Test: Matrices- 1 - JEE MCQ

2^3x3 = 2⁹ = 512.

The number of elements in a 3 X 3 matrix is the product 3 X 3=9.

Each element can either be a 0 or a 1.

Given this, the total possible matrices that can be selected is 2⁹=512

In linear algebra, the identity matrix, or sometimes ambiguously called a unit matrix, of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. It is denoted by I_n, or simply by I if the size is immaterial or can be trivially determined by the context

The product of any matrix with itself can be found only when it is a square matrix.i.e. m = n.

If A and B are square matrices of same order , then , product of the matrices is not commutative.Therefore , the given result is true only when AB = BA.

The given system is:

Therefore, the correct answer is: λ=2.

, therefore matrix A has 4 elements in each row

Let matrix A is of order m x n , and matrix B is of order p x q . then , the product AB is defined only when n = p. that’s why, If A and B are any two matrices, then AB may or may not be defined.

Adj.(KA) = Kⁿ⁻¹ Adj.A , where K is a scalar and A is a n x n matrix.

Here, matrix P is of order 2 × 3 and matrix Q is of order 2 × 2 , then , the product PQ is defined only when : no. of columns in P = no. of rows in Q. And the order of resulting matrix is given by : rows in P x columns in Q.

Lower triangular matrix is given by : 
 ,

here , a_ij = 0
if i is less than j.and a_ij ≠ 0, if i is greater than j.

3^2x2 = 3⁴ = 81

Let B = I + A + A² +…+ A^{k − 1} 
Now multiply both sides by (I − A), we get B(I − A) = (I + A + A² + …+ A^{k − 1})(I − A)
= I − A + A − A² + A² − A³ +… − A^k−1+A^k−1−A^k
= I − A^k = I, since A^k = 0 ⇒ B = (I − A) − 1
Hence (I−A)⁻¹ = I + A + A² +…+ A^k−1 
Thus p = −1

Upper Triangular matrix is given by  :
.
 Here, a_ij=0 , if i is greater than j.and a_ij ≠ 0, if I is less than j.

Let A and B be two matrices such that AB = A and BA = B Now, consider AB = A Take Transpose on both side (AB)^T = A^T 
⇒ A^T = B^T ⋅ A^T ...(1)
Now, BA = B 
Take, Transpose on both side (BA)^T = B^T 
⇒ B^T = A^T⋅B^T…(2)
Now, from equation (1) and (2). we have A^T = (A^T . B^T)A^T 
A^T=A^T(B^TA^T)
= A^T(AB)^T(∵(AB)^T = B^T = B^TA^T)
= A^T ⋅ A^T 
Thus, A^T = (A^T)²

For every square matrix (A + A’) is always symmetric.

The given system of equations does not have solution if : 

- 0 ⇒ 1(-14) - 4(-7) -2(7) = 0

(In general) and in a diagonal matrix non-diagonal elements

For a non trivial solution : 

⇒ bc + ab - 2ac = 0 ⇒  ∴ there, a , b ,c, are in H.P

We have,

and