Questions asked
Eduard Sanchez
Numerade educator
Question 20 (5 points) Find the inverse of A = egin{bmatrix} 9 & -2 \ -10 & 7 end{bmatrix}, if it exists. egin{bmatrix} frac{7}{43} & frac{2}{43} \ frac{10}{43} & frac{9}{43} end{bmatrix} egin{bmatrix} frac{12}{11} & frac{42}{11} \ frac{-1}{11} & frac{5}{11} end{bmatrix} egin{bmatrix} frac{5}{42} & frac{-1}{42} \ frac{11}{42} & frac{12}{42} end{bmatrix} A^{-1} does not exist
Oswaldo Jiménez
Question 19 (5 points) Choose the phrase that best describes the matrix. [ left[egin{array}{ccc:c} 1 & -4 & 3 & 5 \ 1 & 1 & 9 & -8 \ 0 & 0 & 1 & 9 end{array} ight] ] augmented matrix in row-echelon form augmented matrix coefficient matrix none of the above
Danielle Fairburn
Question 18 (5 points) Use an inverse matrix to solve the system of equations, if possible. x+5y-3z=-10 -5x+6y-5z=-21 -x+8y-8z=-25 (1,-1,2) (-6,-5,2) (-6,-2,-2) no solution
Use an inverse matrix to solve the system of equations, if possible. x+5y-3z=-10 -5x+6y-5z=-21 -x+8y-8z=-25 (1,-1,2) (-6,-5,2) (-6,-2,-2) no solution
Nick Johnson
Use an inverse matrix to solve the system of equations, if possible. x + 5y - 3z = -10 -5x + 6y - 5z = -21 -x + 8y - 8z = -25 (1, -1, 2) (-6, -5, 2) (-6, -2, -2) no solution
Question 18 (5 points) Use an inverse matrix to solve the system of equations, if possible. [ egin{array}{l} x+5 y-3 z=-10 \ -5 x+6 y-5 z=-21 \ -x+8 y-8 z=-25 end{array} ] ( (1,-1,2) ) ( (-6,-5,2) ) ( (-6,-2,-2) ) no solution
James Kiss
Melissa Munoz
Question 17 (5 points) Write a matrix equation for the given systems of equations. 2x - 6y - 2z = 1 3y - 2z = -5 2y + 2z = -3 [2 -6 -2] [x] [ 1] [ 3 -2] [y] = [-5] [ 2 +2] [z] [-3] [2 -6 -2] [x] [ 1] [0 3 -2] [y] = [-5] [0 2 2] [z] [-3] [2 -6 -2] [x] [ 1] [3 -2 0] [y] = [-5] [2 2 0] [z] [-3] [2 -6 -2] [ 1] [x] [0 3 -2] [-5] = [y] [0 2 2] [-3] [z]
Gregory Higby
Question 17 (5 points) Write a matrix equation for the given systems of equations. 2x - 6y - 2z = 1 3y - 2z = -5 2y + 2z = -3 [2 -6 -2; 3 -2; 2 + 2] * [x; y; z] = [1; -5; -3] [2 -6 -2; 0 3 -2; 0 2 2] [x; y; z] = [1; -5; -3] [2 -6 -2; 3 -2 0; 2 2 0] * [x; y; z] = [1; -5; -3] [2 -6 -2; 0 3 -2; 0 2 2] [1; -5; -3] = [x; y; z]
Krystal K
Question 16 (5 points) Find the inverse of R = [0 0; 1 3], if it exists. [0 0; 3 1] [1 3; 0 0] [0 0; 1 1/3] R^-1 does not exist.
