Linear Algebra with Applications (3rd Edition)

Quiz 2 Solutions
February 28, 2005
Problem 1 For the first part: the equation of the plane orthogonal to a
given ~v ∈ R
3
is ~x · ~v = 0. Applying this formula to our problem and writing
~x = ( x, y, z ), we get
x + 2 y + 3 z = 0 .
For the second part, we have a formula for the reflection of ~w in a plane P :
ref
P
( ~w ) = ~w - 2 ~w
⊥
where ~w
⊥
is the component of ~w perpendicular to P . But this is just
~w - 2( ~w · ˆ v )ˆ v,
where ˆ v is a unit vector normal to P . Applying this formula with ˆ v =
1
√
14
[123], ~w = [456], we get
ref
P
( ~w ) =
1
7


- 2
- 25
- 48


Problem 2 First we find the elementary row operations that reduce A to
reduced row echelon form (if possible):


1 0 1
2 - 1 0
0 1 1


---------→
- 2( I ) + ( II )
*


1 0 1
0 - 1 - 2
0 1 1


---------→
( II ) + ( III )
*


1 0 1
0 - 1 - 2
0 0 - 1


---------------------→
- 2( III ) + ( II )
*
, ( III ) + ( I )
*


1 0 0
0 - 1 0
0 0 - 1


-----------→
- ( II )
*
, - ( III )
*


1 0 0
0 1 0
0 0 1


1

Since A can be reduced to the identity matrix by elementary row operations,
A is invertible. Applying these same row operations to the 3 × 3 identity
matrix gives us