.

This is because the outlet cannot have profit before it was open. Therefore, the growth must be from year 0 to present. If they give a year as starting, you can have an upper limit too, but there is not enough information here to determine that information.

4 0

2 years ago

Read 2 more answers

A painter built a ladder using 18 rungs. The rungs on the ladder were 5.7 inches apart and 1.1 inches thick. What is the distanc

xeze [42]

18(1.1)+18(5.7)
18.8+102.6
121.4
The distance is 121.4 inches

4 0

2 years ago

the diagonal length of a square window is 40 cm. find the area of the window. Btw any help is appreciated ! :)

7nadin3 [17]

Ok so area is length times width. A square has the same sides all around, which is called a quadrilateral. So you would plug in 40 for length and width. 40x40= 1,600 cm^2 <———— is your answer.

Hope this helped :)

7 0

2 years ago