Solution

Let the basis of 6-dimensional vector space be {e1, e2, e3, e4, e5, e6}. In order for

V1 ∩ V2 to have smallest possible dimension V1 and V2 could be, say, {e1, e2, e3,e4}

and {e3, e4, e5, e6} respectively. The basis of V1 ∩ V2 would then be {e3, e4}. = > Smallest

possible dimension = 2.

Solution

The set A ={(x,y,z): x+y = 0} can be rewritten as

A = {(x,-x,z): x,z ∈ R} This set satisfies the properties of a vector space, as shown in the question.

Similarly, the set

B={(x,y,z) ∣ x−y = 0} also satisfies the properties of a vector space.

However, the set C = {(x,y,z) ∣ x+y = 1} does not contain the origin (0,0,0), which is a requirement for a set to be a vector space. Therefore, C cannot be a vector space.

Similarly, the set

D = {(x,y,z)∣ x − y = 1} does not contain the origin and is also not a vector space.

Solution

Solution

Solution

Consider first 3×3 minors, since maximum possible rank is 3

Since all 3×3 minors are zero therefore rank ≠ 3, now try 2×2 minors.

Solution

Solution

Solution

|A| = -416 = -13 * 32 A-1 does not have integer entries as there are entries in adj(A) which are not divisible by 13.

Solution

Solution

Solution

Solution

Solution

Solution

The augmented matrix for given system is

by Gauss elimination Process

Solution

Solution

Solution

Concepts to use:

1) V1 + V2 is a subset of V.

2) dim(V1 + V2) = dimV1 + dimV2 − dim(V1 ∩ V2).

Analysis:

dim(A) = 2, dim(B) = 1, dim(A ∩ B) = 0

dim(A+B) = 3

all elements of (A + B) are contained in V. (as A + B is the subset of V). Hence, V = A+B

Solution

Solution

Now we can also do column operations, and can stop as soon as we see where the pivots will be.

This is enough for us to see there are three non-pivot rows, so nullity(A) = 3.

Solution

The augmented matrix for given system is Using gauss-elimination method on above matrix we get

Rank ([A|B]) = 3, Rank ([A]) = 3, Since Rank ([A|B]) = Rank ([A]) = number of variables,

The system has a unique solution.

Solution

Solution

Solution

Solution

Solution

Solution

Solution

To find the eigenvalue of A.

Solution

Solution

Solution

Solution

[A]=[L][U]

The [U] matrix is the same as found at the end of the forward elimination of the Naïve Gauss elimination method, that is

To find l₂₁ and l₃₁ find the multiplier that was used to make the a₂₁ and a₃₁ elements zero in the first step of forward elimination of the Naïve Gauss elimination method. It was

l₂₁ = 64/25 = 2.56

l₃₁ = 144/25 = 5.76

To find l₃₂, what multiplier was used to make a₃₂ element zero? It must be noted that a₃₂ element was made zero in the second step of the forward elimination method. The [A] matrix at the beginning of the second step of forward elimination was

So

l₃₂ = -16.8/-4.8 = 3.5

Hence

Confirm [L][U]=[A].

Solution

Solution

Expanding along R1 ,

∆ = (x + 4) [(x – 1) (x – 3) – 0].

Thus, ∆ = 0 implies x = – 4, 1, 3