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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
`" if " |{:(sin x,,sin y,,sin z),(cos x,,cos y,,cos z),(cos^(3) x,,cos^(3)y,,cos^(3)z):}|=0` then which of the following is`//`are possible ?A. `x=y`B. `y=z`C. `x=z`D. `x+y+z=pi//2` |
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Answer» Correct Answer - A::B::C::D `|{:(sin x ,,sin y,,sin z),(cos x,,cos y,,cos z),(cos^(3)x,,cos^(3)y,,cos^(3)z):}|` Taking cos x cos y and z common from `C_(1) ,C_(2) " and " C_(3)` respectively. `Delta = " cos x cos y cos z " |{:(tan x ,,tan y,,tan z),(1,,1,,1),(cos^(2),,cos^(2) y,,cos^(2)z):}|` Applyin `C_(1) to C_(1)-C_(2),C_(2) to C_(2) -C_(3)` we get `Delta = " cos x cos y cos z " |{:(tan x - tan y,,tan y - tan x,,tan z),(0,,0,,1),(cos^(2)x-cos^(2)y,,cos^(2)y-cos^(2)z,,cos^(2)z):}|` `=-" cos x cos y cos z "|{:(tan x-tan y,,tan y- tan z),(cos^(2) x-cos^(2) y,,cos^(2) y -cos^(2) z):}|` `=- cos x cos y cos z` `|{:((sin(x-y))/(cos x cos y),,(sin(y-z))/(cos ycos x)),(-sin (x-y) sin (x+y),,-sin (y-z) sin (y+z)):}|` `=sin (x-y) sin (y-z) |{:(cos z,,cos x ),(sin (x+y),,sin (y+z)):}|` `=sin (x-y) sin(y-z) [cos z sin (y+z)- cos x sin (x+y)]` `=sin (x-y) sin (y-z) [sin y cos^(2) z+sin z cos y cos z- sin x cos y cos x -sin y cos^(2) x]` `=sin (x-y) sin (y-z) [sin y(cos^(2) z-cos^(2)x) +cos y(sin zcos z-sin x cos x)]` `=sin (x-y) sin (y-z)[sin y sin (x-z) sin (x-z) +cos y (sin 2z-sin 2x)//2]` `=sin (x-y) sin (y-z)[sin y sin (x+z) sin (x-z) + cos y sin (z-x) cos (z+x)]` `=sin (x-y) sin (y-z) sin (z-x) -sin y sin (x+z) + cos y cos (z+x)]` `=sin (x-y) sin (y-z) sin (z-x) cos (x+y+z)` For `Delta =0, x=y " or " y=z " or " x+y+z=pi//2` |
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| 552. |
`|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z)C_(r+2)):}|` is equal to |
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Answer» Correct Answer - A::B::C::D `Delta =|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z)C_(r+2)):}|` Applying `C_(2) to C_(2) +C_(1)` we get `Delta = |{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z)C_(r+2)):}|` so (4) is correct option In (1) applying `C_(3) to C_(3) +C_(2)` we get `Delta =|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x+1)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y+1)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z+1)C_(r+2)):}|` So (3) is correct option In above applying `C_(2) to C_(2) +C_(1)` we get `Delta =|{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x+1)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y+1)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z+1)C_(r+2)):}|` So (1) is correct option In above applying `C_(3) to C_(3)+C_(2)` we get `Delta = |{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x+2)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y+2)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z+2)C_(r+2)):}|` so (2) is correct option |
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| 553. |
`" If " |{:(x^(2)+x,,x+1,,x-2),(2x^(2)+3x-1,,3x,,3x-3),(x^(2)+2x+3,,2x-1,,2x-1):}|=xA +B` thenA. `|{:(1,,1,,1),(-1,,-3,,3),(4,,0,,0):}|`B. `|{:(0,,1,,2),(1,,-2,,3),(-4,,0,,0):}|`C. `|{:(1,,1,,-2),(-3,,-2,,3),(4,,0,,1):}|`D. `|{:(0,,1,,-2),(-1,,-3,,3),(4,,0,,0):}|` |
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Answer» Correct Answer - A::D Let `Delta= |{:(x^(2)+x,,x+1,,x-2),(2x^(2)+3x-1,,3x,,3x-3),(x^(2)+2x+3,,2x-1,,2x-1):}|` Applying `R_(2) to R_(2) -2R_(1) " and " R_(3) to R_(3) -R_(1)` we get `Delta = |{:(x^(2)+x,,x+1,,x-2),(x-1,,x-2,,x+1),(x+3,,x-2,,x+1):}|` `=|{:(x^(2),,x+1,,x-2),(0,,x-2,,x+1),(0,,x-2,,x+1):}| + |{:(x,,x+1,,x-2),(x-1,,x-2,,x+1),(x+3,,x-2,,x+1):}|` `=0 +|{:(x,,x+1,,x-2),(x-1,,x-2,,x+1),(x+3,,x-2,,x+1):}|` Applying `R_(2) to R_(2)-R_(1)" and " R_(3) to R_(3)-R_(2)` we get `Delta= |{:(x,,x+1,,x-2),(-1,,-3,,3),(4,,0,,0):}|` `=|{:(x,,x,,x),(-1,,-3,,3),(4,,0,,0):}|+|{:(0,,-1,,-2),(-1,,-3,,3),(4,,0,,0):}|` `=|{:(1,,1,,1),(-1,,-3,,3),(4,,0,,0):}|x + |{:(0,,1,,-2),(-1,,-3,,3),(4,,0,,0):}|` |
