Consider the matrix A =k −k 3 00 k + 1 1 kk −3 5 k − 50 −3 0 −k.(a) Find det(A).(b) Find all values of k such that A is singular.

(a) To find the determinant of a 4x4 matrix, we can use the method of cofactor expansion. However, in this case, the matrix has a lot of zeros, which simplifies the process. We can expand along the first row:

det(A) = k * det(3x3 submatrix) - (-k) * det(3x3 submatrix) + 3 * det(3x3 submatrix) + 0

The 3x3 submatrix for the first element k is:

[ k+1 1 k -3 5 -5 -3 0 -k ]

The determinant of this 3x3 matrix is (k+1)5(-k) - (k+1)(-5)(-3) - 1*(-3)(-k) + 153 + k(-3)(-5) + k1*3 = -5k^2 - 15k + 15k - 3k + 15 + 15k - 15k = -5k^2 + 15

The 3x3 submatrix for the second element -k is:

[ 0 1 k k 5 -5 0 0 -k ]

The determinant of this 3x3 matrix is 0 - 0 - kk0 + k5k + 0 + 0 = 5k^2

The 3x3 submatrix for the third element 3 is:

[ 0 k+1 k k -3 -5 0 -3 -k ]

The determinant of this 3x3 matrix is 0 - 0 - k*(-3)k + k(-3)*(-5) + 0 + 0 = 3k^2 + 15k

So, det(A) = k*(-5k^2 + 15) - (-k)5k^2 + 3(3k^2 + 15k) = -5k^3 + 15k^2 + 5k^3 + 9k^2 + 45k = 9k^2 + 45k

(b) A matrix is singular if and only if its determinant is zero. So, we set the determinant equal to zero and solve for k:

9k^2 + 45k = 0 k(9k + 45) = 0

So, the values of k that make A singular are k = 0 and k = -5.