Number of matrices having unique solution

[utkarshakash]

Homework Statement

Let A be the set of all 3x3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. The number of matrices B in A for which the system of linear equations B \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} 1 \\ 0 \\0 \end{array} \right] has a unique solution is

The Attempt at a Solution

I started by finding total number of matrices in A which comes out to be 12. Now, instead of finding matrices having unique solution I tried finding out inconsistent solutions. Now, the problem arises that how do I find these matrices. Is it by writing out all possible 12 matrices and then checking manually or applying some clever methods (which I unfortunately don't know)?

 

[pasmith]

Homework Statement

Let A be the set of all 3x3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. The number of matrices B in A for which the system of linear equations B \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} 1 \\ 0 \\0 \end{array} \right] has a unique solution is

This equation has a unique solution if and only if \det B \neq 0.

 

[AlephZero]

Find out some facts about determinants. For example, what happens to the determinant if you swap two rows or two columns of the matrix? What happens if two rows or columns of the matrix are the same?

Using results like the above, you won't need to check all 12 matrices separately.

 

[utkarshakash]

This equation has a unique solution if and only if \det B \neq 0.

Why is it so?

 

[haruspex]

Why is it so?

Determinant is zero iff matrix is singular iff rows are linearly dependent.
Consider the matrix equation AX=Y. Linear dependence means there is a nontrivial linear combination of the rows, a r₁+b r₂ + c r₃ = 0. If it is also the case that a y₁ + a y₂ + a y₃ = 0 then the equation represented by one row of AX=Y can be deduced from the other two, so there are not enough distinct equations for a unique solution. If a y₁ + a y₂ + a y₃ ≠ 0 then an equation can be deduced from two of the rows which contradicts the third, and there are no solutions at all.

But you don't really need to worry about determinants for this problem. The point to note is that the existence of a unique solution does not depend on the vector on the right hand side of the equation. If you think about how matrix multiplication works, you can see that shuffling the columns of A is equivalent to shuffling the rows of X, and shuffling the rows of A is equivalent to shuffling the rows of Y. This allows you to reduce the set of matrices to a much smaller canonical set.