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| 554. |
`" if " |{:(1,,1,,1),(a,,b,,c),(bc,,ca,,ab):}|=|{:(1,,1,,1),(a,,b,,c),(a^(3),,b^(3),,c^(3)):}|` where a,b,c are distinct positive reals then the possible values of abc is`//`areA. `(1)/(18)`B. `(1)/(63)`C. `(1)/(27)`D. `(1)/(18)` |
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Answer» Correct Answer - A::B we have `(a-b)(b-c)(c-a)=(a-b)(b-c)(c-a)(a+b+c)` `rArr a+b+c =1` As a , b, and c are positive using `AM gt GM` we get `(a+b+c)/(3) gt (abc)^(1/3)` `:. Abc lt (1)/(27)` |
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| 555. |
If `A=[1 3 2 1]`, find the determinant of the matrix `A^2-2Adot` |
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Answer» `A^2=A.A=[[1,3],[2,1]][[1,3],[2,1]]` `[[1+6,3+3],[2+2,6+1]]=[[7,6],[4,7]]` `A^2-2A=[[7,6],[4,7]]-2[[1,3],[2,1]]` `=[[7-2,6-6],[4-4,7-2]]` `=[[5,0],[0,5]]` `|A^2-2A|=[[5,0],[0,5]]` `=25-0` `=25` `|A^2-2A|=25` |
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| 556. |
Let A(1, 3), B(0, 0) and C(k, 0) be three points such that `ar(triangle ABC) =3` sq units. Find the value of k. |
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Answer» We have `ar(triangle ABC) = 3` sq units `hArr (1)/(2) * |[1, 3, 1], [0, 0, 1], [k, 0, 1]| = +- hArr |[1, 3, 1],[0, 0, 1],[k, 0, 1] = +-6` `hArr (-1) * |[1, 3],[k, 0]|= +-6 hArr 3k = +-6 hArr k = +-2` Hence, `k = +-2` |
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| 557. |
prove that `|[1,a,b+c],[1,b,c+a],[1,c,a+b]|=0` |
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Answer» The given determinant `=|{:(1, a, b+c), (1, b, c+a), (1, c, a+b):}|` `=|{:(1, a, a+b+c), (1, b, a+b+c), (1, c, a+b+c):}| " "["applying "C_(3) to C_(3) + C_(2)]` ` =(a+b+c) * |{:(1, a, 1), (1, b, 1), (1, c, 1):}| ["taking (a+b+c) common from "C_(3)]` ` = (a+b+c) xx 0 = 0 " "[because C_(1) "and "C_(3) " are identical"].` |
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| 558. |
If area of triangle is 35 sq units with vertices `(2, - 6), (5, 4) a n d (k , 4)`. Then k is(A) 12 (B) `-2` (C) ` 12 , 2` (D) `12 , 2` |
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Answer» Area of a triangle is given by, `A = 1/2|[x_1,y_1,1],[x_2,y_2,1],[x_3,y_3,1]|` For the given triangle, `35 = 1/2|[2,-6,1],[5,4,1],[k,4,1]|` Applying `R_2->.R_2 - R_3` on the determinant `=>|[2,-6,1],[5-k,0,0],[k,4,1]| = 70` `=>2(0-0)-((-6)(5-k)-0)+1(4(5-k) = +-70` `=>30-6k+20-4k = +-70` `=>-10k = 20 and -10k = -120` `=>k = -2 and k = 12` |
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| 559. |
Evaluate : \(\begin{vmatrix}a\,+ & c\,+ \\[0.3em]-c\,+ & a\,- \\[0.3em]\end{vmatrix}\) |
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Answer» (a + ib)(a - ib) - (c + id)(- c + id) = [a2 - i2b2] - [i2d2 - c2] = (a2 + b2) - (- d2 - c2) [∵ i2 = -1] = a2 +b2 + d2 + c2 = a2 + b2 + c2 + d2 |
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| 560. |
`|[1,bc,bc(b+c)],[1,ca,ca(c+a)],[1,ab,ab(a+b)]|=0` |
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Answer» The given determinant `=|{:(1, bc, a(b+c)), (1, ca, b(c+a)), (1, ab, c(a+b)):}|` `=|{:(1, bc, ab+bc+ca), (1, ca, ab+bc+ca), (1, ab, ab+bc+ca):}| " "["applying"C_(3) to C_(3) + C_(2)]` ` =(ab + bc +ca) * =|{:(1, bc, 1), (1, ca, 1), (1, ab, 1):}| " "["taking (ab+bc+ca) common from" C_(3)]` ` = (ab + bc+ca) xx 0 =0 " " [because C_(1)" and "C_(3) are identical"]. ` |
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| 561. |
Find the value of x from the following :\(\begin{vmatrix}x & 4 \\[0.3em]2 & 2x \\[0.3em]\end{vmatrix}=0\) |
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Answer» Here, \(\begin{vmatrix}x & 4 \\[0.3em]2 & 2x \\[0.3em]\end{vmatrix}=0\) ⇒ 2x2 - 8 = 0 ⇒ x2 = 4 ⇒ x = ± 2 |
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| 562. |
