PROBLEM 3
(a) Since the matrix is sparse we can easily find the determinant by
expanding by minors. In this case any row or column works equally well.
Using the first row, we get
[ c^2-2 0 0 ] [ 1 c^2-2 0 ]
det(A) = 1 * det [ 1 c^2-2 0 ] - c^4 * det [ 0 1 c^2-2 ]
[ 0 1 1 ] [ 0 0 1 ]
[using the common text abbreivations c^2 = c squared, c^4 = c to the power 4]
and now each 3x3 matrix is triangular so their determinants are just
(c^2-2)^2 and 1, giving finally
det(A) = 1 * (c^2-2)^2 - c^4 * 1 = (c^4 - 4 c^2 + 4) - c^4 = 4 - 4 c^2.
(b) A matrix is invertible if and only if its determinant is not zero.
By (a) we see that det(A)=0 if and only if c^2 = 1, that is,
if and only if c = 1 or c = -1. Therefore A is invertible for all
c other than 1 and -1.
(c) A matrix and its transpose have the same determinant; the determinant
of a product of square matrices is the product of the matrices' determinants.
Hence the product of A with the transpose of A has determinant equal to
the square of det(A). This is zero if and only if det(A) itself is zero,
that is (by the computation in (b)), if and only if c=1 or c=-1.