Tiêu đề Matrices Varsity Daily MCQ (Set-B)-With Solve.pdf ✅

1 Daily-01 [B (Solve Sheet)] wm‡jevm : g ̈vwUa· I wbY©vqK c~Y©gvb: 30 †b‡MwUf gvK©: 0.25 mgq: 20 wgwbU 1. wb‡Pi †KvbwU mgNvwZ g ̈vwUa·? [Which of the following is a cognate matrix?]     2 2 1 – 1     – 2 2 1 – 1     2 – 2 1 – 1     2 – 2 1 1 DËi:     2 – 2 1 – 1 e ̈vL ̈v: A 2 =     2 – 2 1 – 1     2 – 2 1 – 1 =     4 – 2 – 4 + 2 2 – 1 – 2 + 1 =     2 – 2 1 – 1 = A mgNvwZ g ̈vwUa· nIqvi kZ©, A 2 = A  A GKwU mgNvwZ g ̈vwUa·| 2. P = [bij]4 2 I Q = [bij]2  3 n‡j, PQ g ̈vwUa·wUi AvKvi- [P = [bij]4  2 and Q = [bij]2  3 then, The size of the PQ matrix is-] 4  2 4  3 2  3 3  3 DËi: 4  3 e ̈vL ̈v: P Gi gvÎv 4  2 Q Gi gvÎv 2  3  PQ Gi gvÎv = 4  3 3.       3 y – 3 5 6 7 4 x – 1 wbY©vq‡Ki (3, 2) Zg Abyivwki gvb 2 n‡j, x I y Gi m¤úK© †KvbwU? [       3 y – 3 5 6 7 4 x – 1 (3,2)th value of minor is 2, What is its relationship?] 4x – 3y = 1 3x – 4y = 2 4x + 3y = 2 3x – 4y = 1 DËi: 3x – 4y = 2 e ̈vL ̈v: (3, 2) Zg Abyivwk,     3 y 4 x = 2  3x – 4y = 2 4.    K + 1  2 12 – 8 g ̈vwUa·wU e ̈wZμgx n‡j, K Gi gvb- [    K + 1  2 12 – 8 If the matrix is exceptional, then the values of K-] 4 – 8 5 – 4 DËi: – 4 e ̈vL ̈v:     K + 1 2 12 – 8 = – 8K – 8 – 24 = 0  – 8K = 32  K = – 4 5. wb‡Pi †KvbwU Ea©ŸwÎfzRvKvi g ̈vwUa·? [Which of the following is an upper triangular matrix?]       1 0 0 3 2 0 5 4 0       1 3 6 0 6 7 0 0 1       1 5 10 0 2 8 0 6 4       1 0 0 4 4 6 2 3 5 DËi:       1 0 0 3 2 0 5 4 0 e ̈vL ̈v: †h eM© g ̈vwUa‡·i cÖavb K‡Y©i wb‡Pi fzw3 k~b ̈, Zv‡K Ea©ŸwÎfzRvKvi g ̈vwUa· e‡j| msÁvbyhvqx mwVK DËi | 6.     x – y 1 1 x + y =     8 7 1 2 n‡j, (x, y) = ? [     x – y 1 1 x + y =     8 7 1 2 then, (x, y) = ?] (2, 6) (5, – 3) (4, 2) (1, – 3) DËi: (5, – 3) e ̈vL ̈v: g ̈vwUa‡·i mgZv Abymv‡i, x – y = 8 x + y = 2  2x = 10  x = 5  Avevi, x – y = 8  y = – 3  (x, y) = (5, – 3) 7. A =     0 – 1 1 0 , X =     – y x nq ,Z‡e A 2X n‡e - [A =     0 – 1 1 0 , X =     – y x then, A2X = ?]     – x – y     x – y     – y – x     y – x DËi:     y – x e ̈vL ̈v: A 2 =     0 – 1 1 0     0 – 1 1 0 =     – 1 0 0 –1  A 2X =     – 1 0 0 –1     – y x =     y – x