Let A be a non-singular square matrix of order 3 `xx`3. Then |adj A| is equal to (A) `|A|` (B) `|A|^2`(C) `|A|^3` (D) `3|A|` |
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Answer» We know, `A*Adj(A) = |A|*I` `=>|A*Adj(A)| = ||A|*I|` `=>|A|*|Adj(A)| = |[|A|,0,0],[0,|A|,0],[0,0,|A|]|` `=>|A|*|Adj(A)| = |A|^3` `=>|Adj(A)| = |A|^2` So, option `B` is the correct option. |
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| 563. |
Find the cofactor of a12 in the following :\(\begin{vmatrix}1 & -3 & 5 \\[0.3em]6& 0 &4 \\[0.3em]1 & 5 & -7\end{vmatrix}\) |
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Answer» a12 = (- 1)1+2 . M12 \(-\begin{vmatrix}6 &4 \\[0.3em]1 &-7 \\[0.3em]\end{vmatrix}\) = - (- 42 - 4) = 46 |
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| 564. |
Without expanding prove that `|[x+y ,z,1],[y+z,x,1],[z+x,y,1]|=0` |
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Answer» The given determinant `=|{:(x+y, y+z, z+x), (z, x, y), (1, 1, 1):}|` `=|{:(x+y+z, x+y+z, x+y+z), (z, x, y), (1, 1, 1):}| [R_(1) to R_(1) + R_(2)]` `=(x+y+z)*|{:(1, 1, 1), (z, x, y), (1, 1, 1):}| ["taking" (x+y+z) "common from "R_(1)]` ` = (x+y+z) xx 0 = 0 " "[because R_(1) " and "R_(3) " are identical"].` |
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| 565. |
Write the value of the following determinant :\(\begin{vmatrix}a-b& b-c & c-a \\[0.3em]b-c & c-a &a-b \\[0.3em]c-a & a-b & b-c\end{vmatrix}\) |
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Answer» Applying C1 \(\longrightarrow\)C1 + C2 + C3, we get \(\begin{vmatrix}0& b-c & c-a \\[0.3em]0 & c-a &a-b \\[0.3em]0 & a-b & b-c\end{vmatrix}=0\) [∵ All elements of C1 are zero] |
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| 566. |
If |A| = 2, where A is 2 × 2 matrix, find |adj A|. |
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Answer» We are given that, Order of matrix A = 2 × 2 |A| = 2 We need to find the |adj A|. Let us understand what adjoint of a matrix is. Let A = [aij] be a square matrix of order n × n. Then, the adjoint of the matrix A is transpose of the cofactor of matrix A. The relationship between adjoint of matrix and determinant of matrix is given as, |adj A| = |A|n-1 Where, n = order of the matrix Putting |A| = 2 in the above equation, ⇒ |adj A| = (2)n-1 …(i) Here, order of matrix A = 2 ∴, n = 2 Putting n = 2 in equation (i), we get ⇒ |adj A| = (2)2-1 ⇒ |adj A| = (2)1 ⇒ |adj A| = 2 Thus, the |adj A| is 2. |
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| 567. |
If \(\begin{vmatrix}x+2 &3 \\[0.3em]x+5 & 4 \\[0.3em]\end{vmatrix}=3\) , then find the value of x. |
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Answer» Here, \(\begin{vmatrix}x+2 &3 \\[0.3em]x+5 & 4 \\[0.3em]\end{vmatrix}=3\) ⇒ 4x + 8 - 3x -15 = 3 ⇒ x - 7 = 3 ⇒ x = 10 |
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| 568. |
Evaluate `|(1,omega,omega^2),(omega,omega^2,1),(omega^2,omega,omega)|` where `omega` is cube root of unity. |
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Answer» The given determinant ` =|{:(1, omega, omega^(2)), (omega, omega^(2), 1), (omega^(2), 1, omega):}|` ` =|{:(1+omega+omega^(2), 1+omega+omega^(2), 1+omega+omega^(2)), (omega, omega^(2), 1), (omega^(2), 1, omega):}| [R_(1) to R_(1) + R_(2) + R_(3)]` ` =|{:(0, 0, 0), (omega, omega^(2), 1), (omega^(2), 1, omega):}| [because 1+omega + omega^(2) =0]` ` =0 " " [because R_(1) "consists of all zeros"]` |
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| 569. |
Prove that `|{:(1, " "1, " "1), (1, 1+x, " "1), (1, " "1, 1+y):}|=xy.` |
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Answer» The given determinant `=|{:(1, " "1, " "1), (1, 1+x, " "1), (1, " "1, 1+y):}|` `=|{:(1, 0, 0), (1, x, 0), (1, 0, y):}|" "["applying"C_(2) to C_(2)-C_(1) " and "C_(3) to C_(3)-C_(1)]` ` = 1 * |{:(x, 0), (0, y):}| = xy " "["expanding by "R_(1)].` |
