Matrices

Step 1: Matrix Definitions
We are given three matrices , , and as follows: We need to verify the following statements:

Step 2: Statement (A) - and
First, we verify if , where is the identity matrix. Let's calculate : Thus, , which is true. Now, let's check if . To verify this, we compute : After performing matrix multiplication, we obtain: Thus, is true.
Therefore, statement (A) is true.

Step 3: Statement (B) -
This statement involves the determinant property. For determinants, we know that: However, the property hold. Therefore, the statement: is valid.
Thus, statement (B) is true.

Step 4: Statement (D) - Sum of the diagonal entries of is equal to the sum of diagonal entries of
This statement involves the trace of matrices. The trace of a matrix is invariant under similarity transformations, i.e., . Therefore: Similarly: Since the trace is the sum of the diagonal entries, the statement is true.
Thus, statement (D) is true.

Final Answer:
The correct answers are:
(A) and
(B)
(D) Sum of the diagonal entries of is equal to the sum of diagonal entries of

Given :

det(A)=−1 ……(i)
So, For A⁷ − (β − 1)A⁶ − βA⁵ to be singular
|A⁵| |(A² − (β − 1)A − β| = 0
⇒ |A⁵| |(A + I) (A − βI)| = 0 …..(ii)
∴|A⁵| |A + I| |A − βI| = 0
As we know, |A| ≠ 0
|A+I| or |A−βI| = 0
{|A + I| ≠ 0}
It is Given that , −1=0 (Rejected)

⇒ 2 − 3(1 − β) = 0
⇒
Therefore, 9β = 3.
So, the correct answer is 3.

Given :
M = (a_ij), i, j ∈ {1, 2, 3},
a_ij = 1 if j + 1 is divisible by i, otherwise a_ij = 0

|M| = 1(-1) - 1(-1)
= -1 + 1 = 0
So, M is not invertible



There exists an infinite number of possible column matrices.

x + y + z = 0
⇒ x + z = 0
y = 0
So, this is possible only.


So, the correct options are (B) and (C).
 

• We need matrices M such that R_i = C_j = 0 for i, j = 1, 2, 3.
• All row and column sums must be zero.
• This is only possible if the matrix is balanced with respect to sign and number of 1’s, α’s, and β’s. In the set T, we can use permutations and combinations to evaluate this.
• Every row and column must be such that they are permutations of (1, α, β) that sum to zero.
• Construct each matrix by exhausting permutations which will give us the number 2.

• We need symmetric matrices M such that C_j = 0.
• A symmetric matrix has elements across the main diagonal that are equal, which limits our choice.
• To maintain column zeros with symmetry: diagonal sums should be zero and off-diagonal pairs should sum to zero. There are only limited arrangements possible without additional restrictions, giving us 4 such matrices.

• For skew-symmetric matrices where M[x y z]^T = [a₁₂ 0 -a₂₃], the problem wants the solution of the vector set.
• As it's skew-symmetric, diagonal elements are zeroes, and elements satisfy condition: a_ij = -a_ji.
• Solving the system provides infinite solutions as the rank of this system is less than the dimension, making (R) infinite.

• We are given that R_i = 0; thus, row sums are zero.
• This matrix should have a determinant that keeps rows linearly dependent. Evaluating the determinant gives the value 0 under these conditions.

• (P) → (2)
• (Q) → (4)
• (R) → (3)
• (S) → (5)

Let , where .
We need to find the number of such matrices with .
The determinant of is: .
Since , the possible values of are . Also, can be .
We want , so .

Case 1: . Then .
If , then . If , then .
In this case, , .
and can take any value in , so we have 2 choices for and 2 choices for .
Therefore, we have matrices.

Case 2: .
Then . .
. :
- : .
Then , which is impossible.
- : or .
If , then , so .
(1, 0, 0, 0) If , then , so .
(0, 1, 0, 0) - : .
Then , so . (1, 1, 1, 0), (1, 1, 0, 1)

: Since must be at least , we must have , , or .
Thus, .
- : impossible since . - : .
Then , so . (0, 0, 1, 0), (0, 0, 0, 1) - :
or .
Then , so . (1, 0, 1, 1), (0, 1, 1, 1) - :
.
Then , impossible.

In total, we have the following values for (d, b, e, a):
(1, 0, 0, 0), (0, 1, 0, 0), (1, 1, 1, 0), (1, 1, 0, 1) (0, 0, 1, 0), (0, 0, 0, 1), (1, 0, 1, 1), (0, 1, 1, 1)
So we have 8 cases. If , there are 8 possibilities for .
Hence the number of elements in S is .

Final Answer: The final answer is