I can't seem to work this problem right....The question is to find the product of [[-1 0 7], [3 -5 2]] multiplied by [[6], [-4], [1]]
[-1 0 7 ] * [[6],[-4],[1]] = -1 * 6 + 0 * (-4) + 7 * 1 = -6 + 7 = 1. This is the product of the first row with the first column, so it goes in first row and first column.
[3 -5 2 ] * [[6],[-4],[1]] = 3 * 6 - 5 * (-4) + 2 * 1 = 18 + 20 + 2 = 40. This is the product of the second row with the first column, so it goes in second row and first column.
The product matrix is therefore [ [1], [40]].
On our last quiz, what did you mean when you wanted us to give the solution of the matrix and verify it? did you want us to prove why that's the answer, or what?
On the last quiz the question sequence was as follows:
The following matrix given below represents a system of simultaneous equations.
Give the set of equations the matrix represents.
Reduce the matrix.
Give the system of equations represented by the reduced matrices.
Give the solution to the system and verify it.
Graph the set of equations represented by the original matrix.
You weren't asked for the solution to the matrix, but to the system, which refers to the system of equations represented by the reduced matrices. This would be something like, for example, x = -3/2 and y = 4/5. To verify the solution you plug it into the original system of equations and see if you get identities.
Do we need to know the properties of Matrix Addiction for our test?
You probably mean Matrix Addition. Matrix Addiction is a rare but serious condition requiring prolonged treatment.
Seriously, yes you do need to know the properties of matrix addition to the extent required to answer questions such as those posed in the text.
With AX=B, do we just find out what number is multiplied to A to give us B?
To solve A X = B you have to multiply both sides on the left by A^-1, giving you
A^-1 * (A * X) = A^-1 * B. This simplifies to
(A^-1 * A) * X = A^-1 * B, which is the same as
I * A = A^-1 * B so that
A = A^-1 * B.
A^-1 is the inverse of the matrix A, obtained by row-reducing the augmented matrix [ A | I ] until you achieve the matrix [ I | A^-1 ]. When the left half becomes I, the right half will be A^-1.
how would you solve a matrix that is 3x3 multiplied by a 3x1?
Multiply each row of the first matrix by the single column of the second. The product matrix will contain each product in the row and column which correspond to the row and column used to obtain that product.
How do you do our end of the year average, is it mostly our test grades, and can asking our questions each week help with our grade...I hope it does
Class participation and asking questions can count as much as 20% of your grade; asking questions counts particularly heavily.