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| 570. |
Show that `|{:(2, 3, 4), (5, 6, 8), (6x, 9x, 12x):}| =0.` |
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Answer» The given determinant `=|{:(2, 3, 4), (5, 6, 8), (6x, 9x, 12x):}|` `=(3x)|{:(2, 3, 4), (5, 6, 8), (2, 3, 4):}| " "["taking 3x common from "R_(3)]` `=(3x) xx 0 =0 " "[because R_(1) " and " R_(3) " are identical"].` |
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| 571. |
Write the value of the following determinant :\(\begin{vmatrix} 2 & 3 & 4 \\[0.3em] 5& 6 &8 \\[0.3em] 6x & 9x & 12x \end{vmatrix}\) |
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Answer» \(\begin{vmatrix}2 & 3 & 4 \\[0.3em]5& 6 &8 \\[0.3em]6x & 9x & 12x\end{vmatrix}\) = \(3x\begin{vmatrix}2 & 3 & 4 \\[0.3em]5& 6 &8 \\[0.3em]2 & 3 & 4\end{vmatrix}\) [Taking out 3x common from R3] = 3x x 0 = 0 [ ∵ R1 = R3] |
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| 572. |
If A is square matrix of order 3 such that |A| = λ, then write the value of |– A|. |
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Answer» ∵ |A| = λ and order of A = 3 ∴ |–A| = (–1)3 .|A| = –1 × λ = – λ |
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| 573. |
Find the value of \(\begin{vmatrix}sin\,A &-sin\,B \\[0.3em]cos\,A & cos\,B \\[0.3em]\end{vmatrix}\) where A = 53°, B = 37°. |
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Answer» Let \(\Delta=\begin{vmatrix}sin\,A &-sin\,B \\[0.3em]cos\,A & cos\,B \\[0.3em]\end{vmatrix}\) Δ = sin A cos B + sin B cos A ⇒ Δ = sin (A + B) Δ = sin (53 + 37) = sin 90° = 1. |
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| 574. |
Mark the tick against the correct answer in the following:\(\begin{vmatrix}sin 23^ \circ & -sin7^ \circ \\[0.3em]cos23^ \circ & cos7^ \circ\end{vmatrix}\) = ?|(sin 23°, -sin27°),(cos23°, cos 7°)| = ?A. \(\frac{\sqrt{3}}{2}\)B. \(\frac{1}{2}\)C. sin 16°D. cos 16° |
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Answer» Correct answer B \(\frac{1}{2}\) To find: value of \(\begin{vmatrix} sin 23^ \circ & -sin7^ \circ \\[0.3em] cos23^ \circ & cos7^ \circ \end{vmatrix}\) Formula used: (i) sin(A + B) = sin A cos B + cos A sin B We have, \(\begin{vmatrix} sin 23^ \circ & -sin7^ \circ \\[0.3em] cos23^ \circ & cos7^ \circ \end{vmatrix}\) on expanding the above, On expanding the above, ⇒ (sin 23°) (cos 7°) – (cos 23°) (-sin 7°) ⇒ (sin 23°) (cos 7°) + (cos 23°) (sin 7°) On applying formula sin(A + B) = sin A cos B + cos A sin B = sin (23 + 7) = sin (30°) = \(\frac{1}{2}\) |
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| 575. |
`|(x^(2)-x+1, x-1),(x+1,x+1)|` |
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Answer» Correct Answer - `x^(3)-x^(2) + 2` `Delta =(x^(2)-x+1)(x-1)-(x+1)(x-1) = (x^(3)+1)-(x^(2)-1) = (x^(3)-x^(2)+2)` |
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| 576. |
Solve : `(3x +4)/(7) - (x +5)/(14) = (x)/(28) + (x +1)/(14)` The following steps are involved in solving the above problem. Arrange them in sequential order (A) `5x + 3 = (3x + 2)/(2) rArr 10 x + 6 = 3x + 2` (B) `rArr 7x = -4` (C) Given `(3x + 4)/(7) - (x + 5)/(14) = (x)/(28) + (x +1)/(14)` `rArr (6x + 8 -x -5)/(14) = (x + 2x + 2)/(28)` (D) `rArr x = -(4)/(7)` |
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Answer» Correct Answer - x=-2 `(3x xx 4) - 7 xx (-2) = 32-42 rArr 12x +14 =-10 rArr 12x = -24 rArr x = -2` |
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| 577. |
Evaluate:`|[a+ib, c+id],[-c+id, a-ib]|` |
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Answer» Correct Answer - `(a^(2) + b^(2)+c^(2)+d^(2))` `Delta = (a+ib)(a-ib) -(c+id) (-c+id) ={a^(2)-(ib)^(2)} -{(id)^(2) -c^(2)} = (a^(2)+b^(2))-(-d^(2)-c^(2)) = (a^(2) + b^(2)+c^(2)+d^(2))` |
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| 578. |
Solve the following system of equations using Cramers rule.`5x-7y+z=11 , 6x-8y-z=15 and 3x+2y-6z=7` |
