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If A and B are square matrices of equal degree, then which one is correct among the followings?
  • a)
    A + B = B + A
  • b)
    A + B = A – B
  • c)
    A – B = B – A
  • d)
    AB = BA
Correct answer is option 'A'. Can you explain this answer?

Jatin Uppal answered
According to Matrices the formula to calculate is X+Y = Y+X,it means that both the variables should be added to each other and it should be equal to other,Then on the other hand we have to add opposite of variable.

If the system of equations x + ay = 0, az + y = 0 and ax + z = 0 has infinite solutions, then the value of a is
  • a)
    –1
  • b)
    1
  • c)
    0
  • d)
    no real values
Correct answer is option 'A'. Can you explain this answer?

Hridoy Shah answered
We can see that the system of equations has infinite solutions if and only if at least one of the variables x, y, z is equal to 0.

If x = 0, then the first equation becomes 0 = 0, which is always true. The second equation becomes az = 0, which means that either a = 0 or z = 0. The third equation becomes 0 = 0, which is always true. If a = 0, then the second equation becomes 0y = 0, which is always true. If z = 0, then the third equation becomes 0 = 0, which is always true. Therefore, if x = 0, then the system has infinite solutions for any value of a.

If y = 0, then the first equation becomes ax = 0, which means that either a = 0 or x = 0. The second equation becomes 0 = 0, which is always true. The third equation becomes az = 0, which means that either a = 0 or z = 0. If a = 0, then the first equation becomes 0 = 0, which is always true. If x = 0 and z = 0, then the system has infinite solutions for any value of a. If x = 0 and z ≠ 0, or x ≠ 0 and z = 0, then the system has no solutions. Therefore, if y = 0, then the system has infinite solutions if and only if either a = 0 or x = 0 and z = 0.

If z = 0, then the first equation becomes ax = 0, which means that either a = 0 or x = 0. The second equation becomes 0y = 0, which is always true. The third equation becomes ax = 0, which means that either a = 0 or x = 0. If a = 0, then the first and third equations become 0 = 0, which is always true. If x = 0 and y = 0, then the system has infinite solutions for any value of a. If x = 0 and y ≠ 0, or x ≠ 0 and y = 0, then the system has no solutions. Therefore, if z = 0, then the system has infinite solutions if and only if either a = 0 or x = 0 and y = 0.

From the above analysis, we can see that the system has infinite solutions if and only if at least one of the variables x, y, z is equal to 0, and either a = 0 or x = 0 and y = 0 or x = 0 and z = 0. Therefore, the value of a can be any real number except 0.

Let A and B be two symmetric matrices of order 3.
Statement-1: A(BA) and (AB)A are symmetric matrices.
Statement-2: AB is symmetric matrix if matrix multiplication of A with B is commutative.
  • a)
    Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
  • b)
    Statement-1 is true, Statement-2 is false.
  • c)
    Statement-1 is false, Statement-2 is true.
  • d)
    Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Correct answer is option 'A'. Can you explain this answer?

Ishan Chatterjee answered
Explanation:



Statement-1: A(BA) and (AB)A are symmetric matrices.


Let C = AB. Then, (AB)A = CAB and A(BA) = CBA. Now, we have to prove that C^T = C.


Since A and B are symmetric matrices, we have:


A^T = A and B^T = B


Now, we can write:


C^T = (AB)^T = B^T A^T = BA = C


Therefore, C^T = C, which implies that (AB)A and A(BA) are symmetric matrices.



Statement-2: AB is a symmetric matrix if matrix multiplication of A with B is commutative.


If AB is a symmetric matrix then AB = (AB)^T = B^T A^T = BA. Hence, the matrix multiplication of A with B is commutative. However, the converse is not true. That is, just because A and B commute, it does not necessarily imply that AB is symmetric.



Option A: Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.


From the above explanation, we can see that both statements are true. However, Statement-2 does not explain why Statement-1 is true. Therefore, option A is the correct answer.

Let P and Q be 3 x 3 matrices P ≠ Q. If P3 = Q3 and  P2Q = Q2P then determinant of (P2 + Q2) is equal to :
  • a)
    –2
  • b)
    1
  • c)
    0
  • d)
    –1
Correct answer is option 'C'. Can you explain this answer?

Harshitha Saha answered
Given P3 = Q3 ...(1)
and P2Q = Q2P ...(2)
Subtracting (1) and (2), we get
P3 – P2Q = Q3 – Q2P
⇒ P2 (P–Q) + Q2 (P – Q) = 0
⇒ (P2 + Q2) (P–Q) = 0 ⇒ |P2 + Q2| = 0 as P ≠ Q

Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr(A), the sum of diagonal entries of a. Assume that A2 = I.
Statement-1 : If A ≠ I and A ≠ –I, then det(A) = –1
Statement-2 : If A ≠ I and A ≠ –I, then tr (A)  ≠ 0.
  • a)
    Statement -1 is false, Statement-2 is true
  • b)
    Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
  • c)
    Statement -1 is true, Statement-2 is true;  Statement -2 is not a correct explanation for Statement-1
  • d)
    Statement -1 is true, Statement-2 is false
Correct answer is option 'D'. Can you explain this answer?

Anand Kumar answered
Let, A11=a, A12=a, A21=a, A22=-a form the required matrix using these entries and then check the condition given you will see Option D is correct.

