# Linear Algebra with Applications (3rd Edition)

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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 v, where ˆ v is a unit vector normal to P . Applying this formula with ˆ v = 1 14 , ~w = , 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
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