System of linear equations ( matrices)
I'm doing this question:
Determine the values of the real constants a and b for which there are infinitely many solutions to the simultaneous equations
2x+3y+z=6
-x+y+2z=7
ax+y+4z=b
I know that the determinant of the corresponding matrix should be set to zero as that would be the case for zero or infinite solutions. However, I don't quite understand why this works. If the determinant is zero, that should mean that the inverse doesn't exist, so there should be no solutions? Why does a singular matrix imply zero or infinite solutions then?
Determine the values of the real constants a and b for which there are infinitely many solutions to the simultaneous equations
2x+3y+z=6
-x+y+2z=7
ax+y+4z=b
I know that the determinant of the corresponding matrix should be set to zero as that would be the case for zero or infinite solutions. However, I don't quite understand why this works. If the determinant is zero, that should mean that the inverse doesn't exist, so there should be no solutions? Why does a singular matrix imply zero or infinite solutions then?
So each equation represents a plane. Let's split thing up into the varies possibilities:
(a) the three planes intersect at a single point and hence the solution is unique;
(b) the three planes intersect in a line, and there are an infinite number of solutions;
(c) the three planes are identical, and again there are an infinite number of solutions;
(d) the three planes are identical but distinct, and there are zero solutions;
(e) any two planes are parallel but distinct, and there are zero solutions;
(f) or one plane is parallel to the line of intersection of the other two, and there are zero solutions.
If you were to write your system of equations in matrix form, the rows of the matrix represent vectors which are normal to the planes. In cases (b), (c), (d), (e) and (f) there is a fourth plane that can be drawn perpendicular to your three planes. So the normals to your three planes are coplanar (i.e. they all can be considered as vectors lying in the same plane). This mean that the cross product of the first two normals dotted with the third normal (=determinant) will be zero.
On the other hand, if there is a unique solution, the planes meet at a point and there is no fourth plane which is perpendicular to your three planes and so their normals are not coplanar and the determinant doesn't vanish.
(a) the three planes intersect at a single point and hence the solution is unique;
(b) the three planes intersect in a line, and there are an infinite number of solutions;
(c) the three planes are identical, and again there are an infinite number of solutions;
(d) the three planes are identical but distinct, and there are zero solutions;
(e) any two planes are parallel but distinct, and there are zero solutions;
(f) or one plane is parallel to the line of intersection of the other two, and there are zero solutions.
If you were to write your system of equations in matrix form, the rows of the matrix represent vectors which are normal to the planes. In cases (b), (c), (d), (e) and (f) there is a fourth plane that can be drawn perpendicular to your three planes. So the normals to your three planes are coplanar (i.e. they all can be considered as vectors lying in the same plane). This mean that the cross product of the first two normals dotted with the third normal (=determinant) will be zero.
On the other hand, if there is a unique solution, the planes meet at a point and there is no fourth plane which is perpendicular to your three planes and so their normals are not coplanar and the determinant doesn't vanish.
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