Test: Linear Algebra - Electrical Engineering (EE) MCQ

All the rows of the given matrix is same. So the matrix has only one independent row.
Rank of the matrix = Number of independent rows of the matrix.
∴ Rank of given matrix = 1

Following are the possibilities for a system of linear equations:
(i) If matrix A and augmented matrix [AB] have same rank, then the system has solution otherwise there is no solution.
(ii) If matrix A and augmented matrix [AB] have same rank which is equal to the number of variables, then the system has unique solution and if B is zero vector then the system have only a trivial solution.
(iii) If matrix A and matrix [AB] have same rank which is less than the number of variables, then the system has infinite solution.
Therefore, option (c) is false because if m - n and B is non-zero vector, then it is not necessary that system has a unique solution, because m is the number of equations (quantity) and not the number of linearly independent equations (quality).

and

Set of equations is

Above set of equations can be written as

Augmented matrix [AB] is given as

Performing gauss-Elimination on the above matrix

The matrix V can be defined as

So above is antisymmetric matrix and the sum of the elements of any antisymmetric matrix is 0.
OR
Alternate Method:

S₁ is true
Consider two singular matrices

Sum of A and B is given as
However (A + B) is a non-singuiar matrix 
So, S₁ is true.
Now, consider two non-singuiar matrices


However (C + D) is a singular matrix. So, S₂ is also true.
Therefore, both S₁ and S₂ are true.

A, B, C, D is n x n matrix.
Given, ABCD = I

The augmented matrix is

Using gauss-efimination on above matrix we get,

Rank [A | B] - 2 (number of non zero rows in [A | B])
Rank [A ] = 2 (number of non zero row s in [A ])
Rank [A | B] = Rank [A] = 2 = number of variables
∴ Unique solution exists.

The augmented matrix for the given system is

Using gauss -elimination method on above matrix we get,


Since Rank ([A | B]) = Rank ([A]) = number of variables, the system has unique solution.

The characteristic equation of this matrix is given by

∴ The eigen values of A are 1 and 6.

Given that Fu= b and Fv= b
If F is non singular, then it has a unique inverse.
Now, u = F^-1 b and v = F^-1 b
Since F^-1 is unique u = v but it is given that u ≠ v .
This is a contradiction. So F must be singular.
This means that
(a) Determinate of F is zero is true. Also
(b) There are infinite number of solution to Fx = b is true since |F| = 0 
(c) There is an X ≠ 0 such that FX = 0 is also true, since X has infinite number of solutions, including the X = 0 solution.
(d) F must have 2 identical rows is false, since a determinant may become zero, even if two identical columns are present. It is not necessary that 2 identical rows must be present for |F| to become zero.

Since P is a square matrix 
Sum (eigen values) = Trace (P)
= a + a + a = 3a 
Product of eigen values

Only choice (a) has sum of eigen values = 3a and product of eigen values = a³ - 2a.

If rank of (5 x 6) matrix is 4, then surely it must have exactly 4 linearly independent rows as will as 4 linearly in dependent columns.

Trace = Sum of principal diagonal element.