JEE Main 35 Year PYQs- Matrices and Determinants - Free MCQ Test with solutions

By R₃ → R₃ – (xR₁ + R₂);

= (ax² + 2bx + c)(b² – ac) = (+)(–) = -ve.

For homogeneous system of equations to have non zero solution, Δ = 0

On simplification, 

∴ a,b,c are  in Harmonic Progression.

Also since, B = A^-1 ⇒  AB = I

Let r be the common ratio, then

Given A² - A +I= 0
A^-1A2 - A^-1A + A^-1.I= A^-1.0
(Multiplying A^-1 on both sides)
⇒ A^-1 +A^-1 = 0 or A^-1 = 1-A .

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

Applying, C₁ → C₁ + C₂ + C₃ we get

[As given that a² + b² + c² = –2]
∴ a² +b²+c² + 2 = 0
Applying R₁ → R₁-R₂ , R₂ → R₂-R₃

f (x) = ( x - 1)² Hence degree = 2.

∵ a₁, a₂ ,a₃ , ....... are in G.P..
∴ Using a_n = ar ^{n -1} ,we get the given determinant, as

Operating C₃ - C₂ and C₂-C₁ and using

= 0 (two columns being identical)

A² - B² = (A -B)(A+B)
A² - B² = A² + AB - BA-B² ⇒ AB = BA

Hence, AB = BA only when a = b
∴ There can be infinitely many B's for which AB = BA

Hence, D is divisible by both x and y

From these four relations,
a² + bc = bc + d² ⇒ a² = d²
and b(a + d) = 0 = c( a + d) ⇒ a = – d
We can take a = 1, b = 0, c = 0, d  = –1 as one possible set of values, then 

Clearly A ≠ I and A ≠ – I  and  det A = –1 ∴ Statement 1 is true.
Also if A ≠ I then tr(A) = 0
∴ Statement 2 is false.

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

∵ All entries of square matrix A are integers, therefore all  cofactors should also be integers.
If det A = ± 1 then A^–1 exists. Also all entries of A^–1 are integers.

We know that | adj (adj A) | = |adj A|^2–1 
= | A | 2–1= |A|

∴ Both the statements are true and statement -2 is  a correct explantion for statement-1 .

(Taking transpose of second determinant)

⇒ n  should be an odd integer.

 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.

⇒ Given system, does not have any solution.
⇒ No solution

∴ A' = A, B' = B
Now (A(BA))' = (BA)'A'
=  (A'B')A' = (AB)A = A(BA)
Similarly ((AB)A)' = (AB)A
So, A(BA) and (AB)A are symmetric matrices.
Again (AB)' = B'A' = BA
Now if BA = AB, then AB is symmetric matrix.

      ..(1)

We know,

Now, from equation (1), we have

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

Consider

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