Augmented Matrices with Variable Entries

Week 6: section 2.3 + (example videos)

If we do not know the value of the variable entries (ex. k), how do we find a solution?

we determine the nature of the solutions for different values of k

What does it mean to “determine the nature of the solutions for different values of k”?

Find what k would need to be for each three situations (unique, infinite and no solutions

Given the following matrix, what would k need to be for there to be NO solutions?

[ 1 2 | 0 ]
[ 0 k-6 | 1 ]

k must be 6

Given the following matrix, what would k need to be (or not be) for there to be a unique solution?

[ 1 2 | 0 ]
[ 0 k-6 | 1 ]

k must not equal 6

Given the following matrix, what would k need to be (or not be) for there to be NO solutions?

[ 1 0 1 | 1 ]
[ 0 1 1 | 2 ]
[ 0 0 k-2 | k-4]

k must be 2

Given the following matrix, what would k need to be (or not be) for there to be a unique solution?

[ 1 0 2 | 2 ]
[ 0 1 2 | 1 ]
[ 0 0 k²+k | k]

k must not equal -1 or 0

Given the following matrix, what would k need to be (or not be) for there to be infinitely many solutions?

[ 1 0 2 | 2 ]
[ 0 1 2 | 1 ]
[ 0 0 k²+k | k]

k must equal 0

Given the following matrix, what would k need to be (or not be) for there to be

NO solutions?

[ 1 0 2 | 2 ]
[ 0 1 2 | 1 ]
[ 0 0 k²+k | k]

k must equal -1