Determinants Test

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Determinants Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The point with the co-ordinates $$(2a,\ 3a),\ (3b,\ 2b)$$ & $$(c,\ c)$$ are collinear _____________ .

A
for no value of $$a,\ b,\ c$$

B
for all values of $$a,\ b,\ c$$

C
if $$a,\ \dfrac {c}{5},\ b$$ are in $$H.P$$.

D
if $$a,\ \dfrac {2}{5}c,\ b$$ are in $$H.P$$.

Solution

$$\Delta=\dfrac {1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\x_3 & y_3 & 1\end{vmatrix}=0$$

$$\Delta=\dfrac {1}{2}\begin{vmatrix}2a & 3a & 1\\ 3b & 2b & 1\\c & c & 1\end{vmatrix}=0$$

$$\Rightarrow (2a-c)(2b-c)-(3a-c)(3b-c)=0$$

$$\Rightarrow 4ab-2ac-2bc+c^2=(9ab-3ac-3bc+c^2)=0$$

$$\Rightarrow ac+bc-5ab=0$$

$$\dfrac {1}{a}+\dfrac {1}{b}=\dfrac {5}{c}\Rightarrow \dfrac {1}{a}+\dfrac {1}{b}=2\left (\dfrac {5}{2c}\right )$$

$$\because$$$$\dfrac {1}{a},\dfrac {5}{2c}\, and\, \dfrac{1}{b} $$ are in $$A.P.$$

$$\therefore a, \dfrac {2c}{5}, b$$ are in H.P.
Question 2
1 / -0

If lines AB, AC, AD and AE are parallel to a line then_______

A
A, B, C, D, E are collinear points

B
A, B, C, D, E are noncolinear points

C
AB & AC are parallel and AD & AE are perpendicular

D
None of these

Solution

Gievn Line $$AB,AC,AD,AE$$ are parallel so they are only and only they are collinear
Because if we draw line AB To point C we find line AC similarly in moving forward we Get line AC.AD and AE
Hence $$A,B,C,D,E $$ are collinear points
Question 3
1 / -0

The adjoint of the matrix $$\begin{bmatrix} 1 & -2 & 3 \\ 0 & 2 & -1 \\ -4 & 5 & 2 \end{bmatrix}$$ is:

A
$$\begin{bmatrix} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{bmatrix}$$

B
$$\begin{bmatrix} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{bmatrix}$$

C
$$\begin{bmatrix} 9 & -19 & -4 \\ -4 & 14 & -1 \\ 8 & -3 & 2 \end{bmatrix}$$

D
none of these

Solution

Cofactor of the matrix is given by
$$\begin{bmatrix} M11 & -M12 & M13 \\ -M21 & M22 & -M23 \\ M31 & -M32 & M33 \end{bmatrix}$$
where$$ M11 $$= $$\begin{bmatrix} 2 & -1\\ 5 & 2 \end{bmatrix} = 4+5=9$$
$$M12$$ = $$\begin{bmatrix} 0 & -1 \\ -4 & 2 \end{bmatrix} = 0-4 =-4$$
$$M13$$ = $$\begin{bmatrix} 0 & 2 \\ -4 & 5 \end{bmatrix} = 0+8=8$$
$$M21$$ = $$\begin{bmatrix} -2 & 3 \\ 5 & 2\end{bmatrix} = -4-15=-19$$
$$M22$$= $$\begin{bmatrix} 1 & 3 \\ -4& 2 \end{bmatrix} = 2+12=14$$
$$M23$$ = $$\begin{bmatrix} 1 & -2 \\ -4 & 5 \end{bmatrix} = 5-8=-3$$
$$M31$$ = $$\begin{bmatrix} -2& 3 \\ 2 & -1 \end{bmatrix} = 2-6=-4$$
$$M32$$ = $$\begin{bmatrix} 1 &3 \\ 0 & -1\end{bmatrix} = -1-0=-1$$
$$M33 $$= $$\begin{bmatrix} 1 & -2\\ 0 & 2 \end{bmatrix} = 2-0=2$$
substituting all the values in the Cofactor of the matrix formula we get,
= $$\begin{bmatrix} 9 & 4 & 8 \\ 19 & 14 & 3\\ -4 & 1 & 2\end{bmatrix}$$
the adjoint of the matrix is transpose of cofactor matrix
$$\therefore$$ Adjoint of the matrix is
$$\begin{bmatrix} 9 & 19 & -4 \\ 4 & 14& 1 \\ 8 & 3 & 2 \end{bmatrix}$$
Question 4
1 / -0

