Answer to Question #240677 in Linear Algebra for bilal

Question

write vector v=(4,9,19) as a linear combination of vectors u1=(1,-2,3),u2=(3,-7,10),u3=(2,1,9)

Answer 1

Given that:

$\vec v=4\hat i +9\hat j+19\hat k, \vec u_1=\hat i-2\hat j+3\hat k,\\ \vec u_2=3\hat i-7\hat j+10\hat k,\vec u_3=2\hat i+\hat j+9\hat k\\$

Let $\vec v=x(\vec u_1)+y(\vec u_2)+z(\vec u_3)\\$

$\Rightarrow 4\hat i +9\hat j+19\hat k=x(\hat i-2\hat j+3\hat k)+y(3\hat i-7\hat j+10\hat k)+z(2\hat i+\hat j+9\hat k)\\ \Rightarrow4\hat i +9\hat j+19\hat k=(x+3y+2z)\hat i+(-2x-7y+z)\hat j+(3x+10y+9z)\hat k\\$

Comparing both sides, we get:

$x+3y+2z=4\ldots(1)\\ -2x-7y+z=9\ldots(2)\\ 3x+10y+9z=19\ldots(3)\\$

Now, solving these $3$ equations:

$D=\left|\begin{array}{ccc} 1 & 3 & 2 \\ -2 & -7 & 1 \\ 3 & 10 & 9 \end{array}\right|=-8\\ D_{x}=\left|\begin{array}{ccc} 4 & 3 & 2 \\ 9 & -7 & 1 \\ 19 & 10 & 9 \end{array}\right|=-32\\ D_{y}=\left|\begin{array}{ccc} 1 & 4 & 2 \\ -2 & 9 & 1 \\ 3 & 19 & 9 \end{array}\right|=16\\ D_{z}=\left|\begin{array}{ccc} 1 & 3 & 4 \\ -2 & -7 & 9 \\ 3 & 10 & 19 \end{array}\right|=-24\\$

Using Cramer's rule, we get:

$\begin{gathered} x=\frac{D_{x}}{D}, y=\frac{D_{y}}{D}, z=\frac{D_{z}}{D} \\ x=\frac{D_{x}}{D}=\frac{-32}{-8}=4 \\ y=\frac{D_{y}}{D}=\frac{16}{-8}=-2 \\ z=\frac{D_{z}}{D}=\frac{-24}{-8}=3 \end{gathered}$

$\therefore \vec v=4\vec u_1-2\vec u_2+3\vec u_3$