matrix question: if\[A=\left[\begin{matrix}-5&-2&-4\\0&-4&4\end{matrix}\right]\]and\[C=\left[\begin{matrix}7&-5\\-4&10\end{matrix}\right]\]then what is\[-5AA^{T}+4C\]
ah much better :)
definitely better
is A even invertible?I don't think so.
oh you want transpose...
how do i do this?
you are not familiar with the transpose part? all you do to find \(A^T\) is change the \(i\) an \(j\) locations of each element switch for instance....
http://tutorial.math.lamar.edu/Classes/LinAlg/MatrixArithmetic.aspx example 10 shows how to do transposes
OK
example:\[\text{ if }A=\left[\begin{matrix}a&b\\c&d\\e&f\end{matrix}\right]\text{ then }A^T=\left[\begin{matrix}a&c&e\\b&d&f\end{matrix}\right]\]
it sort of "flips" the matrix
OK AND THEN ITS FLIPPED, HOW WOULD I FIND THE ANS?
DO I HAVE TO TRANSPOSE C TOO?
could you please not use all caps, it makes it seem as though you are shouting you matrix A by it's transpose do you not know how to multiply the two matrices?
you matrix multiply A by...*
Ok no caps...sorry
multiplu matrix a when transposed by -5
is that correct?
no, -5 is a scalar there is no such thing as transposing "by" some number there is only just the transpose
ok so i just multiply a by -5?
right first take the transpose then multiply \(AA^T\) then multiply by 5 then subtract 3C
ok lets see
its + 4c / not subtract 3 c
well I'm glad someone's paying attention ;)
-5*a*a^t = (-125, -50,-100) (-50,-100,40) (-100,40,160)
you should get a 2x2 matrix from the multiplication
how do i add 4 c...the dimensions dont match
ok when u say " then multiply A*AT" - i will have to multiply the transposed by A - which is a 2 x 3 * a 3 x 3 - which gives a 3 x 3
\[A_{i\times j}\implies A_{j\times i}^T\]where \(i\) is the number of rows and \(j\) is the number of columns, so\[A_{i\times j}A_{j\times i}^T=B_{i\times i}\]
that means you final matrix will be square, having the the same number of columns as it had rows in this case its\[A_{2\times3}A_{3\times2}^T=B_{2\times2}\]so the dimensions are \(2\times2\)
ok
i see
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