Question 1006746
{{{B =(matrix (3,3,1,2,3,2,k-3,3,3,4,k-4))}}}


I'll do the first part, part (a), to get you started


I'm using the method shown below


<img src = "https://upload.wikimedia.org/math/5/7/6/576a5e4bd5c1b73733ada8b15a204c6c.png">
(Image Source: <a href = "https://en.wikipedia.org/wiki/Determinant">Wikipedia</a>)


Using that method, we can say


{{{det(B) = 1*abs(matrix (2,2,k-3,3,4,k-4))+(-2)*abs(matrix (2,2,2,3,3,k-4))+3*abs(matrix (2,2,2,k-3,3,4))}}}


Let's compute the sub-determinants (the determinants of the 2x2 submatrices)


{{{abs(matrix (2,2,k-3,3,4,k-4)) = (k-3)(k-4)-4*3}}}


{{{abs(matrix (2,2,k-3,3,4,k-4)) = k^2-7k+12-12}}}


{{{abs(matrix (2,2,k-3,3,4,k-4)) = k^2-7k}}}
----------------------------------------------------
{{{abs(matrix (2,2,2,3,3,k-4))=2*(k-4)-3*3}}}


{{{abs(matrix (2,2,2,3,3,k-4))=2k-8-9}}}


{{{abs(matrix (2,2,2,3,3,k-4))=2k-17}}}
----------------------------------------------------
{{{abs(matrix (2,2,2,k-3,3,4))=2*4-3*(k-3)}}}


{{{abs(matrix (2,2,2,k-3,3,4))=8-3k+9}}}


{{{abs(matrix (2,2,2,k-3,3,4))=-3k+17}}}
----------------------------------------------------


Now let's return to the determinant of the 3x3 matrix.


{{{det(B) = 1*abs(matrix (2,2,k-3,3,4,k-4))+(-2)*abs(matrix (2,2,2,3,3,k-4))+3*abs(matrix (2,2,2,k-3,3,4))}}}


{{{det(B) = 1*(k^2-7k)+(-2)*(2k-17)+3*(-3k+17)}}} Make the proper substitutions (see the sub-determinants above)


{{{det(B) = k^2-7k-4k+34-9k+51}}} Distribute


{{{det(B) = k^2-20k+85}}} Combine like terms. This is the final answer for part (a)