Let B 1 2 3 1 B is obtained when one applies the two elemen
Solution
16. Let A =
a
b
c
d
After the elementary row operation r2(3) , A changes to
a
b
3c
3d
Also, after the elementary row operation r21(-1) ( interpreted as -1times 2nd row added to the 1st row) , the matrix A changes further to B=
a-3c
b-3d
3c
3d
Hence a-3c = -1, b-3d = 2, 3c = 3 and 3d = 1. Then a = 3c-1= 3-1= 2, b = 3d+2 = 1+2 = 3, c = 3c/3 = 3/3 = 1 and d = 3d/3 = 1/3.
Thus A =
2
3
1
1/3
Option A is the correct answer.
17. The augmented matrix of the given linear system is A =
1
1
1
2
2
1
-1
4
3
2
0
a
We will reduce A to its RREF as under:
Add -2 times the 1st row to the 2nd row
Add -3 times the 1st row to the 3rd row
Multiply the 2nd row by -1
Add-1 time the 2nd row to the 1st row
Add 1 times the 2nd row to the 3rd row
Then the RREF of A is
1
0
-2
2
0
1
3
0
0
0
0
a-6
Apparently, the given linear system will be consistent only if a = 6. Then it will have infinite solutions. The solutions will be x = 2+2z, y = -3z , z = z so that (x,y,z)T = (2+2z, -3z,z)T = (2,0,0)T +z(2,-3,1)T where z is an arbitrary real number. The 3rd option is the correct answer.
| a | b |
| c | d |
![Let B = [-1 2 3 1] B is obtained when one applies the two elementary row operations r^(3)_2 and then r^(-1)_21 successively to A. Find A A. [2 3 1 1/3] B. [0 7 Let B = [-1 2 3 1] B is obtained when one applies the two elementary row operations r^(3)_2 and then r^(-1)_21 successively to A. Find A A. [2 3 1 1/3] B. [0 7](/WebImages/36/let-b-1-2-3-1-b-is-obtained-when-one-applies-the-two-elemen-1109232-1761587735-0.webp)
![Let B = [-1 2 3 1] B is obtained when one applies the two elementary row operations r^(3)_2 and then r^(-1)_21 successively to A. Find A A. [2 3 1 1/3] B. [0 7 Let B = [-1 2 3 1] B is obtained when one applies the two elementary row operations r^(3)_2 and then r^(-1)_21 successively to A. Find A A. [2 3 1 1/3] B. [0 7](/WebImages/36/let-b-1-2-3-1-b-is-obtained-when-one-applies-the-two-elemen-1109232-1761587735-1.webp)