Question

Find the dimension and a basis of the solution space
W of the system
x+2y+ 2z-S+ 3t = 0,
x+2y+32+ S+t=0
3x+6y+8Z+S+ 5t=0
​

Answer 1

Answer:

what do u mean

Step-by-step explanation:

Answer 2

Answer:

The required answer is $\left\{v_{1}, v_{2}, v_{3}\right\}$.

Step-by-step explanation:

Augmented matrix: a matrix with the constant terms of the equations put in an additional column and the elements being the coefficients of a series of simultaneous linear equations.

Given: a basis of the solution space W of the system.

To find: Its dimension.

Solution:

The augmented matrix is

$\left[\begin{array}{ccccc}1 & 2 & 2 & -1 & 3 \\1 & 2 & 3 & 1 & 1 \\3 & 6 & 8 & 1 & 5\end{array}\right]$

Streamline the system into an echelon:

$R_{3} \rightarrow R_{3}-3 R_{1} \Longrightarrow\left[\begin{array}{ccccc}1 & 2 & 2 & -1 & 3 \\ 1 & 2 & 3 & 1 & 1 \\ 0 & 0 & 2 & 4 & -4\end{array}\right]$

$R_{2} \rightarrow R_{2}-R_{1} \Longrightarrow\left[\begin{array}{ccccc}1 & 2 & 2 & -1 & 3 \\ 0 & 0 & 1 & 2 & -2 \\ 0 & 0 & 2 & 4 & -4\end{array}\right]$

$R_{3} \rightarrow R_{3}-2 R_{2} \Longrightarrow\left[\begin{array}{ccccc}1 & 2 & 2 & -1 & 3 \\ 0 & 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]$

This is consistent with the system:

$z+2 s-2 t=0, x+2 y+2 z-s+3 t=0$

Two nonzero equations in five unknowns make up the system in echelon form. As a result, the system's $5-2=3$ free variables, $y, s, t$, are available.

Thus,

$dim W=3$

The foundation for $\mathrm{W}$ is as follows:

a, $y=1, s=0, t=0$

$v_{1}=(-2,1,0,0,0)$

$b. y=0, s=1, t=0$

$v_{2}=(5,0,-2,1,0)$

$c. y=0, s=0, t=1$

$v_{3}=(-7,0,2,0,1)$

The set $\left\{v_{1}, v_{2}, v_{3}\right\}$ is a foundation for the problem space $W$.

#SPJ2