hey, if we multiply the first row of a matrix and add some constant to it to construct the second row, would that matrix be singular or non-singular? my instructor explained that singular systems have linear dependent rows. what I think is adding a constant to it won’t make it non-linear and hence the rows will be linearly dependent and form a singular system, am I right?
Hello @piyush41y08h
Let’s look at the slide together, what happen if c=d, then under the transformation described by you, we will get a and b that depend on c, d, a muliplication constant, and an addition constant, in this case, would the determinant be zero or not?
Raymond
If you say c=d, then con comparison from the linear function, we’ll also have a=b, and thus the determinant will be zero.
Here’s what I found -
On the left, it’s the case of only scaling the rows by a constant, and on the right, a constant is also added (which is what I wanted to know about).
What I found was that the transformation I described makes the matrix singular iff a=b, or more generally if the input row has all the numbers equal. Am I right?
Yes, @piyush41y08h 
Raymond
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