How do I make a column vector to add to my original matrix?

Im trying to solve this: You want to add 4 to each element in the first row of A, subtract 1 from each element in the second row of A, and keep the third row as-is. Create a column vector that you can add to A to perform this task. Call your column vector B.

I have the matrix: A = [1 3 5; -10 -8 -6; (sin(pi/2)) 5^3 (exp(-2))]

I then did:

A = [1 3 5; -10 -8 -6; (sin(pi/2)) 5^3 (exp(-2))]

A = 3×3

1.0000 3.0000 5.0000 -10.0000 -8.0000 -6.0000 1.0000 125.0000 0.1353

B_one = A(1,:) + 4

B_one = 1×3

5 7 9

B_two = A(2,:) - 1

B_two = 1×3

-11 -9 -7

B_three = [(sin(pi/2)) 5^3 (exp(-2))]

B_three = 1×3

1.0000 125.0000 0.1353

B = [B_one B_two B_three]'

B = 9×1

5.0000 7.0000 9.0000 -11.0000 -9.0000 -7.0000 1.0000 125.0000 0.1353

I am trying to make the column vector but my column vector is never the right size and I keep getting the error: Variable B must be of size [3 1]. It is currently of size [9 1]. Check where the variable is assigned a value.

Answer 1

[Torsten]

What about

B = [4;-1;0]

?

Answer 2

[Voss]

A = [1 3 5; -10 -8 -6; (sin(pi/2)) 5^3 (exp(-2))];
B_one = A(1,:) + 4;
B_two = A(2,:) - 1;
B_three = [(sin(pi/2)) 5^3 (exp(-2))];

In constructing B from B_one, B_two, B_three, use vertical concatenation by separating the rows with semicolons, and avoid transposing the result:

B = [B_one; B_two; B_three]

B = 3×3

5.0000 7.0000 9.0000 -11.0000 -9.0000 -7.0000 1.0000 125.0000 0.1353

You can construct a column vector the same way, using vertical concatenation, which can then be added to A. Example:

A = magic(3)

A = 3×3

8 1 6 3 5 7 4 9 2

to_add = [2; 5; -9] % this column vector will be used to add 2 to the first row of A, 5 to the 2nd, -9 to the 3rd

to_add = 3×1

2 5 -9

B = A + to_add

B = 3×3

10 3 8 8 10 12 -5 0 -7