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Answer» `D= |(5,-7,1),(6,-8,-1),(3,2,-6)| = 55` `D_1 = |(11,-7,1),(15,-8,-1),(7,2,-6)|= 55` `D_2 = |(5,11,1),(6,15,-1),(3,7,-6)|= -55` `D_3 = |(5,-7,11),(6,-8,15),(3,2,7)|= -55` `x= D/D_1 = 55/55 = 1` `y= 55/-55 = -1` `z= 55/-55=-1` `x=1; y=-1; z=-1` Answer |
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| 579. |
If A and B are square matrices of order 3 such that `absA=-1,absB=3," then "abs(3AB)` equalsA. `-9`B. `-81`C. `-27`D. 81 |
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Answer» Correct Answer - A We have, `|A| = -1, |B| =3` `:. |3AB| = 3| A|| B| = 3xx -1 xx 3 = -9` |
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| 580. |
If A and B are square matrices of order 3 such that |A| = -1, |B| = 3, then find the value of |3AB|. |
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Answer» We are given that, A and B are square matrices of order 3. |A| = -1, |B| = 3 We need to find the value of |3AB|. By property of determinant, |KA| = Kn|A| If A is of nth order. If order of A = 3 × 3 And order of B = 3 × 3 ⇒ Order of AB = 3 × 3 [∵, Number of columns in A = Number of rows in B] We can write, |3AB| = 33|AB| [∵, Order of AB = 3 × 3] Now, |AB| = |A||B|. ⇒ |3AB| = 27|A||B| Putting |A| = -1 and |B| = 3, we get ⇒ |3AB| = 27 × -1 × 3 ⇒ |3AB| = -81 Thus, the value of |3AB| = -81. |
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| 581. |
Let A, B be two square matrices such that A + B = AB, then(A) AB = BA (B) AB = -BA (C) AB + 2BA = 0 (D) None of these |
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Answer» Answer is (A) A + B = AB A + B – AB = 0 A + B – AB + I = 0+1 (A – I)(B – I) = 1 ⇒ A -1 inverse of B -1 (B -1) (A -1) = I BA – B – A + = I BA – (B + A) ⇒ BA = B + A = A + B ∴ AB = BA |
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| 582. |
Find y \(\begin{bmatrix}x & x - y \\[0.3em]2x + y& 7 \\[0.3em]\end{bmatrix}\) = \(\begin{bmatrix}3 &1 \\[0.3em]8 &7 \\[0.3em]\end{bmatrix}\) |
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Answer» x = 3 x - y = 1 3 - y = 1 y = 2 |
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| 583. |
The system of equation `kx+y+z=1, x +ky+z=k and x+y+kz=k^(2)` has no solution if k equals |
| Answer» Correct Answer - D | |
| 584. |
the sum of values of p for which the equations x+y+z=1x+2y +4z=p and x+4y +10z =`p^(2)` have a solution is `"____"` |
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Answer» Correct Answer - 3 `x+y+z=1` `x+2y +4z=p` `x+4y+10z=p^(2)` `Delta = |{:(1,,1,,1),(1,,2,,4),(1,,4,,10):}|` `R_(1) to R_(1)-R_(2),R_(2),R_(2) to R_(2)-R_(3)` `= |{:(0,,-1,,-3),(-,,-2,,-6),(1,,4,,10):}|=0` Since `Delta =0` solutions is not unique The system will have infinite solutions if `Delta_(1)=0 ,Delta_(2) =0 Delta_(3)=0` `Delta_(1)= |{:(1,,1,,1),(p,,2,,4),(p^(2),,4,,10):}|=0` `C_(3) to C_(3)-C_(2)` `Delta_(1)= |{:(1,,1,,0),(p,,2,,2),(p^(2),,4,,6):}|=0` `" or " 1(12 -8) -1 (6p -2p^(2)) =0` `" or " 4-6p +2p^(2) =0` `" or " 2(p^(2) -3p+2) =0` `" or " p^(2) -3p +2=0` `rArr p=1" or "2` Also for these values of `p,Delta_(2),Delta_(3)=0` |
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| 585. |
If `a_1, a_2, a_3,54,a_6,a_7, a_8, a_9`are in H.P., and `D=|a_1a_2a_3 5 4a_6a_7a_8a_9|`, then the value of `[D]i sw h e r e[dot]`represents the greatest integer function |
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Answer» Correct Answer - 2 We have `D=|{:(a_(1),,a_(2),,a_(3)),(5,,4,,a_(6)),(a_(7),,a_(8),,a_(9)):}|` Since `a_(n)=(20)/(n): d=(1)/(20)` `" Hence " D= |{:(20,,(20)/(2),,(20)/(3)),((20)/(4),,(20)/(5),,(20)/(6)),((20)/(7),,(20)/(8),,(20)/(9)):}|=((20)^(3))/(4xx7) |{:(1,,(1)/(2),,(1)/(3)),(1,,(4)/(5),,(2)/(3)),(1,,(7)/(8),,(7)/(9)):}|` `R_(1) to R_(1)-R_(2) " and " R_(2) to R_(2)-R_(3)` `=((20)^(3))/(4xx7) |{:(0,,(-3)/(10),,(-1)/(3)),(0,,(-3)/(40),,(-1)/(9)),(1,,(7)/(8),,(7)/(9)):}|=(50)/(21)` `rArr [D] =2` |
