Certainly! Let's go through each part of this question step-by-step.
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### Part 1: Finding the Equation of the Plane
We need to find the equation of the plane passing through points [tex]\( P(0,0,1) \)[/tex], [tex]\( Q(2,0,0) \)[/tex], and [tex]\( R(0,3,0) \)[/tex].
1. Find two vectors in the plane:
- Vector [tex]\( \mathbf{PQ} \)[/tex] going from [tex]\( P \)[/tex] to [tex]\( Q \)[/tex]:
[tex]\[
\mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (2,0,0) - (0,0,1) = (2,0,-1)
\][/tex]
- Vector [tex]\( \mathbf{PR} \)[/tex] going from [tex]\( P \)[/tex] to [tex]\( R \)[/tex]:
[tex]\[
\mathbf{PR} = \mathbf{R} - \mathbf{P} = (0,3,0) - (0,0,1) = (0,3,-1)
\][/tex]
2. Find the normal vector to the plane by taking the cross product of [tex]\( \mathbf{PQ} \)[/tex] and [tex]\( \mathbf{PR} \)[/tex]:
[tex]\[
\mathbf{n} = \mathbf{PQ} \times \mathbf{PR} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 0 & -1 \\
0 & 3 & -1 \\
\end{vmatrix} = \mathbf{i} (0 \cdot (-1) - 3 \cdot (-1)) - \mathbf{j} (2 \cdot (-1) - 0 \cdot (-1)) + \mathbf{k} (2 \cdot 3 - 0 \cdot 0)
\][/tex]
Simplifying this, we get:
[tex]\[
\mathbf{n} = (0 + 3)\mathbf{i} + (2)\mathbf{j} + (6)\mathbf{k} = 3\mathbf{i} + 2\mathbf{j} + 6\mathbf{k} = (3,2,6)
\][/tex]
3. Form the equation of the plane:
The equation of a plane is given by:
[tex]\[
a x + b y + c z = d
\][/tex]
where [tex]\( (a, b, c) \)[/tex] is the normal vector [tex]\( \mathbf{n} \)[/tex]. Using the normal vector [tex]\( (3,2,6) \)[/tex]:
[tex]\[
3x + 2y + 6z = d
\][/tex]
4. Find the constant [tex]\(d\)[/tex] by substituting one of the points on the plane, say [tex]\( P(0,0,1) \)[/tex], into the plane equation:
[tex]\[
3(0) + 2(0) + 6(1) = d \implies 6 = d
\][/tex]
Thus, the equation of the plane is:
[tex]\[
3x + 2y + 6z = 6
\][/tex]
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### Part 2: Proof by Mathematical Induction
We need to prove that if [tex]\( B \)[/tex] is a square matrix such that [tex]\( B(B-1)=0 \)[/tex], then [tex]\( (I+B)^n-I=(-1+2^n) B \)[/tex], where [tex]\( I \)[/tex] is the identity matrix of the same size as [tex]\( B \)[/tex].
#### Base Case (n = 1):
For [tex]\( n=1 \)[/tex]:
[tex]\[
(I+B)^1 - I = I + B - I = B \\
(-1 + 2^1)B = B \\
\)
This satisfies the base case.
#### Inductive Step:
Assume that the statement holds for \( n = k \). That is:
\[
(I+B)^k - I = (-1 + 2^k) B
\][/tex]
We need to prove that the statement holds for [tex]\( n = k+1 \)[/tex]. Consider [tex]\( (I+B)^{k+1} \)[/tex]:
[tex]\[
(I+B)^{k+1} = (I+B)^k (I+B)
\][/tex]
By the inductive hypothesis:
[tex]\[
(I+B)^k = I + (-1+2^k) B \rightarrow \text{Let this be } X
\][/tex]
Then:
[tex]\[
(I + B)^{k+1} = (I + (-1 + 2^k) B) (I + B) = I + B + (-1 + 2^k) B + (-1 + 2^k) B^2
\][/tex]
Using the property [tex]\( B(B-1)=0 \)[/tex]:
[tex]\[
B^2 = B
\][/tex]
So:
[tex]\[
(I + B)^{k+1} - I = B + (-1 + 2^k) B + (-1 + 2^k) B = B + (-1 + 2^k) B + (-1 + 2^k) B
\][/tex]
Simplify and combine similar terms:
[tex]\[
B + (-1 + 2^k) B + (-1 + 2^k) B = B + (-1 + 2^k + 2^k) B = B + (2 \times 2^k - 1) B = B + (2^{k+1} - 1) B = (-1 + 2^{k+1}) B
\][/tex]
Thus, the inductive step is proven and hence, the statement is true for all [tex]\( n \geq 1 \)[/tex]:
[tex]\[
(I+B)^n - I = (-1 + 2^n) B
\][/tex]
This completes the proof using mathematical induction.
