a) Suppose that B is an n ¡Á n matrix, and k is a scalar. Give a formula for det (kB) in terms of det B .
b) Show that your formula from (a) is true for all n and for any k.
det (kB) = k^n det B
This is because the determinant is a multilinear function of it's columns (or rows) considered as vectors to the reals. If you multiply one particular colomn of the matrix by k, then the entire determinat get's multiplied by k.
So, if you muliply the whole matrix by k, then that amounts to multiplying each column by k, and thus the determinant picks up a factor k^n.
To show that the formula det(kB) = k^n det(B) is true for all n and any k, we can use mathematical induction.
For n = 1, we have a 1x1 matrix B. The determinant of a 1x1 matrix is just the single entry itself. So, det(kB) = kB and det(B) = B. Therefore, the formula holds for n = 1 since kB = k^1 * B.
Now, let's assume that the formula holds for an n × n matrix, i.e., det(kB) = k^n det(B).
For an (n + 1) × (n + 1) matrix C, we can consider it as a matrix partitioned into an n × n matrix B and an additional column vector b.
C = [ B | b ]
To compute the determinant of C, we can expand it along the last column using cofactor expansion:
det(C) = ∑ (-1)^(i+j) c_ij det(A_ij)
where c_ij is the entry in the i-th row and j-th column of C, det(A_ij) is the determinant of the (n × n) submatrix obtained by removing the i-th row and j-th column from C.
However, all entries in the last column are multiplied by k, including the determinant of the submatrix B. So, we can rewrite the determinant of C as:
det(C) = ∑ (-1)^(i+j) k^n c_ij det(B_ij)
Notice that this is equal to k^n times the determinant of (n + 1) × (n + 1) matrix C' obtained by dividing all entries in the last column of C by k:
C' = [ B | b/k ]
Now, since B has size n × n, we can apply our induction hypothesis to det(C'):
det(C') = k^n det(B)
Substituting back, we have:
det(C) = k^n ∑ (-1)^(i+j) c_ij det(B_ij)
But the sum on the right-hand side is just the determinant of the matrix C without the last column, which is B. So, we can simplify further:
det(C) = k^n det(B)
This completes the proof by induction.
Therefore, we have shown that the formula det(kB) = k^n det(B) is true for all n and any scalar k.