Answered

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let a = (a1, a2) and b = (b1, b2) and c = (c1,c2) be three non zero vectors. if a1b2 - a2b1 is not equal to 0. then show three are two scalars, α and ß, such that c = αa + ßb.

Sagot :

LammettHash

Consider the contrapositive of the statement you want to prove.

The contrapositive of the logical statement

p ⇒ q

is

¬q ⇒ ¬p

In this case, the contrapositive claims that

"If there are no scalars α and β such that c = αa + βb, then a₁b₂ - a₂b₁ = 0."

The first equation is captured by a system of linear equations,

[tex]\begin{cases}c_1 = \alpha a_1 + \beta b_1\\ c_2 = \alpha a_2 + \beta b_2\end{cases}[/tex]

or in matrix form,

[tex]\begin{pmatrix}c_1\\c_2\end{pmatrix} = \begin{pmatrix}a_1&b_1\\a_2&b_2\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}[/tex]

If this system has no solution, then the coefficient matrix on the right side must be singular and its determinant would be

[tex]\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} = a_1b_2-a_2b_1 = 0[/tex]

and this is what we wanted to prove. QED

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