Question

Let A be the set of all 3×3 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 A in d for which the system of linear equations A\left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right] has a unique solution, is

A. less than 4
B. at least 4 but less than 7
C. Atleast 7 but less than 10
D. at least 10

Answer 1

In order to solve this problem we need to assume the number of matrices with five 1 and four 0 such that the determinant of any of it is not zero, here we will assume the variables and then we will solve to get the right answer.

Complete step-by-step answer :
We know that there is a matrix of A or order 3x3 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 A in d for which the system of linear equations A\left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right] has a unique solution we need to find.
We know that if the determinant of a matrix is not equal to zero then it is not singular and can be used to solve the system of linear equations which has a unique solution.
The equation we have A\left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right]
The determinant of the matrix present in the system of equations must not be zero to have a solution.

The matrix A have five 1 and four 0 and have the unique solution:
So, we do,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 0&a;&b; \\\ a&0&c; \\\ b&c;&1 \end{array}} \right] here in this matrix either c = 0 or b = 0 this implies $|A| \ne 0$.
So from here we got two matrices.
Now by changing the place of 1 we can have another set as,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 0&a;&b; \\\ a&1&c; \\\ b&c;&0 \end{array}} \right] here in this matrix either c = 0 or a = 0 this implies $|A| \ne 0$.
So from here also we got two matrices.
Again by changing the place of 1 we can have another set as,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&a;&b; \\\ a&0&c; \\\ b&c;&0 \end{array}} \right] here in this matrix either b = 0 or a = 0 this implies $|A| \ne 0$.
We got two matrices from here as well.
Therefore we can see that we got 6 matrices.
So, the correct option is the number of matrices is at least 4 but less than 7 and that is option B.

Note : When you get to solve such problems you need to know that If D is not equal to 0, and if atleast one of Dx, Dy and Dz is not equal to 0, then the system of equations is Consistent and has a Unique solution. If D = 0 and if Dx, Dy and Dz = 0 but if atleast one of the the Constituents of the Coefficient matrix (aij) or at least one of the 2 x 2 minors is not equal to 0, then the system of equations is Consistent and has Infinitely many solutions. If D = 0 and at least one of Dx, Dy and Dz is not zero, then the system of equations is inconsistent (No solution). Knowing this will solve your problem.