#$&* course mth 164 2pm 11/29 015. Matrix Algebra; Solving the matrix equation A * X = C by finding the Inverse Matrix*********************************************
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Given Solution: `aMultiplying with the identity matrix on the left we have: 1st row * 1st column = [ 1 0 ] [ a c ]` = a 1st row * 2nd column = [ 1 0 ] [ b d ]` = b 2nd row * 1st column = [ 0 1 ] [ a c ]` = c 2nd row * 2nd column = [ 0 1 ] [ b d ]` = d Placing these results in the appropriate positions in the product matrix we get the result indicated in the first equation of Figure 63. Multiplying with the identity matrix on the right we have: 1st row * 1st column = [ a b ] [ 1 0 ]` = a 1st row * 2nd column = [ a b ] [ 0 1 ]` = b 2nd row * 1st column = [ c d ] [ 1 0 ]` = c 2nd row * 2nd column = [ c d ] [ 0 1 ]` = d. Placing these results in the appropriate positions in the product matrix we get the result indicated in the second equation of Figure 63. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. The inverse of a 2 x 2 matrix A is the matrix denoted by A^-1 which when multiplied by A gives the identity I = [ [1 0], [0 1] ]. It is tempting to conjecture that the inverse of the matrix A = [ [ a b ] , [ c d ] ] is the matrix [ [ a^-1 b^-1 ] , [ c^-1 d^-1 ] ] = [ [ 1/a 1/b ] , [ 1/c 1/d ] ]. Show that this matrix is in fact not generally equal to the inverse of A = [ [ a b ] , [ c d ] ]. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t know how to show that. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf [ [ 1/a 1/b ] , [ 1/c 1/d ] ] is the inverse matrix then [ [ 1/a 1/b ] , [ 1/c 1/d ] ] * [ [ a b ] , [ c d ] ] = [ [ 1 0 ], [0 1] ]. However, for example, the first row of the first matrix and the first column of the second matrix is [ 1/a 1/b] [ a c ]` = 1 + c/b. Unless c = 0, which cannot be assumed, this is not the first-row first-column element 1 of the identity matrix and this alone shows that the product matrix cannot be the identity [ [ 1 0 ], [0 1] ]. Figure 71 shows that the result of this multiplication is [[ 1 + c/b, d/b + b/a], [c/d + a/c, 1 + b/c] ], which except in certain special cases is not equal to the identity matrix. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t understand how [1/a 1/b] [a c] = 1 + c/b. ------------------------------------------------ Self-critique Rating:2
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Given Solution: `aThe first row is [ 2 3 | 1 0 ]; doubling the first row we get [ 4 6 | 2 0 ], which we add to the second row [ -4 1 | 0 1 ] to get [ 0 7 | 2 1 ]. Inserting this into the original matrix in place of the second row we get the matrix [ [ 2 3 | 1 0 ] , [ 0 7 | 2 1 ] ]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. In the preceding problem you should have obtained the matrix [ [ 2 3 | 1 0 ] , [ 0 7 | 2 1 ] ]. Proceed by first multiplying the second row by 1/7 to get the second row into the desired form. Then do whatever is necessary to get the first row into the desired form. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After we multiply the 2nd row by 1/7 we get [0 1| .2857 .1428] which gives us a new matrix [2 3| 1 0],[0 1|.2857 .1429] Then multiply the 2nd row by 3 and subtract it from the 1st row to get [2 0|.1429 -.4286] [2 0|.1429 -.4286],[0 1|.2857 .1429] as the new matrix so we then multiply the 1st row by .5 and get [1 0|.0714 -.2143] so the inverse matrix is [.0714 -.2143],[.2857 .1429] It is so much easier to do this with a calculator will I to work this out by hand on a test. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aMultiplying the second row by 1/7 we get the matrix [ [ 2 3 | 1 0 ] , [ 0 1 | 2/7 1/7 ] ]. We obtain the desired 0 in the second position of the first row by adding -3 * second row to the first row and replacing the first we get [ [ 2 0 | 1/7 -3/7 ] , [ 0 1 | 2/7 1/7 ] ]. We now want the 1 in the first position of the second row. 1/2 * first row accomplishes this and gives us the matrix [ [ 1 0 | 1/14 -3/14 ] , [ 0 1 | 2/7 1/7 ] ]. This matrix has the desired form, with the identity matrix formed by the first two columns. The last two columns therefore constitute the inverse of the original matrix. These steps are illustrated in Figure 92. