Question

Determine whether b can be written as a linear combination of Bold a Subscript Bold 1a1​, Bold a Subscript Bold 2a2​, and Bold a Subscript Bold 3a3. In other​ words, determine whether weights x 1x1​, x 2x2​, and x 3x3 ​exist, such that x 1x1Bold a Subscript Bold 1a1plus+x 2x2Bold a Subscript Bold 2a2plus+x 3x3Bold a Subscript Bold 3a3equals=b. Determine the weights x 1x1​, x 2x2​, and x 3x3 if possible.

Answer 1

Complete  Question

The complete question is shown on the first uploaded image

Answer:

 a

    b can be written as a linear combination of  $a_1 \ and \ a_2$

b

   The values of   $x_1 = 4 \ and \ x_2 = 2$

Step-by-step explanation:

From the question we are told that

      $x_1 a_1 +x_2 a_2 = b$

Where $a_ 1 = (4, 5,-4)$,  $a_2 = (-4 , 3, 3)$ and  $b = (8,26 , -10)$

   So  

      $x_1 ( 4, 5,-4) + x_2 (-4 , 3, 3) = (8,26 , -10)$

     $4x_1, 5x_1,-4x_1 + -4x_2 , 3x_2, 3x_2 = (8,26 , -10)$

=>   $4x_1 -4x_2 =8$

     $x_1 -x_2 =2 ---(1)$

=>   $5x_1 + 3x_2 = 26 --- (2)$

=>   $-4x_1 + 3x_2 = -10 ---(3)$

Now multiplying equation 1 by 3 and adding the product to equation 2

          $.\ \ \ 3x_1 -3x_2 = 6\\+ \ \ 5x_1 + 3x_2 = 26 \\=> \ \ \ 8x_1 = 32$

=>        $x_1 = 4$

 substituting $x_1$ into equation 1

        $4 - x_2 =2$

        $x_2 =2$

Now to test substitute $x_1 \ and \ x_2$ into equation 3

       $-4(4) + 3(2) = -10$

      $-10 = -10$

Since LHS = RHS then there exist values $x_1 = 4 \ and \ x_2 = 2$  such that

       $x_1 a_1 +x_2 a_2 = b$

Hence b can be written as a linear combination of  $a_1 \ and \ a_2$