Answer on Question #78477 – Math – Linear Algebra
Question
Give examples, with justification, of the following:
two non-singular 2 × 2 2 \times 2 2 × 2 matrices C C C and D D D , with ∣ C ∣ = 2 ∣ D ∣ |C| = \sqrt{2} |D| ∣ C ∣ = 2 ∣ D ∣
Solution
Since we have to satisfy only ∣ C ∣ = 2 ∣ D ∣ |C| = \sqrt{2} |D| ∣ C ∣ = 2 ∣ D ∣ , we can choose matrix D D D arbitrary. Let it be equal to unity matrix E 2 E_2 E 2 :
D = E 2 = ( 1 0 0 1 ) D = E_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} D = E 2 = ( 1 0 0 1 )
Then
∣ D ∣ = ∣ 1 0 0 1 ∣ = 1 |D| = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1 ∣ D ∣ = ∣ ∣ 1 0 0 1 ∣ ∣ = 1
To satisfy the given condition we can create matrix C C C from matrix D D D by changing upper left element to 2 \sqrt{2} 2 :
C = ( 2 0 0 1 ) C = \begin{pmatrix} \sqrt{2} & 0 \\ 0 & 1 \end{pmatrix} C = ( 2 0 0 1 )
Indeed,
∣ C ∣ = ∣ 2 0 0 1 ∣ = 2 ⋅ 1 = 2 = ∣ D ∣ |C| = \begin{vmatrix} \sqrt{2} & 0 \\ 0 & 1 \end{vmatrix} = \sqrt{2} \cdot 1 = \sqrt{2} = |D| ∣ C ∣ = ∣ ∣ 2 0 0 1 ∣ ∣ = 2 ⋅ 1 = 2 = ∣ D ∣
Answer: C = ( 2 0 0 1 ) C = \begin{pmatrix} \sqrt{2} & 0 \\ 0 & 1 \end{pmatrix} C = ( 2 0 0 1 ) , D = ( 1 0 0 1 ) D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} D = ( 1 0 0 1 ) .
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