1.
Discuss the following matrix A() as a function of . That is, when you plug in a number for
lambda, out comes the rank of the matrix where lambda is replaced with that number:
Solution
Take one 2x2 minor, say
1 1
4 10
This has determinant 6, non-zero
If all 2x2 had determinant 0, than the matrix would had rank 2. But we found one nonzero.
So the rank is bigger than 2; then it is 3 or 4
The rank would be 4 if the determinant were non-zero and conversely, the rank would be 3 if the
determinant were 0.
Let's calculate the determinant of the matrix.
we expand via the first row (see http://en.wikipedia.org/wiki/Laplace_expansion )
det A=3*D 11 +(-1)*D 12 +D 13 +4*(-1)*D 14
D 11 = 4 10 1
7 17 1
2Â Â 4Â Â 3
=4*(51-4)-10*(21-2)+(28-34)=-8
D 12 = 10 1
1Â Â 17 3
2Â Â 4Â Â 3
=(51-12)-10(3-6)+(4-34)
=39+30-30=39
D 13 = 4 1

2.
1 7 3
2Â Â 2 3
=(21-6)-4(3-6)+2-14=15+12-12=15
D 14 = 4 10
1 7Â Â Â 17
2Â Â 2Â Â Â 4
=(28-34)-4(4-34)+10(2-14)=-6
So we have
det A=3*(-8)-39+15-4(-6)=
=-24-39+15+24=-24
The determinant cannot nonzero, so the matrix has rank 4
In other words, no matter what is, the rank is always 4.