If $$A$$ and $$B$$ are square matrices of same orders then adj. $$(AB)$$ equals

A
adj $$A$$. adj $$B$$

B
adj $$B$$/adj $$A$$

C
adj $$A+$$ adj $$B$$

D
adj $$A-$$adj $$B$$

Solution

$$(AB)^{-1}=\displaystyle\frac{adj(AB)}{|AB|}$$
$$\Rightarrow B^{-1}A^{-1}=\displaystyle\frac{adj(AB)}{|AB|}$$
$$\Rightarrow \displaystyle\frac{adj B}{|B|}\frac{adj A}{|A|}=\frac{adj(AB)}{|A||B|}$$
$$\therefore adj(AB)=(adj B)(adj A)$$
Hence, option B.
Question 5
1 / -0

If $$A = (a_{ij})$$ is a $$4\times 4$$ matrix and $$C_{ij}$$ is the co-factor of the element $$a_{ij}$$ in Det (A), then the expression $$a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + a_{14}C_{14}$$ equals

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$Det. (A)$$

Solution
Question 6
1 / -0

Let $$A = [a_{ij}]_{n\times n}$$ be a square matirx and let $$c_{ij}$$ be cofactor of $$a_{ij}$$ in A. If $$C = [c_{ij}]$$, then

A
$$|C|=|A|$$

B
$$|C|=|A|^{n-1}$$

C
$$|C|=|A|^{n-2}$$

D
none of these

Solution

$$C\rightarrow$$ Cofactor matrix
$$AdjA= \left (C \right )^{^{T}}$$
But det of $$AdjA= Det \quad of \quad C$$
Because they are transpore of each other .
$$\Rightarrow\left | AdjA \right | = \left | C \right |= \left | A \right |^{n-1} $$
Option-B
Question 7
1 / -0

The minors and cofactors of -4 and 9 in determinant $$\begin{vmatrix} -1 & -2 & 3 \\ -4 & -5 & -6 \\ -7 & 8 & 9\end{vmatrix}$$ are respectively

A
$$42, 42 ; 3, 3$$

B
$$-42, 42 ; -3, -3$$

C
$$42, -42 ; 3, -3$$

D
$$42, 3 ; 42, 3$$

Solution

Minor of -4 is $$\begin{vmatrix} -2\quad & 3 \\ 8 & 9 \end{vmatrix}=-42$$
Cofactor of -4 is $${ \left( -1 \right) }^{ 1+2 }\left( 42 \right) =42$$

Minor of 9 is $$\begin{vmatrix} -1\quad & -2 \\ -4 & -5 \end{vmatrix}=-3$$
Cofactor of 9 is $${ \left( -1 \right) }^{ 3+3 }.\left( 3 \right) =-3$$
Question 8
1 / -0

If in the determinant $$\Delta=\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}, A_i, B_i, C_i$$ etc. be the co-factors of $$a_i, b_i, c_i$$ etc., then which of the following relations is incorrect?

A
$$a_1A_1 + b_1B_1 + c_1C_1=\Delta$$

B
$$a_2A_2 + b_2B_2 + c_2C_2=\Delta$$

C
$$a_3A_3 + b_3B_3 + c_3C_3=\Delta$$

D
$$a_1A_2 + b_1B_2 + c_1C_2=\Delta$$

Solution

$$\Delta=\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}, A_1, B_1, C_1$$ etc. be the co-factors of $$a_1, b_1, c_1$$ etc.
Sum of product of any row with its corresponding cofactors is Determinant.
$$\Rightarrow a_1A_1+b_1B_1+c_1C_1=a_2A_2+b_2B_2+c_2C_2=a_3A_3+b_3B_3+c_3C_3=\Delta $$
Sum of product of any row with another row's cofactors is $$0$$.
$$\Rightarrow a_1A_2+b_1B_2+c_1C_2=0$$
Hence, option D is incorrect.
Question 9
1 / -0

If $$A+B+C=\pi$$, then $$\begin{vmatrix}sin(A+B+C) & sin B & cos C\\ -sinB & 0 & tan A\\ cos(A+B) & -tanA & 0\end{vmatrix}$$ equals

