Solve a matrix in R (inverse of a covariance table)

Question

I'm new to R and portfolio optimization. I try to produce an inverse matrix of a co-variance table. I have 10 stocks with a rate of returns over 5 days. (I reduced the actual size for the example) Here is the Return Table:

             ID1    ID2     ID3     ID4     ID5     ID6      ID7      ID8   ID9      ID10
3/29/2017   0.0654 -0.0027  0.0025  0.0168  0.4676  0.8515  -0.1152 -0.0298 0       -0.3897
3/30/2017  -0.0326  0      -0.0076  0      -0.5556  0.4907   0.0142 -0.0097 0        0.1392
3/31/2017   0      -0.0054  0.0152 -0.0335  0.2336  0.5856  -0.0577  0     -0.0054  -0.219
4/3/2017    0.0643  0.0027  0.0152  0       0.6429  1.0852   0.0289  0.0287 0        0.267
4/4/2017    0.0318  0       0.005   0.0169 -0.1906 -0.0313  -0.145  -0.0096 0.0027  -1.1001

I then apply the co-variance function:

CovTable <- cov(Returns)

Which produce this table:

     ID1        ID2         ID3         ID4         ID5         ID6         ID7         ID8         ID9         ID10
ID1 0.001791    0.000034    0.000192    0.000408    0.017139    0.007902    -0.000916   0.000108    0.000039    -0.003565
ID2 0.000034    0.000009    -0.000004   0.000034    -0.000036   0.000172    0.000101    0.000034    0.000007    0.000387
ID3 0.000192    -0.000004   0.000092    -0.000096   0.003535    0.001392    0.000027    0.000128    -0.000013   0.000273
ID4 0.000408    0.000034    -0.000096   0.000423    -0.000804   -0.001490   -0.000610   -0.000166   0.000057    -0.004438
ID5 0.017139    -0.000036   0.003535    -0.000804   0.240015    0.152353    0.000960    0.003544    -0.000363   0.056594
ID6 0.007902    0.000172    0.001392    -0.001490   0.152353    0.177324    0.018715    0.003369    -0.000409   0.177301
ID7 -0.000916   0.000101    0.000027    -0.000610   0.000960    0.018715    0.005890    0.001099    -0.000057   0.038787
ID8 0.000108    0.000034    0.000128    -0.000166   0.003544    0.003369    0.001099    0.000454    -0.000009   0.005793
ID9 0.000039    0.000007    -0.000013   0.000057    -0.000363   -0.000409   -0.000057   -0.000009   0.000009    -0.000623
ID10 -0.003565  0.000387    0.000273    -0.004438   0.056594    0.177301    0.038787    0.005793    -0.000623   0.290340

Now if i try to have the reverse with

minverse <- solve(CovTable)

I get this error:

Error in solve.default(covTable) : system is computationally singular: reciprocal condition number = 1.67973e-21

Now, if I use only 5 stocks or less, or any number of stocks that is smaller or equal to the number of days I have in the Return Table, that works perfectly.

So my question is how can I have an inverse matrix of a co-variance table with more stocks than the number of days observed in the return table?

Here is the full code that generates the error:

library(RODBC)
ch <- odbcDriverConnect('driver={SQL Server};server=MyServername\\SQLEXPRESS;database=MyDatabase;trusted_connection=true')
Returns <- sqlFetch(ch, "Returns") # Import Returns Table from SQL to R
Returns <-cast(Returns, Date ~ Id) # Organize Returns Table in Pivot Table
rownames(Returns) <- Returns[,1] # Use the Date Column as Row Headers
Returns <- Returns[,-1] # Remove the Date Column 
CovTable <- cov(Returns) # Create Covariance Matrix
minverse <- solve(CovTable) # Solve CovTable