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. The matrix [ [ 1 0 | 1/14 -3/14 ] , [ 0 1 | 2/7 1/7 ] ] obtained by reducing the matrix [ [ 2 3 | 1 0 ] , [ -4 1 | 0 1 ] ] consists of the identity matrix augmented by the matrix [ [ 1/14 -3/14 ], [ 2/7 1/7] ]. This matrix is the inverse A^-1 = [ [ 1/14 -3/14 ], [ 2/7 1/7] ]. Show that this matrix is indeed the inverse by showing that A * A^-1 = I, the identity matrix, and also that A^-1 * A = I. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [2 3]*[.0714 -.2143] = [1 0] [-4 1] [.2857 .1429] [0 1] confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFigure 96 shows both products, each obtained by the usual method of matrix multiplication (each row of the first matrix multiplied by each column of the second with the results positioned as usual in the product matrix). We see that both product give us the identity matrix. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. Write the system 2 x + 3 y = -6, -4 x + y = 8 as a matrix equation A * X = C. Give the matrices A, X and C. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):[2 3]*[x]=[-6] [-4 1] [y] [8] ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. The inverse of the matrix A = [ [ 2 3 ], [ -4 1 ] ] is the matrix A^-1 = [ [ -1/14 -3/14 ], [ 2/7 1/7] ], as found in the preceding series of questions. If we multiply both sides of the matrix equation A * X = C by this inverse we get A^-1 * A * X = A^-1 * C; since A^-1 * A = I we thus have X = A^-1 * C. Calculate A^-1 * C. Then using this result write out the matrices in the equation X = A^-1 * C. What does this tell you about the values x and y that satisfy the given system of equations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [.0714 -.2143]*[-6] =[-2.1428] [.2857 .1429] [8] =[-.571] So x = -2.1428 and y = -.571 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFigure 98 shows the original matrix equation translated into the symbolic algebraic form A X = C. This equation is then multiplied on sides A^-1, resulting in the equation A^-1 * A * X = A^-1 * C. Since A^-1 * A = I, the identity matrix, the left-hand side becomes I * X; and since I is the identity matrix we know that I * X = X so we have X = A^-1 * C. Translating this back into matrix form we have [ 1/14, - 3/14], [ 2/7, 1/7] ] * [ -6, 8 ], with the result as shown in the figure. Our solution is [ -15/7, -4/7 ]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. Apply the techniques of this assignment to find the solution of the system 3 x - 2 y = 10, -5 x + 3 y = -13. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Just for my own sake I solved this equation by substitution just to get an answer shouldn’t the answer be the same as the answer using the techniques of the current problem. Because when I solved using sub. I got 3(-4) - 2(-11) = 10 and -5(-4) +3(-11) = -13 which makes x = -4 and y = -11.But the given solution says x = 2.95 and y = .58 If you plug these into the equations 3(2.95) - 2(.58) = 7.69 and -5(2.95) + 3(.58) = 16.49. Am I missing something here or is this answer wrong. confidence rating #$&*: ? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe system can be expressed as the product [ [ 3 -2 ], [ -5 3 ] ] * [ x y ]` = [ 10 -13 ]` which is of form A X = C with A = [ [ 3 -2 ], [ -5 3 ] ] and C = [ 10 -13 ]`. The solution of the equation A X = C will be X = A^-1 * C, easily obtained if we can find A^-1. The inverse of A is obtained by reducing the augmented matrix [ A | I ] = [ [ 3 -2 | 1 0 ], [ -5 3 | 1 0 ] ] to the form [ I | * ] = [[ 1 0 | * * ], [ 0 1 | * * ] ], with the * matrix being A^-1. Figure 113 illustrates the reduction of the original matrix to the desired form, with the result that the inverse is A^-1 = [ [3/19, -2/19], [5/19, 3/19] ]. We now calculate our solution X = A^-1 * C: X = [ [3/19, -2/19], [5/19, 3/19] ] * [ 10 -13 ]` = [56/19, 11/19]` = [x, y']`. Thus x = 56/19 and y = 11/19, or in approximation x = 2.95, y = .58. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!