The system of equations
α x + y  + z = α – 1
x + α y + z = α – 1
x + y + α z = α – 1
has infinite solutions, if α is
  • a)
    – 2
  • b)
    either – 2 or 1
  • c)
    not – 2
  • d)
    1
Correct answer is option 'A'. Can you explain this answer?

Madhurima Mehta answered
α x + y + z = α -1
x + α y + z = α – 1;
x + y + z α = α – 1
= α(α2 - 1) - 1(α - 1) + 1(1 - α)    
= α (α-1)(α+1) - 1(α-1) - 1(α-1)
For infinite solutions, Δ = 0
⇒ (α - 1)[α2 + α - 1 - 1] =  0
⇒ (α - 1)[α2 + α - 2] = 0  ⇒ α = – 2,1;
But a ≠ 1 . ∴ α = – 2

Let X and Y be two arbitrary, 3 × 3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 × 3, non zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?
  • a)
    Y3Z4 – Z4Y3
  • b)
    X44 + Y44
  • c)
    X4Z3 – Z3X4
  • d)
    X23 + Y23
Correct answer is option 'C,D'. Can you explain this answer?

Devika Choudhury answered
X ' = –X, Y ' = –Y, Z ' = Z
(Y3Z4 – Z4Y3)' = (Z4)'(Y3)' – (Y3)'(Z4)'
= (Z')4(Y')3 – (Y')3(Z')4
= –Z4Y3 + Y3Z4 = Y3Z4 – Z4Y3
∴ Symmetric matrix.
Similarly X44 + Y44 is symmetric matrix and X4Z3 – Z3X4 and X23 + Y23 are skew symmetric matrices.

Let M and N be two 3 × 3 matrices such that MN = NM. Further, if M ≠ N2 and M2 = N4, then
  • a)
    determinant of (M2 + MN2) is 0
  • b)
    there is 3 × 3 non-zero matrix U such that (M2 + MN2)U is the zero matrix
  • c)
    determinant of (M2 + MN2) > 1
  • d)
    for a 3 × 3 matrix U, if (M2 + MN2)U equals the zero matrix then U is the zero matrix
Correct answer is option 'A,B'. Can you explain this answer?

Kajal Dey answered
Given MN = NM, M ≠ N2 and M2 = N4.
Then M2 = N4 ⇒ (M + N2) (M – N 2) = 0
⇒ (i) M + N2 = 0 and M – N2 ≠ 0
(ii) |M + N2| = 0 and |M – N2| = 0
In each case |M + N2| = 0
∴ |M2 + MN2| = |M| |M + N2| = 0
∴ (a) is correct and (c) is not correct.
Also we know if |A| = 0, then there can be many matrices U, such that AU = 0
∴ (M2 + MN2)U = 0 will be true for many values of U.
Hence (b) is correct.
Again if AX = 0 and |A| = 0, then X can be non-zero.
∴ (d) is not correct.

PASSAGE - 2
Let  be the set of all 3× 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.
Q. The number of matrices A in for which the system of linear equations
  • a)
    0
  • b)
    more than 2
  • c)
    2
  • d)
    1
Correct answer is option 'B'. Can you explain this answer?

Aman Chauhan answered
For the given system to be inconsistent |A| = 0. The matrices for which | A | = 0 are



We find for A = (i)
By Cramer’s rule D1 = 0 = D2=D3
∴ infinite many solution
For A  = (ii)
By Cramer ’s rule D1 ≠ 0
⇒ no solution i.e. inconsistent.
Similarly we find the system as inconsistent in cases (iii), (v) and (vi).
Hence for four cases system is inconsistent.

Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx, and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to
  • a)
    2
  • b)
    –1
  • c)
    0
  • d)
    1
Correct answer is option 'D'. Can you explain this answer?

Abhishek Majumdar answered
The given equations are
–x + cy + bz = 0
cx –y + az = 0
bx + ay – z = 0
∵ x, y, z are not all zero
∴ The above system should not have unique (zero) solution
⇒ –1(1– a2) – c(– c – ab) + b(ac + b) = 0
⇒–1 + a2 + b2 + c2 + 2abc = 0
⇒ a2 + b2 + c2 + 2abc = 1

The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is
  • a)
    5
  • b)
    6
  • c)
    at least 7
  • d)
    less than 4
Correct answer is option 'C'. Can you explain this answer?

Siddharth Roy answered
 are 6 non-singular matrices because 6 blanks will be filled by 5 zeros and 1 one. 
Similarly,  are 6 non-singular matrices.
So, required cases are more than 7, non-singular 3 × 3 matrices.

The number of 3 ×  3 matrices A whose entries are either 0 or1 and for which the system  has exactly twodistinct solutions, is
  • a)
    0
  • b)
    29–1
  • c)
    168
  • d)
    2
Correct answer is option 'A'. Can you explain this answer?

Arpita Shah answered
Then the given system is equivalent to
Which represents three distinct planes. But three planes can not intersect at two distinct points, therefore no such system exists.

If A is a 3 × 3 non-singular matrix such that AA' = A'A and B = A–1A', then BB' equals:
  • a)
    B –1
  • b)
    (B -1)'
  • c)
    I +B
  • d)
    I
Correct answer is option 'D'. Can you explain this answer?

Sparsh Majumdar answered
BB' = B(A-1 A')' = B(A ')'(A-1)' = BA (A–1)'
= (A -1 A')(A(A-1)')
= A–1A .A'.(A–1)' {as AA' = A'A}
= I(A–1A)'
= I.I = I2 = I

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