Evaluate the determinants
(i) \(\begin{vmatrix}3 & -1 & -2\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}\) (iii) \(\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}\)
(ii) \(\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}\)(iv) \(\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}\)
(i) Let A = \(\begin{vmatrix}3 & -1 & -2\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}\)
It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.
|A| = -0\(\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}\) + 0\(\begin{vmatrix}3 & -2\\3 & 0\end{vmatrix}\) – (-1)\(\begin{vmatrix}3 &-1\\3 &-5\end{vmatrix}\) = (-15 + 3) = -12
(ii) Let A = \(\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}\)
By expanding along the first row, we have:
|A| = 3\(\begin{vmatrix}1 & -2\\3 & 1\end{vmatrix}\) + 4\(\begin{vmatrix}1 & -2\\2 & 1\end{vmatrix}\) + 5\(\begin{vmatrix}1 &1\\2 &3\end{vmatrix}\)
= 3 (1+6) + 4(1+4) + 5(3-2)
= 3 (7) + 4 (5) + 5 (1)
= 21 + 20 + 5
= 46
(iii) Let A = \(\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}\)
By expanding along the first row, we have:
|A| = 0\(\begin{vmatrix}0 & -3\\3 & 0\end{vmatrix}\) - 1\(\begin{vmatrix}-1 & -3\\-2 & 0\end{vmatrix}\) + 2\(\begin{vmatrix}-1 & 0\\-2 &3\end{vmatrix}\)
= 0 – 1(0 – 6) + 2 (-3 - 0)
= -1 (-6) + 2(-3)
= 6 – 6
= 0
(iv) Let A = \(\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}\)
By expanding along the first column, we have:
|A| = 2\(\begin{vmatrix}2 & -1\\-5 & 0\end{vmatrix}\) - 0\(\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}\) + 3\(\begin{vmatrix}-1 & -2\\2 & -1\end{vmatrix}\)
= 2(0 – 5) – 0 + 3(1 + 4)
= -10 + 15 = 5
Popular Questions of Class 12 Mathematics
- Q:- Given an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive. - Q:- Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3,13, 14} defined as
R = {(x, y): 3x − y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x − y is as integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y} - Q:- Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
- Q:- Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. - Q:- Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.
- Q:-
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
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- Q:-
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
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Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
- Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
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