Is it possible to solve a matrix by only having one row and one column of data?

Question

Is it possible to solve a matrix by only having one row and one column of data? 

 

For example: 

                                                    Consequences

 

                            Insignificant   Minor  Moderate  Major   Catastrophic

                                     1             2          3            4             5

Likelihood            -----------------------------------------------------------        

Almost Certain  5|                                     15

Likely               4|                                     12

Possible            3|          3             6           9          12           15

Unlikely            2|                                      6

Rare                 1|                                      3

 

Can the rest of this be completed by using an equation(s)?

 

 

Answer 1

Here is another way to see that there is no solution:

 

The determinant of the matrix must not be zero.  Try expanding the determinant by the minors method (Are you familiar with it?) along the top row, and see that the determinant is zero.  Another way to know the determinant is zero is to compare rows.  If one row is a multiple of another, that is sufficient (but not necessary) to make the determinant equal to zero.  Note that rows 1,2,4, and 5 are multiples of each other  By inspection, the determinant is zero and hence you do not have a solution.

Answer 2

well there are technicaly 5 columns, which means there are 5 unknown vars

Lets call the unknowns: V,W,X,Y,Z

 

there are also 5 rows, which means you have five equations.

 

The first problem is there is no column vector containing the 

constant coefficients that each of the 5 equations are equal to.

(Unless we are to assume they are all zero)

Let's call this vector column of constants (k1,k2,k3,k4,k5)

 

THe biggest problem is 4 of the 5 equations are all in terms of x: 

 15x = k1

 12x = k2

  6x  = k4

  3x = k5

 

So x = k1/15 = k2/12 = k4/6 = k5/3

which are contradicting statements.

 

Then we have 3V + 6W + 9X + 12Y + 12Z = k3

 

The system has no solution