Let A be a 3 × 3 matrix whose entries are 17 or 0. What is the largest possible value for det(A) ?

- anonymous

Let A be a 3 × 3 matrix whose entries are 17 or 0. What is the largest possible value for det(A) ?

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Since it say's OR I am actually not sure how I would do this. Setting all the entries to 17 simply makes the determinant 0.

- anonymous

true because all would be 0

- anonymous

|dw:1359528317295:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

That's such a stupid question then...

- anonymous

Are you sure though? What if some of them are 17 and some of them are 1?

- anonymous

0* now 1 sorry.

- anonymous

not*

- anonymous

Like if I change 2 of those to 0 the determininant become 4913 all of a sudden.

- anonymous

maybe thinking off this as M or C matrices i forgot what they're called

- anonymous

Cofactor?

- anonymous

|dw:1359528522846:dw|

- anonymous

now each one of those cofactors i think they're called =|dw:1359528606154:dw|

- anonymous

that's why you'll get zero if you put 17's all in it

- anonymous

Okay but what if I put SOME of them as 17 and SOME of them as 0. Then it a lot bigger.

- anonymous

there was also a theory i think if i remember that if you somehow prove that two vectors are just combinations of the other... then you can prove it's zero also

- anonymous

because if two are the combinations of each other than you can reduce them down to [17 17 17] and then if you can easily get a row of zeroes

- anonymous

|dw:1359528822331:dw|

- anonymous

THat's 4913.

- anonymous

SO clearly, there is a maximum.

- anonymous

|dw:1359528911003:dw|

- anonymous

Nvm, I got it.

- anonymous

|dw:1359528962928:dw|

- anonymous

is this what you got?

- anonymous

so i'm thinking the rule is the largest determinant comes from having vectors that are dependent of one anothre

- anonymous

i mean independent

- anonymous

so the largest happens when all are independent of one another

Looking for something else?

Not the answer you are looking for? Search for more explanations.