A
0

B
2 sin B tan A cos C

C
1

D
none of these

Solution

$$\Delta =\begin{vmatrix} \sin { \left( A+B+C \right) } & \sin { B } & \cos { C } \\ -\sin { B } & 0 & \tan { A } \\ \cos { \left( A+B \right) } & -\tan { A } & 0 \end{vmatrix}$$

$$ =\begin{vmatrix} \sin { \left( \pi \right) } & \sin { B } & \cos { C } \\ -\sin { B } & 0 & \tan { A } \\ \cos { \left( \pi -C \right) } & -\tan { A } & 0 \end{vmatrix}$$

$$ =\begin{vmatrix} 0 & \sin { B } & \cos { C } \\ -\sin { B } & 0 & \tan { A } \\ -\cos { \left( C \right) } & -\tan { A } & 0 \end{vmatrix}$$

$$ =-\sin { B } \left( \tan { A } \cos { C } \right) +\cos { C } \left( \sin { B\tan { A } } \right) =0$$
Question 10
1 / -0

If $$f(x)=\begin{vmatrix} cos x & 1 & 0 \\ 1 & cos x & 1 \\ 0 & 1 & cos x\end{vmatrix}$$ the $$f'(\dfrac \pi 3)$$ equals

A
$$\dfrac {11\sqrt 3}{8}$$

B
$$\dfrac {5\sqrt 3}{8}$$

C
$$-\dfrac {5\sqrt 3}{8}$$

D
none of these

Solution

Given, $$f\left( x \right) =\begin{vmatrix} \cos { x } & 1 & 0 \\ 1 & \cos { x } & 1 \\ 0 & 1 & \cos { x } \end{vmatrix}\\ =\cos { x } \left( \cos ^{ 2 }{ x-1 } \right) -1\cos { x } \\ =\cos ^{ 3 }{ x } -\cos { x } -\cos { x } \\ =\cos ^{ 3 }{ x } -2\cos { x } $$
Now
$$f^{ ' }\left( x \right) =-3\cos ^{ 2 }{ x } \sin { x } +2\sin { x } $$
$$\displaystyle { f }^{ ' }\left( \frac { \pi }{ 3 } \right) =\frac { 5\sqrt { 3 } }{ 8 } $$

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Determinants Test - 39

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Determinants Test - 39

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Question 1

for no value of $$a,\ b,\ c$$

for all values of $$a,\ b,\ c$$

if $$a,\ \dfrac {c}{5},\ b$$ are in $$H.P$$.

if $$a,\ \dfrac {2}{5}c,\ b$$ are in $$H.P$$.

Solution

Question 2

A, B, C, D, E are collinear points

A, B, C, D, E are noncolinear points

AB & AC are parallel and AD & AE are perpendicular

None of these

Solution

Question 3

$$\begin{bmatrix} 9 & 19 & -4 \\ 4 & 14 & 1 \\ 8 & 3 & 2 \end{bmatrix}$$

$$\begin{bmatrix} 9 & 4 & 8 \\ 19 & 14 & 3 \\ -4 & 1 & 2 \end{bmatrix}$$

$$\begin{bmatrix} 9 & -19 & -4 \\ -4 & 14 & -1 \\ 8 & -3 & 2 \end{bmatrix}$$

none of these

Solution

Question 4

adj $$A$$. adj $$B$$

adj $$B$$/adj $$A$$

adj $$A+$$ adj $$B$$

adj $$A-$$adj $$B$$

Solution

Question 5

$$0$$

$$-1$$

$$1$$

$$Det. (A)$$

Solution

Question 6

$$|C|=|A|$$

$$|C|=|A|^{n-1}$$

$$|C|=|A|^{n-2}$$

none of these

Solution

Question 7

$$42, 42 ; 3, 3$$

$$-42, 42 ; -3, -3$$

$$42, -42 ; 3, -3$$

$$42, 3 ; 42, 3$$

Solution

Question 8

$$a_1A_1 + b_1B_1 + c_1C_1=\Delta$$

$$a_2A_2 + b_2B_2 + c_2C_2=\Delta$$

$$a_3A_3 + b_3B_3 + c_3C_3=\Delta$$

$$a_1A_2 + b_1B_2 + c_1C_2=\Delta$$

Solution

Question 9

0

2 sin B tan A cos C

1

none of these

Solution

Question 10

$$\dfrac {11\sqrt 3}{8}$$

$$\dfrac {5\sqrt 3}{8}$$

$$-\dfrac {5\sqrt 3}{8}$$

none of these

Solution

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