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| 586. |
If `alpha,beta,gamma`are the angles of a triangle and system of equations`cos(alpha-beta)x+cos(beta-gamma)y+cos(gamma-alpha)z=0``cos(alpha+beta)x+cos(beta+gamma)y+cos(gamma+alpha)z=0``sin(alpha+beta)x+sin(beta+gamma)y+sin(gamma+alpha)z=0`has non-trivial solutions, then triangle is necessarilya. equilateral b. isoscelesc. right angled`""`d. acute angledA. equiliateralB. isosocelesC. right angledD. acute angled |
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Answer» Correct Answer - B Let `Delta = |{:(cos (alpha -beta),,cos (beta-gamma),,cos (gamma-alpha)),(coa(alpha+beta),,cos(beta+gamma),,cos(gamma+alpha)),(sin(alpha+beta),,sin(beta+gamma),,sin(gamma+alpha)):}|` It is clear that either `alpha =beta " or " beta = gamma " or " gamma =alpha ` is suffieicient to make `Delta =0 ` . It is not necessary that triangle is equilateral. Also isosceles traingle can be obtuse one. |
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| 587. |
`a ,b ,c`are distinct real numbers not equal to one. If `a x+y+z=0,x+b y+z=0,a n dx+y+c z=0`have nontrivial solution, then the value of `1/(1-a)+1/(1-b)+()/(1-c)`is equal to`1`b. `-1`c.`z e ro`d. none of theseA. `-1`B. `1`C. zeroD. none of these |
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Answer» Correct Answer - B since the system has non-trivial solution we have `|{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}|=0` Applying `R_(1) toR_(1)-R_(2),R_(2) to R_(2) -R_(3) `we get `Delta =|{:(a-1,,1-b,,0),(0,,b-1,,1-c),(1,,1,,c):}|=0` `" or " c(1-a)(1-b)+(1-b)(1-c)-(1-c) (a-1)=0` Dividing throughout by (1-a)(1-b)(1-c) we get `(c ) /(1-c) +(1)/(1-a)+(1)/(1-b) =0` `" or " -1 + (1)/(1-c)+(1)/(1-a)+(1)/(1-b)=0` `" or " (1)/(1-c)+(1)/(1-a)+(1)/(1-b) =1` |
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| 588. |
If the system of equations `x + ay + az = 0` `bx + y + bz = 0` `cx + cy + z = 0` where a, b and c are non-zero non-unity, has a non-trivial solution, then value of `(a)/(1 -a) + (b)/(1- b) + (c)/(1 -c)` is |
| Answer» Correct Answer - C | |
| 589. |
Consider the system of equations `x+y+z=6` `x+2y+3z=10` `x+2y+lambdaz =mu` The system has no solution ifA. `lambda ne 3`B. `lambda =3, mu =10`C. `lambda =3, mu ne 10`D. none of these |
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Answer» Correct Answer - C `Delta = |{:(1,,1,,1),(1,,2,,3),(1,,2,,lambda):}|=2lambda +3+2 -2-lambda -6=lambda -3` `Delta_(1) = |{:(6,,1,,1),(10,,2,,3),(mu,,2,,lambda):}|=12lambda +3mu +20 -2mu -10lambda -36` `=2lambda +mu -16` `Delta_(2) =|{:(1,,6,,1),(1,,10,,3),(1,,mu,,lambda):}|=10lambda +18 +mu -10 -3mu -6lambda =4lambda -2mu +8` `Delta_(3) = |{:(1,,1,,6),(1,,2,,10),(1,,2,,mu):}|=2mu +10 +12 -12 -mu -20 =mu -10` Thus the system has unique solutions if `Delta ne 0 " or " lambda ne 3` and the system has infinite solution if `Delta=Delta_(1) =Delta_(2) =Delta_(3) =0 " ot " lambda =3` and `mu=10`. System has no solution if `Delta =0` and at least one of `Delta_(1) ,Delta_(2) ,Delta_(3)` is nonzero of `lambda =3 " and " mu ne =10` |
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| 590. |
consider the system of equations : ltbr. `3x-y +4z=3` `x+2y-3z =-2` `6x+5y+lambdaz =-3` Prove that system of equation has at least one solution for all real values of `lambda`.also prove that infinite solutions of the system of equations satisfy `(7x-4)/(-5)=(7y+9)/(13)=z` |
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Answer» `Delta= |{:(3,,-1,,4),(1,,2,,-3),(6,,5,,lambda):}|=7lambda +35` `" If "7lambda +35ne 0 i.e., lambda ne -5` then system has a unique solution (As `Delta ne 0` gives unique solution). But if `lambda =-5` then we have `Delta =0` . Solution exists in this case if `Delta_(x) =Delta_(y) =Delta_(z)=0` `" For " lambda =-5` `Delta_(x)= |{:(3,,-1,,4),(-2,,2,,-3),(-3,,5,,5):}|=0` `Delta_(y)= |{:(3,,-3,,4),(1,,-2,,-3),(6,,-3,,-5):}|=0` `" and "Delta_(z)= |{:(3,,-1,,3),(1,,2,,-2),(6,,5,,-3):}|=0` Thus `Delta =Delta_(x)=Delta_(y)=Delta_(z)=0` and hence there exists infinite number of solutions. Now. eliminating x from the equations .we get `7y-13z =-9` Let us put `z = k in R. :. y= (13k -9)//7" and " so x=(4-5k)//7` Where k is any real number . `" Thus " .(7x-4)/(-5) =(7y+9)/(13)=z` |
