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Mid2Sol_1

Mid2Sol_1 - ⎠ ⎛ ⎝ 2 1 ⎞ ⎠ ⎛ ⎝ − 2 − 1 ⎞...

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Math 33A, Midterm 2 solutions 1. We have to solve the linear system of 1 equation in 3 variables. Its augmented coefficient matrix is ( 2 11 | 0 ) , and after dividing by 2 we get its RREF ( 1 1 2 1 2 | 0 ) . Alternatively, we can say that we are ±nding the kernel of the matrix ( 2 11 ) . The solution is x 1 x 2 x 3 = s 1 2 1 0 + t 1 2 0 1 where s, t are arbitrary parameters, i.e. the subspace (the plane) can be written as span { 1 2 1 0 , 1 2 0 1 } . The above two vectors are linearly independent (which is always the case whenever we compute the kernel from the RREF), and thus 1 2 1 0 , 1 2 0 1 is a basis for the given subspace. 2. Denote the matrices by A and B respectively. image( A ) = span { 1 1 1
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Unformatted text preview: ⎠ , ⎛ ⎝ 2 1 ⎞ ⎠ , ⎛ ⎝ − 2 − 1 ⎞ ⎠ } = span { ⎛ ⎝ 1 1 1 ⎞ ⎠ , ⎛ ⎝ 2 1 ⎞ ⎠ } , because ⎛ ⎝ − 2 − 1 ⎞ ⎠ = − 2 ⎛ ⎝ 1 1 1 ⎞ ⎠ + ⎛ ⎝ 2 1 ⎞ ⎠ . Also image( B ) = span { ⎛ ⎝ 3 1 2 ⎞ ⎠ , ⎛ ⎝ 1 − 1 ⎞ ⎠ , ⎛ ⎝ − 4 k ⎞ ⎠ } = span { ⎛ ⎝ 1 1 1 ⎞ ⎠ , ⎛ ⎝ 2 1 ⎞ ⎠ , ⎛ ⎝ − 4 k ⎞ ⎠ } . In the last equality we have used that ⎛ ⎝ 3 1 2 ⎞ ⎠ = ⎛ ⎝ 1 1 1 ⎞ ⎠ + ⎛ ⎝ 2 1 ⎞ ⎠ , ⎛ ⎝ 1 − 1 ⎞ ⎠ = − ⎛ ⎝ 1 1 1 ⎞ ⎠ + ⎛ ⎝ 2 1 ⎞ ⎠ 1...
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