give answer in 2 step all parts with explanation at the end of each step and final answer at the end:b
or dependent. If they are linearly dependent, find scalars 4 |,
3 4 ic not all zero such that a

Question

give answer in 2 step all parts with explanation at the end of each step and final answer at the end:b
or dependent. If they are linearly dependent, find scalars 4 |,
3 4 ic not all zero such that au + bv + cw =0. bl
Ro. E=(2,0,1),v= EE) w= ©0,-2,-1)
20. u=(3.45.V= 2.3.3), w= (2,51)
21 n= (9, 2,5), y= (-2, =3 9, a= (-1, 3,0)
2. u=(1,1,0,v=061 3), w = (0,1,2)
P=1(2.0,3), v= (5.4,-2),w=02.-11
oU u=(1,45,v=*2 5), w= (=3,3,-1)
In Problems 25-28, express the vector tas a linear combina-
tion of the vectors u, V, and Ww.
%.1=(2.-7,9,u=(1,-2.2.v=G0 1, w=(1,-12
b t: (9,—54,44),u = (4,3, a Ni—(1,6, =4),
~ wW=(4,-5,6)
t= 0,0,19),u = (1,4,3), v= (-1, _2.2),w= (44 1)
t=(17.7),u=(2.53),v=* 1,-1), w= (1.15
; ob lems 29-32, show that the given set Vis closed we
i and under multiplication by scalars and is ther for
space of R3,
+4 gheserof all (x,y,z) such that x = 0- "
— setof all (x, y, z) such that x + dud”
ps thesetof all (x, y, z) such that 5x = 2.
he set of ll (x, y, 2) such that 2 = 3% “=
ea. as 0

VIPSolutions · Accepted Answer

Let's solve **Problem 23** step-by-step as requested:

---

### **Problem 23**

Given:
- **u** = (2, , 3)
- **v** = (5, 4, -2)
- **w** = (2, -1, 1)

Are these vectors linearly dependent? If so, find scalars **a**, **b**, **c** (not all zero) such that:

$$ a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = \mathbf{} $$

---

## **Step 1: Set Up the Equation**

Write the linear combination and set equal to zero:
$$
a(2, , 3) + b(5, 4, -2) + c(2, -1, 1) = (, , )
$$

This gives us a system:
$$
\begin{align*}
2a + 5b + 2c &=  \
a + 4b - 1c &=  \
3a - 2b + 1c &= 
\end{align*}
$$

**Explanation:**  
We express the requirement for linear dependence as a homogeneous system of linear equations.

---

## **Step 2: Solve the System**

From the second equation:
$$
4b - c =  \implies c = 4b
$$

Substitute $c = 4b$ into the other equations:

**First equation:**
$$
2a + 5b + 2c =  \implies 2a + 5b + 2(4b) =  \implies 2a + 13b =  \implies a = -\frac{13}{2}b
$$

**Third equation:**
$$
3a - 2b + c = 
$$
Substitute $a = -\frac{13}{2}b$ and $c = 4b$:
$$
3(-\frac{13}{2}b) - 2b + 4b =  \implies -\frac{39}{2}b - 2b + 4b = 
$$
$$
-\frac{39}{2}b + 2b =  \implies -\frac{39}{2}b + \frac{4}{2}b =  \implies -\frac{35}{2}b =  \implies b = 
$$
If $b = $, then $a = $ and $c = $.

**But we are looking for nontrivial solutions! Let's check the calculations:**

#### Let's try using matrices:

$$
\begin{bmatrix}
2 & 5 & 2 \
 & 4 & -1 \
3 & -2 & 1
\end{bmatrix}
\begin{bmatrix}
a \ b \ c
\end{bmatrix}
=
\begin{bmatrix}
 \  \ 
\end{bmatrix}
$$

Row 2: $4b - c =  \to c = 4b$

Plug into Row 1:
$2a + 5b + 2(4b) =  \implies 2a + 13b =  \implies a = -\frac{13}{2}b$

Plug into Row 3:
$3a - 2b + 4b =  \implies 3a + 2b =  \implies a = -\frac{2}{3}b$

Now, equate the two expressions for a:
$$
-\frac{13}{2}b = -\frac{2}{3}b \implies \frac{13}{2}b = \frac{2}{3}b
$$

Multiply both sides by 6:
$$
39b = 4b \implies 39b - 4b =  \implies 35b =  \implies b = 
$$
So, $b =  \implies a = , c = $.

---

## **Final Answer**

### The only solution is the trivial solution: $ a = , b = , c =  $.

**Conclusion:**  
The vectors **u**, **v**, and **w** are **linearly independent**.  
There are **no nontrivial scalars** $ a, b, c $ (not all zero) such that $ a\mathbf{u} + b\mathbf{v} + c\mathbf{w} =  $.

---

**Let me know if you want the next problem solved!**

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers