含虚数的行最简形应用：求使y属于v、w张成平面的h值

Alright, let's break this down to find the value of ( h ) that puts ( \mathbf{y} ) in the plane spanned by ( \mathbf{v} ) and ( \mathbf{w} ). Here's the step-by-step solution:

Key Condition #

For ( \mathbf{y} ) to lie in ( \text{span}{\mathbf{v}, \mathbf{w}} ), it must be a linear combination of ( \mathbf{v} ) and ( \mathbf{w} ). That means there exist scalars ( a ) and ( b ) (complex, since our vectors have imaginary components) such that:
[
a\mathbf{v} + b\mathbf{w} = \mathbf{y}
]

Expand to a System of Equations #

Substitute the given vectors ( \mathbf{v} = \begin{bmatrix} 1 \ 0 \ -2 \end{bmatrix} ), ( \mathbf{w} = \begin{bmatrix} -3 \ i \ 8 \end{bmatrix} ), and ( \mathbf{y} = \begin{bmatrix} h \ -5i \ -3 \end{bmatrix} ) into the equation above. This gives us three equations:

• First component: ( a - 3b = h )
• Second component: ( 0 \cdot a + i \cdot b = -5i )
• Third component: ( -2a + 8b = -3 )

Solve for Scalars ( a ) and ( b ) #

1. Start with the second equation—it's the simplest:
   [
   i b = -5i
   ]
   Divide both sides by ( i ) (since ( i \neq 0 )) to get ( b = -5 ).

2. Plug ( b = -5 ) into the third equation to solve for ( a ):
   [
   -2a + 8(-5) = -3 \
   -2a - 40 = -3 \
   -2a = 37 \
   a = -\frac{37}{2}
   ]

Calculate ( h ) #

Now substitute ( a = -\frac{37}{2} ) and ( b = -5 ) into the first equation:
[
h = -\frac{37}{2} - 3(-5) = -\frac{37}{2} + 15 = -\frac{37}{2} + \frac{30}{2} = -\frac{7}{2}
]

Verify with Augmented Matrix (Optional) #

If you prefer using the augmented matrix approach you mentioned, set up the matrix for the system:
[
\left[\begin{array}{cc|c}
1 & -3 & h \
0 & i & -5i \
-2 & 8 & -3
\end{array}\right]
]
Perform row operations (add 2×Row 1 to Row 3):
[
\left[\begin{array}{cc|c}
1 & -3 & h \
0 & i & -5i \
0 & 2 & 2h - 3
\end{array}\right]
]
From Row 2, we still get ( b = -5 ). Substitute into Row 3:
[
2(-5) = 2h - 3 \
-10 = 2h - 3 \
2h = -7 \
h = -\frac{7}{2}
]

Same result—perfect, that confirms our answer.

内容的提问来源于stack exchange，提问作者sk58