Let A Show that A is singular Show that Ax 3 1 3 has no so

Let A = Show that A is singular. Show that Ax = [3 1 3] has no solutions. Show that Ax = [3 0 3] has infinitely many solutions.

Solution

a) The matrix is singular if the determinant of the matrix is equal to zero

Det(A) = 2(4-1) + -1(1 + 1) + 1(-1-2)

=> 6 - 3 - 3

=> 0

Since determinant is zero hence the matrix is singular

b) Using the solution, we can make the x1,x2 and x3 as 3X1 vector

2x1 - x2 + x3 = 3

-x1 + 2x2 + x3 = 1

x1 + x2 + 2x3 = 3

Hence the matrix has no solution since there is no value of x1,x2 and x3 satisfy this equation

c)

2x1 - x2 + x3 = 3

-x1 + 2x2 + x3 = 0

x1 + x2 + 2x3 = 3

The equation will have infinie solution x1 = x3 and x2 = x3

Hence there are infinite solutions

2	-1	1
-1	2	1
1	1	2