Let A = [1 2; 8 18], b1 = [-5; -36], b2 = [3; 18], b3 = [1;...

Let A = [1 2; 8 18], b1 = [-5; -36], b2 = [3; 18], b3 = [1; 4], and b4 = [2; 4]. (a) Find A^-1 and use it solve the four equations Ax = b1, Ax = b2, Ax = b3, and Ax = b4. (b) The four equations in part (a) can be solved by the same set of operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix [A b1 b2 b3 b4]. Find A^-1. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice A. A^-1 = [ ; ] (Simplify your answers.) B. The matrix is not invertible.

Transcript

00:01 In this question we are given a matrix a and we are given four vectors and we are asked to solve the system of equations a x equals to b1 a x equals to b2 and so on in two different ways one way is by finding a inverse and another way is by reducing an augmented matrix first let's find the inverse of the matrix of the matrix a inverse equals to 1 over the determinant of a times the matrix obtained from a by interchanging the entries on the main diagonal 18 1 and multiplying the entries on the other diagonal by negative 1 where the determinant of a equals to the product of the entries on the main diagonal which is 1 times 18 minus the product of the entries on the other diagonal which is 2 times 8 this equals to 18 minus 16 equals to 2 therefore a inverse equals to 1 half times 18 negative 2 negative 8 1 or after multiplying each entry by 1 half we are going to get 9 negative 1 in the first row and negative 4 1 in the second row now to find the value of x to find the solutions in each system of equations we need to multiply each vector by a inverse right the first solution is going to be x1 equals to a inverse times b1 equals to a inverse times negative 36 and this equals to 9 times negative 5 minus 1 times negative 36 and the second coordinate is going to be negative 4 times negative 5 plus 1 half times negative 36 and this simplifies to negative 45 plus 36 and 20 minus 18 and equals to negative 9 2 this is a solution to the first system of equations x2 equals to a inverse times b2 and b2 equals to 318 and after multiplication we are going to get 9 times we are first multiplying the first row by the by the vector we are going to get 27 minus 18 and then after multiplying the second row by the vector b2 we are going to get negative 12 plus 1 half times 18 equals to 9 and this simplifies to 9 and negative 3 this is x2 now x3 equals to a inverse times b3 and b3 equals to 1 4 therefore x3 equals to 9 minus 1 times 4 and negative 4 times 1 plus 1 half times 4 which simplifies to 5 and negative 4 plus 2 equals to negative 2.

05:31 This is x3.

05:32 Finally x4 equals to a inverse times b4 and a b4 equals to 2 4 and this equals to 9 times 2 minus 1 times 4 and negative 4 times 2 plus 1 half times 4 and equals to 18 minus 4 is 14 and negative plus 2 is negative 6 this is a solution of the fourth system of equations all right so we found all three solutions x1 is negative 2 negative 9 2 x2 is 9 negative 3 x 3 is 5 negative 2 and x 4 is 14 negative 6 now we can find same solutions by solving by solving by reducing an augmented matrix.

06:55 By reducing an augmented matrix consisting of the matrix a on the left -hand side and the vectors b1, b2, b3 and b4 to the right of the vertical bar.

07:22 1 4 and 2 4.

07:26 Now we need to reduce this matrix to the row reduced echelon form.

07:30 To do that we will multiply the first row by negative 8 and add it to the second row, to get rid of this 8 because 1 times negative 8 plus 8 equals to 0 right what we are going to do now is we will rewrite the first row without change we will rewrite the first row and in the second row we are going to get 0 below 1 next 2 times negative 8 is negative 16 negative 16 plus 18 is 2 negative 5 times negative 8 is 40 40 minus 36 equals to 4 3 times negative 8 is negative 24, negative 24 plus 18 is negative 6.

08:32 1 times negative 8 is negative 8, negative 8 plus 4 is negative 4, and 2 times negative 8 is negative 16, negative 12.

08:49 Next we will divide the second row by 2.

08:55 We will rewrite the first row.

09:00 Or before doing that, what we are going to do is we will subtract the first row, the second row from the first row.

09:09 We will rewrite the second row.

09:17 We will rewrite the second row.

09:20 And in the first row, 1 minus 0 is 1, 2 minus 2 is 0.

09:28 We are going to get 1 0...

Question

Please give Ace some feedback