GATE Computer Science Engineering(CSE) 2027 Test: Linear Algebra- 2 Free

Characteristic equation is

(1)

Eigen values are: 0,0,0
The eigen vector should satisfy the equation: αz = 0

Let the eigen values be a,b
Sum of Eigen Values = Trace(Diagonal Sum)
⇒ a+b = 2+5 = 7
Product of Eigen Values = Det(A)  
⇒ a.b = 6
Solving these we  get 1 and 6.. So, Option(B) is Correct ..

Sum of eighen values is equal to trace(sum of diagonal elements),and product of eighen values is equal to det of matrix 
So 2+y=8+4 y =10
2y-3x=32
Solving this we get x =-4
Option d is ans

5(a) the eigen value for upper triangular/lower triangular/diagonal matrices are the diagonal elements of the matrix

R2->R2 - R1
R3 -> R3 - R2
you will gt det = (a-b)*(a-c)*(b+c)
in matrix operations, you cannot multiply rows or columns. That will not yield the same matrix. So abc is not correct

Given A^K = 0
I  -  A^k = I
I  -  A^k  = (I - A) (I + A + A² + A³  +.........+A^k-1)
I  = (I - A) (I + A + A² + A3  +.........+A^k-1)
(I - A)^-1 =  (I + A + A² + A³  +.........+A^k-1)
Hence (D) is the Answer.

(A) : Correct. We are given


Since  so we have a non-zero solution  to homogeneous equation  Now any vector  is also a solution of and so we have infinitely many solutions of and so determinant of F is zero.
(B) : Correct. Consider a vector


So there are infinitely many vectors of the form   which are solutions to equation Fx = b.
(C) : Correct. In option (a), we proved that vector
(D) : False. This is not necessary.
So option (D) is the answer.

Answer :- 0, because it is easy to see that first column and third column are multiple of each other.
Third column = First column * 15.
So rank is  < 3, so Determinant must be 0.
It stays zero as row & column transformations don't affect determinant.

Given 

We need to find l₂₁ and u₁₂ l₂₁ *  1 + l₂₂ * 0  = 4 l₂₁ = 4

l₁₁ * U₁₂ + 0 * 1 = 2 l₁₁ = 2 U₁₂ = 1 Putting value of l₂₁ and u₁₂ in (1), we get 4 * 1 + l₂₂ * 1 = 9 l₂₂ = 5

I think we can solve using Cayley- Hamilton Theorem

Yes A is correct option!, Both statements are True

whenever 1st row is 0 then its determent is 0 , and similarly if any 2 or more rows are linearly dependent then its |det|=0 In order to find the odd determinant the
1st row must be non zero --> totally(2^4-1) possibilities |0/1  0/1  0/1 0/1| like totally=16-1
2nd row must be non zero and not linearly depends on 1st row so--> totally (2^4-2) possibilities
for 3rd row it must be non-zero as well as not linearly depends on first 2 rows(not start with 0) --->totally (2^4-4)
for 4th row -->(2^4-8) :: total possibilities=(2^4-2^0) * (2^4-2^1) * (2^4-2^2) *(2^4-2^3)=15*14*12*8=20160 possible

Rank of this matrix is 1 as the determint of 2nd order matrix is 0 and 1st order matrix is non zero so rank is 1

Best part of this question  is dont solve by mathematical procedures. Verification is very easy in this question. just put n=1 , u ll get a matrix like [3].. find itdeterminant.. determinant = 3. now check options. 
by putting n=1, i am getting following results..

A. 5
B. 7
C. 3
D. 3

A,B cant be answer..
now check for n=2.
determinantdeterminant = 9-1 =8
put n=2 in C,D.
C = 7
D = 8

so D is answer..

Given, X² - X + I  = O
=>   X² = X - I
=> X^-1 (X²)  = X-1(X - I)    {Multiplying X^-1 both sides..}
=> X = I - X^-1 
 => X^-1 = I - X   

Which gives Option (B)..

In symmetric matrix, So, we have choice only for either the upper triangular elements or the lower triangular elements. Number of such elements will be  Now, each element being either 0 or 1 means, we have 2 choices for each element and thus for  elements we have possibilities. 

we can eliminate all other rows using row 1. in the last only 1 row will be left.

rank = no of non zero rows = 1

Given A = A' and B = B'
(AB)' = B'A' = BA

Determinant comes out to be 0. So, rank cannot be 3. The minor 

maximum number of non-zero entries, such that no two are on the same row or column

Any entry will be a member of some row and some column. So, with  rows we can have maximum  elements such that no row has a repeated element. Same is applicable for b columns also. So, combining both, answer should be .
We can also apply pigeonhole principle here. Let
P = min(a.b) be the number of holes. So, we can place up to
P non-zero entries (pigeons) and as soon as
(P + 1)^th entry comes it must be making two entries in some column or row. 

The multiplication will commute if

Since M has zero determinant, its rank is not full i.e. if M is of size 3*3, then its rank is not 3. So there is a linear combination of rows which evaluates to 0 i.e.

and there is a linear combination of columns which evaluates to 0 i.e.

Now any row Ri can be written as linear combination of other rows as :
 
Similar is the case for columns.
Now MX = 0 always has one solution : X = 0 (which is called trivial solution). Now if |M| = 0, then MX = 0 has non-trivial solutions also.
So (S1), (S2), and (S3) are true. So option D is correct.

Determinant=0. Therefore apply reduction method on (A|B)

 obtain the resultant matrix

or infinitely many solutions, we must have 2+1.5a=0 i.e., a=-4/3 so for only 1 value of a, this system has infinitely many solutions. So option (B) is correct.

rank of matrix = rank of augmented matrix = no of unknown = 3 so unique solution..

rank = r(A) = r(A|B) = 2
rank = total number of variables Hence, unique solution

There are no solutions.
If we multiply 1st equation by 4, we get
4x + 8y = 20
But 2nd equation says
4x + 8y = 12
Clearly, there can not be any pair of (x,y), which satisfies both equations.