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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

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