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| 591. |
If the system of linear equation `x+y+z=6,x+2y+3c=14 ,a n d2x+5y+lambdaz=mu(lambda,mu R)`has a unique solution, then`lambda=8`b. `lambda=8,mu=36`c.`lambda=8,mu!=36""`d. none of theseA. `lambda ne 8`B. `lambda =8,mu ne 36`C. `lambda=8 , mu =36`D. none of these |
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Answer» Correct Answer - A The given system of linear equations has a unique solution if `|{:(1,,1,,1),(1,,2,,3),(2,,5,,lambda):}| ne 0` `i.e., " if " lambda- 8 ne 0 " or " lambda ne 8.` |
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| 592. |
Consider the system of equations `x+y+z=6` `x+2y+3z=10` `x+2y+lambdaz =mu` the system has infinite solutions ifA. `lambda ne 3`B. `lambda =3,mu =10`C. `lambda =3,mu ne 10`D. `lambda =3,mu ne 10` |
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Answer» Correct Answer - B `Delta = |{:(1,,1,,1),(1,,2,,3),(1,,2,,lambda):}|=2lambda +3+2 -2-lambda -6=lambda -3` `Delta_(1) = |{:(6,,1,,1),(10,,2,,3),(mu,,2,,lambda):}|=12lambda +3mu +20 -2mu -10lambda -36` `=2lambda +mu -16` `Delta_(2) =|{:(1,,6,,1),(1,,10,,3),(1,,mu,,lambda):}|=10lambda +18 +mu -10 -3mu -6lambda =4lambda -2mu +8` `Delta_(3) = |{:(1,,1,,6),(1,,2,,10),(1,,2,,mu):}|=2mu +10 +12 -12 -mu -20 =mu -10` Thus the system has unique solutions if `Delta ne 0 " or " lambda ne 3` and the system has infinite solution if `Delta=Delta_(1) =Delta_(2) =Delta_(3) =0 " ot " lambda =3` and `mu=10`. System has no solution if `Delta =0` and at least one of `Delta_(1) ,Delta_(2) ,Delta_(3)` is nonzero of `lambda =3 " and " mu ne =10` |
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| 593. |
Given that the system of equations `x=cy+bz ,y=az+cx , z=bx +ay` has nonzero solutions and and at least one of the a,b,c is a proper fraction. System has solution such thatA. `x,y,z -= (1-2a^(2)):(1-2b^(2)):(1-2c^(2))`B. `x.y.z -= (1)/(1-2a^(2)):(1)/(1-2b^(2)):(1)/(1-2c^(2))`C. `x.y.z -= (a)/(1-a^(2)):(b)/(1-b^(2)):( c)/(1-c^(2))`D. `x.y.z -= sqrt(1-a^(2)):sqrt(1-b^(2)):sqrt(1-c^(2))` |
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Answer» Correct Answer - D The system of equation `-x+cy+bz=0` `cx-y+az=0` `bx+ay-z=0` has a nonzero solution if `Delta = |{:(-1,,c,,b),(c,,-1,,a),(b,,a,,-1):}|=0` then Clearly the system has infinitely many solutions .From (1) and (2) we have `(x)/(ac+b) =(y)/(bc+a)=(z)/(1-c^(2)) ` `" or " (x^(2))/((1-a^(2))(1-c^(2)))=(y^(2))/((1-b^(2))(1-c^(2)a)) =(z^(2))/((1-c^(2))^(2))`[From (4)] `" or " (x^(2))/(1-a^(2))=(y^(2))/(1-b^(2)) =(z^(2))/(1-c^(2))` from (5) we see that `1-a^(2),1-b^(2),1-c^(2)` are all positive or all negative .Given that one of a,b,c is proper fraction so `1-a^(2) gt ,1-b^(2) gt 0,1-c^(2) gt 0` which gives `a^(2) +b^(2)+c^(2) lt 3` using (4) and (6) we get `1lt 3+2 abc` `" or " abc gt -1` |
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| 594. |
Consider the system of equations `x+y+z=6` `x+2y+3z=10` `x+2y+lambdaz =mu` the system has unique solution ifA. `lambda ne 3`B. `lambda =3, mu =10`C. `lambda =3 , mu ne 10`D. none of these |
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Answer» Correct Answer - A `Delta = |{:(1,,1,,1),(1,,2,,3),(1,,2,,lambda):}|=2lambda +3+2 -2-lambda -6=lambda -3` `Delta_(1) = |{:(6,,1,,1),(10,,2,,3),(mu,,2,,lambda):}|=12lambda +3mu +20 -2mu -10lambda -36` `=2lambda +mu -16` `Delta_(2) =|{:(1,,6,,1),(1,,10,,3),(1,,mu,,lambda):}|=10lambda +18 +mu -10 -3mu -6lambda =4lambda -2mu +8` `Delta_(3) = |{:(1,,1,,6),(1,,2,,10),(1,,2,,mu):}|=2mu +10 +12 -12 -mu -20 =mu -10` Thus the system has unique solutions if `Delta ne 0 " or " lambda ne 3` and the system has infinite solution if `Delta=Delta_(1) =Delta_(2) =Delta_(3) =0 " ot " lambda =3` and `mu=10`. System has no solution if `Delta =0` and at least one of `Delta_(1) ,Delta_(2) ,Delta_(3)` is nonzero of `lambda =3 " and " mu ne =10` |
