# Dividing the values of a vector

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Anna Mogilevskaja 2019 年 12 月 3 日
Answered: Star Strider 2019 年 12 月 3 日
Hello,
I have the following question:
this is my code:
A=[1 1.1; 1.1 1]
l= [sym(2)/sym(3); sym(1)/sym(3)]
w=1
B=[1.09 1.44; 1.44 0.99]
r=linspace(0,1,100)
for k = 1:numel(r)
p(:,k)=(inv(B-(1+r(k))*A))*l
end
The p vector consists out of p1 and p2. After the loop is finished I would like to calculate p1/p2 (also in a loop as I would like to plot p1/p2 against r). Unfortunately, I could not find any references to this case.
Does anyone have a solution for dividing the values of a vector.
Greetings,
Anna

#### 2 件のコメント

Stephane 2019 年 12 月 3 日
Maybe try p(1,:) ./ p(2,:), and then plot(r, p(1,:) ./ p(2,:))
Anna Mogilevskaja 2019 年 12 月 3 日
I habe tried this but it did not work out yet:
A=[1 1.1; 1.1 1]
l= [sym(2)/sym(3); sym(1)/sym(3)]
w=1
B=[1.09 1.44; 1.44 0.99]
r=linspace(0,1,100)
for k = 1:numel(r)
p(:,k)=(inv(B-(1+r(k))*A))*l
end
for k = 1:numel(r)
q(:,k)=(p(1,:)(k)./ p(2,:)(k)
end
figure
plot(r,q)
grid
xlabel('r')
ylabel('q(r)')

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### 件の回答 (2)

Shota Kato 2019 年 12 月 3 日
q can be calculated as follows
q = p(1,:) ./ p(2,:)
or
q(:,k) = p(1,k) / p(2,k)
Then, you can plot q against r.

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Star Strider 2019 年 12 月 3 日
Try this:
A=[1 1.1; 1.1 1];
l = [2; 1]/3; % Create As Column Vector
w=1;
B=[1.09 1.44; 1.44 0.99];
r=linspace(0,1,100);
for k = 1:numel(r)
p(:,k)=(B-(1+r(k))*A)\l; % Replace ‘inv’ Call With ‘\’
end
p_div = p(1,:)./p(2,:);
figure
plot(r, p_div)
grid
xlabel('$r$', 'Interpreter','latex')
ylabel('$\frac{p_1}{p_2}$', 'Interpreter','latex')

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