Question

A basis for C is [1] [3] [1 B= 21- 11 [-1] 2] [1] Compute PB compe-() -9 PB -9

Answer 1

Let us consider a relation x*
121
+y*
ܚ .
+z*$\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ = $\begin{bmatrix} -4\\ -9\\ -3 \end{bmatrix}$ , where x,y,z are real number.

Then we have,

x+3y+z = -4

2x-y+z = -9

-x+2y+z = -3

Here, the augmented matrix is : B =

「1 3 1 -4 | 2 - 11-9 |-1 21 -3

Now we apply elementary row operations on B.

Step I : R₂-2R₁=R₂, R₃+R₁=R₃

Then B becomes =

| |0 lo 3 -7 5 1 -1 2 -4 -1 -7

Step II : (-1/7)R₂=R₂

Then B becomes =

|1 31 -4 |0 117 17 |0 5 2 -7

Step III : R₁-3R₂=R₁, R₃-5R₂=R₃

The B becomes =

|1 0 47 -31/7 |0 117 17 |0 09/7 -547

Step IV : (7/9)R₃=R₃

Then B becomes =

|1 0 47 -31/7 |0 117 17 |0 0 1 -6

Step V : R₁-(4/7)R₃=R₁, R₂-(1/7)R₃=R₂

Then B becomes =

「100 -1] |0 10 1 |0 01-6

This gives, x = -1

  y = 1

z = -6

Therefore, $\rho_{B}\left (\begin{bmatrix} -4\\ -9\\ -3 \end{bmatrix} \right )=\begin{bmatrix} -1\\ 1\\ -6 \end{bmatrix}$ .