Matrix determinants

[soft prairie]

Question is show how M is non singular for all values of k

?

I don’t understand that bit 12b for mark scheme

That’s the only bit of the mark scheme tackling that but of the question

[pliant epoch]

basically for a matrix to be singular, the matrix must always be equal to 0

for a matrix to be non-singular the matrix dterminant must be not 0

therefore the matrix must be above 0 always or below 0 always

this gives us the idea of a function or in this case an expression which is always positive no matter what value of k

completing the square is useful here because the bit that is squared is always equal to 0 or positive

therefore when u complete the square, the +9.75 always means even when the squared bracket is 0

the function itself is still above 0

that means the determinant is always above 0

and so the matrix is always non singular

[soft prairie]

No I totally understand this yes but in the mark scheme where did k^2 + 3k + 9/4 come from

And then the -9/4 at the end

[pliant epoch]

they completed square in a weird way

(k + 3/2)^2 - 9/4 + 12

so (k + 3/2)^2 + 39/4

thats how u complete the square

[soft prairie]

But be cause there’s a -9/4 at the end wouldn’t that cancel the one in the bracket and leave us with js +12

[ebon leaf]

can u send the question

[soft prairie]

Yes give me a couple of minutes

[ebon leaf]

yup thats exactly what completing the square is

you cant not spawn in a term and not account for it

[dull moth]

All your doing is showing that
k^2 + 3k +12 is greater than 0
completeing the square helps you do that because you know any squared term >=0 so it makes it easy since u will have a term thats >=0 plus a constant