JEE Advanced Level Test: Matrices & Determinants - JEE MCQ

(1,12),(2,6),(3,4(,(4,3),(6,2),(12,1)
6 different matrices.

x²+x=0
x(x+1)=0
therefore x=0 or -1
3-x+1=5
x=-1.
Hence, x = -1

No, we cannot multiply 3*1 by 3*3, So it doesnt exist.

(A+B)² = (A+B) (A+B)
= A(A+B) +B(A+B)
A² + AB + BA + B²

We know that for a skew-symmetric matrix the sum of diagonal elements is zero.
a[ij] = 0∀i = j
So,tr(A) = 0

Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.

As we know that, A² = I
On multiplying both sides by A^-1
(A²)A^-1 = I A^-1
A = I A-¹
A = A^-1

∣AB∣=∣A∣∣B∣
Also for a square matrix of order 3
∣KA∣=K³∣A∣ because each element of the matrix A is multiplied by k and hence in this case we will have k³ common.
∴∣3AB∣ = (3)³
 ∣A∣∣B∣ = 27(−1)(3)
= −81

If we are given 2 matrices of with order a×b and c×d, in order to be multiplicable, b must be equal to c and the resultant matrix will have order a×d. So using this formula in this question, We have p = p and so the resultant matrix will have m×n order. So, order of matrix AB is m×n.

D = 0
{(-2,1,1) (1,-2,1) ( 1,1, λ) = 0
= (4λ +1 +1) - (-2 - 2 + λ) = 0
4λ + 2 + 4 - λ = 0
3λ + 6 = 0
λ = -2

For a system of linear equations having unique solution, value of the determinant should not be 0
⇒ {(1,1,-1) (1,2,-3) (2,5,-λ)} not equal to o
⇒ [−2λ+15−1(−λ+6)−(5−4)] ≠ 0
⇒ −λ+8 ≠ 0
⇒ λ ≠ 8

 A = {(2,-1) (-7,4)} B = {(4,1) (7,2)}
A’ = {(2,-7) (-1,4)} B’ = {(4,7) (1,2)}
B’A’ = {(1,0) (0,1)}
Therefore, identity matrix.

AB = A and BA = B
B² = B.B
= (BA).B
= B.(AB)
= B.A
= B

As A and B are symmetric matrices, therefore
A’ = A & B’ = B
So we have
(ABA)’ = A’ (AB)’= A’B’A’
= ABA ie symmetric matrix

A → skew symmetric matrix
n → positive integer
An → is symmetric if n is an odd integer and is skew symmetric if n is an even integer.

As we know that, |A| is not equal to zero
|A’| is also not equal to zero
So |A| + |A’| will not be equal to zero. Hence option d is correct.

If A is an n×n square matrix and n is odd, then 
det(−A)=−det(A)
Proof:
det(−A)=(−1)n det(A) = −det(A)

Let ∣A∣ is not equal to 0
∴A ⁻¹ exists.
Given AB=AC
Now, pre multiplying by A⁻¹ on both sides. we get,
A⁻¹(AB) = A⁻¹(AC)
⇒ IB = IC
⇒ B = C
So, A is non singular
 

Matrix multiplication is not commutative in general i.e., AB is not equal to BA.
Matrix multiplication is distributive over matrix addition i.e.,
(i) A(B+C)=AB+AC
(ii) (A+B)C=AB+AC, whenever both sides of equality are defined.
Matrix multiplication is associative i.e., (AB)C=A(BC), whenever both sides are defined.

Let λI = {(a,b) (c,d)}
{(λ,0) (0,λ)} - {(a,b) (c,d)}
= {(λ-a,0) (0,λ-d)} = 0
A =  {(λ-a,b) (c,λ-d)}
= λ² - (a+d)λ + (ad+bc)
= A² - (a+d)A + (ad+bc)
⇒ k = ad - bc

An n×n matrix A is called nilpotent if Ak=O, where O is the n×n zero matrix.
(a) The matrix A is nilpotent if and only if all the eigenvalues of A is zero. 
(b) The matrix A is nilpotent if and only if An=O.
Therefore, option c gives zero matrix.