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| 595. |
The system of linear equations `x+y-z=6, x+2y-3z=14` and `2x+5y-lambda z=9` has a unique solution (A) `lambda=8` (B) `lambda !=8` (C) `lambda=7` (D) `lambda !=7` |
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Answer» For a system of linear equations having unique solution, value of the determinant should not be `0`. Here, `Delta = |[1,1,-1],[1,2,-3],[2,5,-lambda]| != 0` `=>[-2lambda+15-1(-lambda+6)-(5-4)] != 0` `=>-lambda+8 != 0` `=>lambda !=8` So, option `B` is the correct option. |
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| 596. |
The system of simulataneous equations `kx + 2y -z = 1` `(k -1) y -2z = 2` `(k +2) z = 3` have a unique solution if k equalsA. `-2`B. `-1`C. 0D. 1 |
| Answer» Correct Answer - B | |
| 597. |
Find the inverse the matrix (if it exists)given in`[0 0 0 0cosalphasinalpha0sinalpha-cosalpha]` |
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Answer» `"Let A="|{:(1,0,0),(0,"cos"alpha,"sin"alpha),(0,"sin"alpha,"-cos"alpha):}|` `rArr" |A| ="|{:(1,0,0),(0,"cos"alpha,"sin"alpha),(0,"sin"alpha,"-cos"alpha):}|` `=1(cos^(2)alpha-sin^(2)alpha)-0=-1ne0` `A_(11)=(-1)^(2)(-cos^(2)alpha-sin^(2)alpha)=-1` `A_(12)=(-1)^(3) (0-0)=0` `A_(13)=9-1)^(4)(0-0)=0` `A_(21)=(-1)^(3)(0-0)=0` `A_(22)=(-1)^(4)(-cos alpha - 0)=-cos alpha` `A_(23)(-1)^(5)(sin alpha-0)=0` `A_(31)=(-1)^(4)(0-0)=0` `A_(32)=(-1)^(5)(sin alpha -0) = - sin alpha`, `A_(33)=(-1)^(6)(cos alpha -0) = cos alpha` `|{:(1,0,0),(0,"-cos"alpha,"-sin"alpha),(0,"-sin"alpha,"cos"alpha):}|=|{:(-1,0,0),(0,"-cos"alpha,"-sin"alpha),(0,"-sin"alpha,"cos"alpha):}|` `"Now A"^(-1)=1/|A|"adj A"=1/-1|{:(-1,0,0),(0,"-cos"alpha,"-sin"alpha),(0,"-sin"alpha,"cos"alpha):}|` `=|{:(1,0,0),(0,"cos"alpha,"sin"alpha),(0,"sin"alpha,"-cos"alpha):}|` |
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| 598. |
Find the inverse the matrix (if it exists)given in`[1-1 2 0 2-3 3-2 4]` |
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Answer» `"Let A ="|{:(1,-1,2),(0,2,-3),(3,-2,4):}|` `rArr" |A|="|{:(1,-1,2),(0,2,-3),(3,-2,4):}|` =1(8-6)-(-1)(0+9)+2(0-6) =2+9-12=-1 `A_(11)=(-1)^(2)(8-6)=2,A_(12)=(-1)^(3)(0+9)=-9` `A_(13)=(-1)^(4)(0-6)=-6` `A_(21)=(-1)^(3)(-4+4)=0,A_(22)=(-1)^(4)(4-6)=-2` `A_(23)=(-1)^(5)(-2+3)=-1` `A_(31)=(-1)^(4)(3-4)=-1,A_(32)=(-1)^(5)(-3-0)=3` `A_(33)=(-1)^(6)(2-0)=2` `therefore" adj A"=|{:(2,-9,-6),(0,-2,-1),(-1,3,2):}|=|{:(2,0,-1),(-9,-2,3),(-6,-1,2):}|` `"Now A"^(1)=1/|A|".adj A "=1/-1|{:(2,0,-1),(-9,-2,3),(-6,-1,2):}|` `|{:(-2,0,1),(9,2,-3),(6,1,-2):}|` |
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| 599. |
Find the inverse the matrix (if it exists) given in`[[2, 1, 3],[ 4,-1, 0],[-7, 2, 1]]` |
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Answer» `"Let A ="|{:(2,1,3),(4,-1,0),(-7,2,1):}|` `rArr" "|A|=|{:(2,1,3),(4,-1,0),(-7,2,1):}|` `=2,(-1-0)-1(4-0)+3(8-7)` `=-2-4+3=-3ne0` `A_(11)=(-1)^(2)(-1-0)=-1` `A_(12)=(-1)^(3)(4-0)=-4`, `A_(13)=(-1)^(4)(8-7)=1,` `A_(21)=(-1)^(3)(1-6)=5`, `A_(22)=(-1)^(4)(2+21)=23` `A_(23)=(-1)^(5)(4+7)=-11,` `A_(31)=(-1)^(4)(0+3)=3` `A_(32)=(-1)^(4)(0+3)=3` `A_(33)=(-1)^(6)(-2-4)=-6` `therefore" adj A"=|{:(-1,-4,1),(5,23,-11),(3,12,-6):}|=|{:(-1,5,3),(-4,23,12),(1,-11,-6):}|` `"Now A"^(-1)=1/|A|"adj A"=-1/3|{:(-1,5,3),(-4,23,12),(1,-11,-6):}|` |
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| 600. |
Choose the correct answer from the following : The inverse of matrix `|{:(1,3),(2,-1):}|is:`A. `1/7|{:(1,3),(2,-1):}|`B. `1/7|{:(-1,3),(2,1):}|`C. `1/7|{:(1,-3),(-2,-1):}|`D. `1/7|{:(-1,-1),(2,1):}|` |
| Answer» Correct Answer - A | |