| 3) a.) Dimension - The number of rows and columns in a matrix. If you were to add two matrices the number of rows must be the same and the number of columns must also be the same. A 2 by 3 matrix, 2 is the number of rows and 3 is the number of columns. Identity Matrix - It is a matrix that if multiplied by another matric it will make variables stand alone. [1 0 0] [0 1 0] [0 0 1] This is an identity matrix. This is what happens if you multiply it by another matrix like [a] [b] [f] 1*a = a 0*b = 0 0*f = 0 0*a = 0 1*b = b 0*f = 0 0*a = 0 0*b = 0 1*f = f Inverse Matrix - An inverse matrix is made from a coefficient matrix [1 1 0 0 0 0] -1 [1 1 1 0 -1 -1] [0 0 1 1 0 1] [0 -1 -1 0 1 1] [0 0 0 0 1 1] = [0 1 1 1 -1 -1] [0 1 1 0 0 0] [0 0 -1 -1 1 1] [0 1 0 1 0 1] [0 -1 -1 -1 1 2] [0 0 1 0 1 0] [0 1 2 1 -1 -2] Coefficient Matrix Inverse Matrix b.) If one of the numbers in your matrix is 7, the inverse for it is -7, because 7 -7 = 0 but this is only if you're adding. If multiplying and one of your numbers in a matrix is 6 the inverse for it would be 1/6 because 6*1/6 = 1 or 6* -1/6 [2 5] [2 1] [4 6] [4 0] + [3 1] = [7 1] To add matrices the dimensions of both matrices must be exactly the same, in this case they are both 2 by 2. First, add the first number in the first row in the first column of the first matrix by the 1st number in the first row in the first column of the 2nd matrix and put the result in the first row in the first column of a new matrix. 2+2 = 4 Add the number in the first row in the 2nd column of the 1st matrix with the number in the first for in the 2nd column of the 2nd matrix. 5+1 = 6 Add the number in the 2nd row in the 1st column of the first matrix with the number in the 2nd row in the 1st column of the 2nd matrix. 4+3 = 7 Add the number in the 2nd row in the 2nd column of the 1st matrix with the number in the 2nd row in the 2nd column of the 2nd matrix. 0+1 = 1 Example 2 - [12 10 6 4] [1 3 5 0] [13 13 11 4] [3 5 0 1] [1 4 2 11] = [4 9 2 12] [0 4 10 4] + [5 7 4 15] [5 11 14 19] |
| c.) When multiplying matrices always multiply row by column. The number o rows in the 1st matrix must be the same as the number of columns in the second matrix. The number in the 1st row of the first matrix is multiplied by the 1st number in the 1st column of the 2nd matrix plus th 2nd number of the 1st row in the 1st matrix times the second number in the 1st column of the second matrix and so on depending on how many rows and columns you have. Example - [2 5] [2 1] = [2*2+5*4 2*1+5*0] = [24 3] [1 3] [4 0] [1*2+3*4 1*1+3*0] [14 1] -The first number in the 1st row of th 1st matrix (2) times the first number in the 1st column in the 2nd matrix (2) plus the second number in the 1st row of the 1st matrix (5) times the second number in the 1st column of the 2nd matrix (4). The answer to that goes into the new matrix you form into the first spot of the first number in the 1st row. (1st row 1st column) -The first number in the 2nd row of the 1st column(1) times the first number in the 1st column of the 2nd matrix(2) plus the second number in the 2nd row of the 1st matrix times the second number in the 1st column of the 2nd matrix. The answer goes into the new matrix in the 1st spot in the 2nd row. ( 2nd row 1st column) -The first number in the 1st row of the 1st matrix(2) times thefirst number in the 2nd column of the second matrix(1) plus the second number in the 1st row of the 1st matrix(5) times the second number in the 2nd column of the 2nd matrix(0) The answer goes in the new matrix in the 1st row 2nd column. -The first number in the 2nd row of the 1st matrix(1) times the first number in the 2nd column of the 2nd matrix(1) plus the second number in the 2nd row of the 1st matrix (3) times the second number in the 2nd column of the 2nd matrix (0) The answer goes in the new matrix in the 2nd row 2nd column Example 2 - [3 2 4] [2] [(3*2+2*1+4*2)] [16] [1 2 3] [1] = [ (1*2+2*1+3*2)] = [10] [2] |
| d.) To solve a system of equations with 3 or more variables using matrices make 3 different matrices. The first is the coefficient, the second is the variable and the third is a constant. 3f+1k+6t=38 f+k+t=13 f-k=0 To put these equations in a matrix form separate all the variables from the coefficients and put them in a matrix. [f ] [k] [t ] Put all the coefficients in a matrix [3 1 6] [1 1 1] Put 1's here because 1*k= k or 1*t = t [1 -1 0] put a negative one here because -1*k= -k and put a 0 here because there is no t Put all the answers in a matrix [38] [13] [ 0 ] So now you have [3 1 6] [f] [38] [1 1 1] [k] = [13] [1 -1 0] [t] [0 ] To get the complete answer get the inverse of the coeficient matrix and multiply it by the constant matrix and you will get an answer like [5] [5] [3] example 2 - 4a-b+c+2d = 44 3a+2b-c-d = 53 a-b+2c+3d = 21 -a+3b+2c+d = 57 [4 -1 1 2] [a] [44] [3 2 -1 -1] [b] [53] [1 -1 2 3] [c] = [21] [-1 3 2 1] [d] [57] Inverse of the coefficient times the constant = |