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The rank of a 3 x 3 matrix C (= AB), found by multiplying a nonzero column matrix A of size 3 x 1 and a nonzero row matrix B of size 1 x 3, is
Characteristic equation is
(1)
In questions 1.1 to 1.7 below, one or more of the alternatives are correct. Write the code letter(s) a, b, c, d corresponding to the correct alternative(s) in the answer book. Marks will be given only if all the correct alternatives have been selected and no incorrect alternative is picked up. 1.1). The eigen vector (s) of the matrix
is (are)
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 ..
Consider the following matrix
If the eigenvalues of A are 4 and 8, then
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
2y3x=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
Consider the following determinant
Which of the following is a factor of Δ?
R2>R2  R1
R3 > R3  R2
you will gt det = (ab)*(ac)*(b+c)
in matrix operations, you cannot multiply rows or columns. That will not yield the same matrix. So abc is not correct
Let A be a matrix such that A^{k} = 0. What is the inverse of I  A?
Given A^{K} = 0
I  A^{k} = I
I  A^{k} = (I  A) (I + A + A^{2} + A^{3} +.........+A^{k1})
I = (I  A) (I + A + A^{2} + A3 +.........+A^{k1})
(I  A)^{1} = (I + A + A^{2} + A^{3} +.........+A^{k1})
Hence (D) is the Answer.
F is an n*n real matrix. b is an n*1 real vector. Suppose there are two n*1 vectors, u and v such that, u ≠ v and Fu = b, Fv = b. Which one of the following statements is false?
(A) : Correct. We are given
Since so we have a nonzero 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.
Perform the following operations on the matrix
i. Add the third row to the second row
ii. Subtract the third column from the first column.
The determinant of the resultant matrix is _____.
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.
In the LU decomposition of the matrix , if the diagonal elements of U are both , then the lower diagonal entry of L is_________________.
Given
We need to find l_{21} and u_{12} l_{21} * 1 + l_{22} * 0 = 4 l_{21} = 4
l_{11} * U_{12} + 0 * 1 = 2 l_{11} = 2 U_{12} = 1 Putting value of l_{21} and u_{12} in (1), we get 4 * 1 + l_{22} * 1 = 9 l_{22} = 5
Let the characteristic equation of matrix M be λ^{2}  λ  1 = 0 . Then
I think we can solve using Cayley Hamilton Theorem
Consider the following statements:
• SI: The sum of two singularn n x n matrices may be nonsingular
• S2: The sum of two n x n nonsingular matrices may be singular
Which one of the following statements is correct?
Yes A is correct option!, Both statements are True
How many 4 x 4 matrices with entries from have odd determinant?
Hint: Use modulo arithmetic.
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(2^41) possibilities 0/1 0/1 0/1 0/1 like totally=161
2nd row must be non zero and not linearly depends on 1st row so> totally (2^42) possibilities
for 3rd row it must be nonzero as well as not linearly depends on first 2 rows(not start with 0) >totally (2^44)
for 4th row >(2^48) :: total possibilities=(2^42^0) * (2^42^1) * (2^42^2) *(2^42^3)=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
Let A be an n × n matrix of the following form.
What is the value of the determinant of A?
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 = 91 =8
put n=2 in C,D.
C = 7
D = 8
so D is answer..
If matrix and X^{2}  X + I = 0 (I is the identity matrix and is the zero matrix), then the inverse of X is
Given, X^{2}  X + I = O
=> X^{2} = X  I
=> X^{1} (X^{2}) = X1(X  I) {Multiplying X^{1} both sides..}
=> X = I  X^{1}
=> X^{1} = I  X
Which gives Option (B)..
The number of different symmetric matrices with each element being either 0 or 1 is: (Note: (2,X) is same as 2^{X})
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.
The rank of the following matrix, where is a real number is
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
Let A and B be real symmetric matrices of size. Then which one of the following is true?
Given A = A' and B = B'
(AB)' = B'A' = BA
Determinant comes out to be 0. So, rank cannot be 3. The minor
Let A, B, C,D be n * n matrices, each with nonzero determinant. If ABCD = I, then B^{1} is
In an M × N matrix all nonzero entries are covered in rows and columns. Then the maximum number of nonzero entries, such that no two are on the same row or column, is
maximum number of nonzero 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 nonzero 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
If M is a square matrix with a zero determinant, which of the following assertion (s) is (are) correct?
S1: Each row of M can be represented as a linear combination of the other rows
S2: Each column of M can be represented as a linear combination of the other columns
S3: MX = 0 has a nontrivial solution
S4: M has an inverse
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 nontrivial solutions also.
So (S1), (S2), and (S3) are true. So option D is correct.
Consider the following system of linear equations
Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of , does this system of equations have infinitely many solutions?
Determinant=0. Therefore apply reduction method on (AB)
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.
Consider the following system of linear equations :
The system of equations has
rank of matrix = rank of augmented matrix = no of unknown = 3 so unique solution..
How many solutions does the following system of linear equations have?
rank = r(A) = r(AB) = 2
rank = total number of variables Hence, unique solution
Consider the following set of equations x + 2y = 54 x+8y = 123x + 6y+3z = 15. This set
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.
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