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Compute [latex]\begin{bmatrix} 1 &3 &0 \\ 2&1 &4 \end{bmatrix} \times \begin{bmatrix} 1 &3 \\ 0&2 \\ -2&-1 \end{bmatrix}[/latex]

1) 2) 3) 4) Solution: (3)... View Article

If A = [latex]\begin{bmatrix} 2 &-1 \\ 1&2 \end{bmatrix}[/latex] then A² + 2A – 3I =

1) 2) 0 3) 4) 5I Solution: (1)... View Article

If A and B are square matrices of equal degree, then which one is correct among the following?

1) A + B = B + A 2) A + B = A – B 3) A – B = B – A 4) AB = BA Solution: (1) A + B = B + A If A and B are square matrices of equal degree,... View Article

If A = [latex]begin{bmatrix} 3 &1 \ -1&2 end{bmatrix}[/latex], then

1) A2 + 7A – 5I = 0 2) A2 – 7A + 5I = 0 3) A2 + 5A – 7I = 0 4) A2 – 5A + 7I = 0 Solution: (4) A2 – 5A + 7I = 0... View Article

Let a, b, c be the real numbers. Then the following system of equations in x, y and z (x² / a²) + (y² / b²) – (z² / c²) = 1, (x² / a²) – (y² / b²) + (z² / c²) = 1,(- x² / a²) + (y² / b²) + (z² / c²) = 1 has

1) no solution 2) unique solution 3) infinitely many solutions 4) finitely many solutions Solution: (4) finitely many solutions... View Article

If AB = A and BA = B, then B² is equal to

1) B 2) A 3) - B 4) B3 Solution: (1) B AB = A and BA = B BA × B = B × B [multiply both sides by B] B (AB) = B2 B (A) = B2 [because BA = B]... View Article

If n ≥ 2 is an integer A = [latex]\begin{bmatrix} cos(2\pi/n) &sin(2\pi/n) &0 \\ -sin(2\pi/n)&cos(2\pi/n) &0 \\ 0&0 &1 \end{bmatrix}[/latex] and I is the identity matrix of order 3. Then

1) An = I and An-1 ≠ 1 2) Am ≠ I for any positive integer m 3) A is not invertible 4) Am = 0 for a positive integer m Solution: (1) An = I... View Article

If A = [latex]\begin{bmatrix} 2-k &2 \\ 1&3-k \end{bmatrix}[/latex] is a singular matrix, then the value of 5k – k² =

1) 0 2) 6 3) – 6 4) –4 5) 4 Solution: (5) 4 A = is singular |A| = 0 (2 - k) (3 - k) - 1 * 2 = 0 6 - 5k + k2 - 2 = 0 k2 - 5k + 4 = 0 5k - k2 = 4... View Article

The symmetric part of the matrix A = [latex]\begin{bmatrix} 1 &2 &4 \\ 6&8 &2 \\ 2&-2 &7 \end{bmatrix}[/latex] =

1) 2) 3) 4) Solution: (4) A = Symmetric part of A = A + A’ Symmetric part of A =... View Article

If A = [latex]\begin{bmatrix} \alpha &2 \\ 2&\alpha \end{bmatrix}[/latex] and |A³| = 125 then the value of ɑ is

1) ± 1 2) ± 2 3) ± 3 4) ± 5 Solution: (3) ± 3 |A| = ɑ2 - 4 |A3| = 125 |A| = 5 ɑ2 - 4 = 5 ɑ2 = 9 ɑ = ± 3 Watch The Below Video for a Detailed... View Article

Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of A consisting of all determinants with value 1. Let C be the subset of A consisting of all determinants with value – 1. Then

1) C is empty 2) B has as many elements as C 3) A = B ⋃ C 4) B has twice as many elements as C Solution: (2) B has as many elements as C A... View Article

State Disaster Response Fund (SDRF)

The State Disaster Response Fund or SDRF is the primary fund available with the state governments of India to respond to disasters in the... View Article

05 May 2021: The Hindu Analysis

UPSC Current Affairs Preparation: The Hindu Analysis Watch expert analysis of ‘The Hindu’ dated 5th May 2021. Important news and views to boost... View Article

In a ∆ABC, if a = 3, b = 4, c = 5, then the distance between its incentre and circumcentre is

1) 1/2 2) √3/2 3) 3/2 4) √5/2 Solution: The distance between incentre and circumcentre = √(R2 - 2rR) R = circumradius r = inradius Given... View Article

In an equilateral triangle of side 2√3 cm, the circumradius is

1) 1 cm 2) √3 cm 3) 2 cm 4) 2√3 cm Solution: Given side = 2√3 Radius of circumcircle of equilateral triangle = side/√3 = 2√3/√3 = 2 Hence... View Article

If in a ∆ABC, r₁ = 2 r₂ = 3 and r₃ = 6, then a equals

1) 4 2) 1 3) 2 4) 3 Solution: Given r1 = 2 r2 = 3 and r3 = 6 We have r1 = ∆/(s - a) 2 = ∆/(s - a) => s = a + ∆/2 …(i) Similarly s = b +... View Article

If a, b and c are the sides of a triangle such that a⁴ + b⁴ + c⁴ = 2c² (a² + b²), then the angles opposite to the side c is

1) 45 or 90 2) 30 or 135 3) 45 or 135 4) 60 or 120 Solution: Given a4 + b4 + c4 = 2c2 (a2 + b2) a4 + b4 + c4 - 2c2a2 - 2c2b2 = 0 Add 2a2b2... View Article

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