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Q. 1. be an identity in λ , where p, q, r, s and t are constants.
Then, the value of t is ................. (1981  2 Marks)
Ans. t = 0
Solution. As given equation is an identity in λ, it must be true for all values of λ.
∴ For λ = 0 also. Putting λ = 0 we get
Q. 2. The solution set of the equation .................. (1981  2 Marks)
Ans. x = –1, 2
Solution. Given equation is,
Clearly on expanding the det. we will get a quadratic equation in x.
∴ It has 2 roots. We observe that R3 becomes identical to R_{1} if x = 2. thus at x = 2 ⇒ Δ = 0
∴ x = 2 is a root of given eq.
Similarly R_{3} becomes identical to R_{2} if x = – 1. thus at x = – 1 Δ D = 0
∴ x = – 1 is a root of given eq.
Hence equation has roots as –1 and 2.
Q. 3. A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of determinant chosen is positive is .................. (1982  2 Marks)
Ans. 3/16
Solution. With 0 and 1 as elements there are 2 × 2 × 2 × 2 = 16 determinants of order 2 × 2 out of which only are the three det whose value is +ve.
∴ Req. prob. = 3/16
Q. 4. Given that x = 9 is a root of the other two roots are ................. and .............. (1983  2 Marks)
Ans. 2, 7
Solution.
Operating R_{1} → R_{1} + R_{2} + R_{3} we get
Expanding along R_{1}
⇒(x + 9) (x – 2) (x – 7) = 0
⇒ x = – 9, 2, 7
∴ Other roots are 2 and 7.
Q. 5. The system of equations
Will have a nonzero solution if real values of l are given by .................. (1984  2 Marks)
Ans. λ = 0
Solution. The given homogeneous system of equations will have non zero solution if D = 0
⇒ λ (λ^{2} + 1) – 1 (–λ + 1) + 1 (1 + λ) = 0 ⇒ λ^{3} + 3λ = 0
⇒ λ (λ^{2} + 3) = 0, but λ^{2} + 3 ≠ 0 for real λ ⇒ λ = 0
Q. 6. The value of the determinant is .................. (1988  2 Marks)
Ans. 0
Solution.
Operating R_{1}→ R_{1} – R_{2}; R_{2}→ R_{2} – R_{3}
Q. 7. For positive numbers x, y and z, the numerical value of the determinant (1993  2 Marks)
Ans. 0
Solution. Given x, y, z and + ve numbers, then value of
Taking common from R_{1}, R_{2} and R_{3} respectively
True / False
Q. 1. The determinants are not identically equal. (1983  1 Mark)
Ans. F
Solutions.
[C_{1} ⇔ C_{3} and then C_{2} ⇔ C_{3}]
∴ Equal. Hence statement is F.
Q. 2. If then the two triangles with vertices (x_{1} , y_{1}), (x_{2} , y_{2}), (x_{3} , y_{3}), and (a_{1} , b_{1}), (a_{2} , b_{2}), (a_{3} , b_{3}) must be congruent. (1985  1 Mark)
Ans. F
Solutions.
Ar (Δ_{1}) = Ar (D_{2})
Where Δ_{1} is the area of triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}); and Δ_{2} is the area of triangle with ; vertices (a_{1}, b_{1}), (a_{2}, b_{2}) and (a_{3}, b_{3}). But two D’s of same area may not be congruent.
∴ Given statement is false.
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