Answer to Question #103242 in Linear Algebra for michael

Question

Let a 4x4 matrix M =
7 5 -1 8 //
7 5 -2 3 //
7 5 5 9 //
7 5 6 10

Find a 3 x 3 matrix A of the form
3 -1 -5
3 * *
3 * *

such that det(M) = 7 det(A)

Answer 1

If we carefuly look through the first matrix we can easily understand that the det(M) = 0. It folows from the properties of the determinant (multiplying the second column by (7/5) and adding to the first column, we get the first column of zeros. It means the determinant is 0).

So that exactly means we need for second matrix determinant = 0 (det(M) = 0 = 7 det(A) $\implies$det(A) = 0) . So there many ways of the solution but it does not say to find all of them. For several of them algorithm is next - from the property that we've already used erlier determinant = 0, when one of the column is combination of the others. So we can put -1 into places (2,2) and (3,2) and we don't care about the others values or we can also put -5 into places (2,3) and (3,3) and as well don't care about the rest.

The answer is $\begin{vmatrix} 3 & -1 & -5\\ 3 & -1 & a1 \\ 3 & -1 & a2 \end{vmatrix}$ or $\begin{vmatrix} 3 & -1 & -5\\ 3 & a1 & -5 \\ 3 & a2 & -5 \end{vmatrix}$ where a1, a2 - parametrs (any number can be used).

As example $\begin{vmatrix} 3 & -1 & -5\\ 3 & -1 & 1 \\ 3 & -1 & 2 \end{vmatrix}$