I am struggling with letter b and keep getting this as an error. Variable b must be of size [1 1]. It is currently of size [3 4]. Check where the variable is assigned a value.

Consider the array (row vector), matrix, and column vector below.

A = [11.3 25.4 -58.6 66.2 82.8 37.9 41.5 3.6 60.7 14.2]

B = [1.256 3.135 5.489 7.236; 8.457 2.236 -4.456 6.128; 0.236 7.458 8.569 9.745]

C = [2.23; 3.58; -5.69; 7.47; 8.23; 5.46; 0.19; -9.73; 1.28; 4.46]

a) Sort A in ascending order and then calculate log base 10 of each element.

b) Calculate log base 10 of each element in B and then find the maximum value for the entire matrix.

c) Round all values down and then calculate log base 10 of each element in C.

here is my code:

A = [11.3 25.4 -58.6 66.2 82.8 37.9 41.5 3.6 60.7 14.2]

B = [1.256 3.135 5.489 7.236; 8.457 2.236 -4.456 6.128; 0.236 7.458 8.569 9.745]

C = [2.23; 3.58; -5.69; 7.47; 8.23; 5.46; 1.19; -9.73; 1.28; 4.46]

% a) Sort A in ascending order and then calculate log base 10 of each element.

A = sort(A)

a = log10(A)

% b) Calculate log base 10 of each element in B and then find the maximum value.

b = log10(B)

B = max(B,[],'all')

% c) Round all values down and then calculate log base 10 of each element in C.

C = floor(C)

c = log10(C)

Answer 1

[Jasvin]

Hi Monique,

You just have to reassign the output to b instead of B as you are doing in this case. 

Also, here’s another way you can accomplish the same thing in one line,  

b = max(max(log10(B)));

[Monique]

thank you!

Answer 2

[Dyuman Joshi]

Variable b has been used to store the log10 values of B matrix and it has not been updated afterwards, You have used B instead b

%In this line of code
B = max(B,[],'all')

There are some negative elements in B (and A and C as well) and by definition the input to log should be a positive value. I would suggest you to clarify from your instructor what is to be done, but the function log() accepts negative values as well

B =  [1.256 3.135 5.489 7.236; 8.457 2.236 -4.456 6.128; 0.236 7.458 8.569 9.745];
logB = log10(B);
b = max(logB,[],'all')

b = 0.6489 + 1.3644i

[Monique]

thank you!

Answer 3

[Sulaymon Eshkabilov]

One typo B instead of b in computing max of log10(B):

A =  [11.3 25.4 -58.6 66.2 82.8 37.9 41.5 3.6 60.7 14.2];
B =  [1.256 3.135 5.489 7.236; 8.457 2.236 -4.456 6.128; 0.236 7.458 8.569 9.745];
C =  [2.23; 3.58; -5.69; 7.47; 8.23; 5.46; 0.19; -9.73; 1.28; 4.46];
% a) Sort A in ascending order and then calculate log base 10 of each element. 
A = sort(A)

A = 1×10

-58.6000 3.6000 11.3000 14.2000 25.4000 37.9000 41.5000 60.7000 66.2000 82.8000

a = log10(A)

a = 
   1.7679 + 1.3644i   0.5563 + 0.0000i   1.0531 + 0.0000i   1.1523 + 0.0000i   1.4048 + 0.0000i   1.5786 + 0.0000i   1.6180 + 0.0000i   1.7832 + 0.0000i   1.8209 + 0.0000i   1.9180 + 0.0000i

% b) Calculate log base 10 of each element in B and then find the maximum value.
b = log10(B)

b = 
   0.0990 + 0.0000i   0.4962 + 0.0000i   0.7395 + 0.0000i   0.8595 + 0.0000i
   0.9272 + 0.0000i   0.3495 + 0.0000i   0.6489 + 1.3644i   0.7873 + 0.0000i
  -0.6271 + 0.0000i   0.8726 + 0.0000i   0.9329 + 0.0000i   0.9888 + 0.0000i

B = max(b,[],'all')

B = 0.6489 + 1.3644i

% c) Round all values down and then calculate log base 10 of each element in C.
C = floor(C)

C = 10×1

2 3 -6 7 8 5 0 -10 1 4

c = log10(C)

c = 
   0.3010 + 0.0000i
   0.4771 + 0.0000i
   0.7782 + 1.3644i
   0.8451 + 0.0000i
   0.9031 + 0.0000i
   0.6990 + 0.0000i
     -Inf + 0.0000i
   1.0000 + 1.3644i
   0.0000 + 0.0000i
   0.6021 + 0.0000i

[Monique]

thank you!

Answer 4

[Sarthak]

According to the question (b) part, you need to find log10 of all the elements in the matrix. Since the size of B is [3 4], when you apply log10(B) and save it in variable b, the size of b will eventually be [3 4] only. The maximum element out of b will have a size of [1 1] or we can say it will be a scalar.