Ex 3.1, 6 - Class 12

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Ex 3.1 , 6 Find the value of x, y, and z from the following equation: (i) [■8(4&3@x&5)] = [■8(y&z@1&5)] [■8(4&3@x&5)] = [■8(y&z@1&5)] Since matrices are equal. Their corresponding elements are equal Hence, 4 = y 3 = z x = 1 Hence x = 1, y = 4 & z = 3 Ex 3.1 , 6 Find the value of x, y, and z from the following equation: (ii) [■8(x+y&2@5+z&xy)] = [■8(6&2@5&8)] Since matrices are equal. Their corresponding elements are equal x + y = 6 5 + z =5 xy = 8 Solving these equations From (2) 5 + z = 5 z = 5 – 5 z = 0 From (1) x + y = 6 x = 6 – y From (3) xy = 8 Putting x = 6 – y (6 – y) y = 8 6y – y2 = 8 y2 – 6y = - 8 y2 – 6y + 8 = 0 y2 – 4y – 2y + 8 = 0 y2 (y – 4) –2(y – 4) = 0 (y – 2) (y – 4) = 0 Hence y = 2 and y = 4 Hence x = 2, y = 4 , z = 0 or x = 4, y = 2, z = 0 Ex 3.1 , 6 Find the value of x, y, and z from the following equation: (iii) [■8(𝑥+𝑦+𝑧@█(𝑥+𝑧@𝑦+𝑧))] = [■8(9@█(5@7))] Since matrices are equal. Their corresponding elements are equal x + y + z = 9 x + z = 5 y + z = 7 Solving these equations From (2) x + z = 5 x = 5 – z From (3) y + z = 7 y = 7 – z Putting value of x & y in (1) x + y + z = 9 (5 – z) + (7 – z) + z = 9 5 – z + 7 – z + z = 9 5 + 7 – z – z + z = 9 12 – z = 9 – z = 9 – 12 – z = – 3 z = 3 Putting z = 3 in (2) x + z = 5 x + 3 = 5 x = 5 – 3 x = 2 Putting z = 3 in (3) y + 3 = 7 y = 7 – 3 y = 4 Hence x = 2, y = 4, & z = 3