Determinants Test

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Determinants Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

If two rows of a determinant are identical, then what is the value of the determinant ?

A
0

B
1

C
-1

D
Can be any real value.

Solution

Let determinant of this matrix is $$x$$, if we interchange the two identical rows of the matrix then by property the determinant of the new matrix is $$-x$$, but overall the matrix will be same as we have interchanged only the two identical rows.
So, $$x=-x$$, we have $$x=0$$ .
Hence, the determinant is zero.
Question 2
1 / -0

The value of k for which $$kx+3y-k+3=0$$ and $$12x+ky=k$$, have infinite solutions, is?

A
$$0$$

B
$$-6$$

C
$$6$$

D
$$1$$

Solution

For infinite many solution $$\displaystyle\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$
$$\Rightarrow \displaystyle\dfrac{k}{12}=\dfrac{3}{k}=\dfrac{-(k-3)}{-k}, \dfrac{3}{k}=\dfrac{(k-3)}{k}$$
$$k^2=36$$ $$k-6=0$$
$$k=\pm 6$$ $$k=6$$.
Question 3
1 / -0

For positive numbers $$x, y$$ and $$z$$ the numerical value of the determinant $$\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}$$ is

A
$$0$$

B
$$1$$

C
$$\log_e xyz$$

D
$$-\log_{e} xyz$$

Solution

Let $$D=\begin{vmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 1 & \log_y z \\ \log_z x & \log_z y & 1 \end{vmatrix}$$

$$=1(1\times 1-\log_yz\times \log_zy)-\log_xy(\log_yx\times 1-\log_zx\times \log_yz)+\log_xz(\log_yx\times \log_zy-1\times \log_zx)$$

Since, $$\log_ab=\dfrac{\log a}{\log b}$$
So simplifying above further we get
$$\log_yz\times \log_zy=\dfrac{\log z}{\log y}\times \dfrac{\log y}{\log z}=1$$

$$\Rightarrow D=1(1-1)-\log_xy(\log_yx-\log_yx)+\log_xz(\log_zx-\log_zx)$$
$$\Rightarrow D=0$$
Question 4
1 / -0

The number of line segments possible with three collinear points is ________.

A
$$1$$

B
$$2$$

C
$$3$$

D
Infinite

Solution

Let three collinear points be $$ A , B , C$$
They can represent three line segments namely, $$AB, BC, AC$$.
Thus namely $$3$$ line segments are possible with three collinear points.
Question 5
1 / -0

If any two adjacent rows or columns of a determinant are interchanged in position, the value of the determinant :

A
Becomes zero

B
Remains the same

C
Changes its sign

D
Is doubled

Solution

If two rows or columns are interchanged
then the value of determinant changes its sign
$$A=\begin{vmatrix} a & b \\ c & d \end{vmatrix}$$
Interchanging column
$$B=\begin{vmatrix} b & a \\ d & c \end{vmatrix}\\ A=-B$$
Similarly, interchanging rows will give
$$C=\begin{vmatrix} c & d \\ a & b \end{vmatrix}\\ A=-C$$
Hence, C is correct.
Question 6
1 / -0

If $$a, b, c$$ are non-zero and different from $$1$$, then the value of $$\begin{vmatrix}\log_a 1 & \log_a b & \log_ac\\ \log_a \left( \dfrac{1}{b} \right ) & \log_b 1 &\log_a \left( \dfrac{1}{c} \right ) \\ \log_a \left( \dfrac{1}{c}\right ) & \log_a c & \log_c 1\end{vmatrix}$$ is

A
$$0$$

B
$$1 + \log_a (a +b + c)$$

C
$$\log_a (ab + bc + ca)$$

D
$$1$$

E
$$\log_a(a + b + c)$$

Solution

$$\Delta =\left| \begin{matrix} \log _{ a }{ 1 } & \log _{ a }{ b } & \log _{ a }{ c } \\ \log _{ a }{ \dfrac { 1 }{ b } } & \log _{ b }{ 1 } & \log _{ a }{ \dfrac { 1 }{ c } } \\ \log _{ a }{ \dfrac { 1 }{ c } } & \log _{ a }{ c } & \log _{ c }{ 1 } \end{matrix} \right| \\ \Rightarrow \Delta =\left| \begin{matrix} 0 & \log _{ a }{ b } & \log _{ a }{ c } \\ -\log _{ a }{ b } & 0 & -\log _{ a }{ c } \\ -\log _{ a }{ c } & \log _{ a }{ c } & 0 \end{matrix} \right| \\ \Rightarrow \Delta =0-\log _{ a }{ b } \{ 0-{ (\log _{ a }{ c) } }^{ 2 }\} +\log _{ a }{ c } \{ -\log _{ a }{ b } \log _{ a }{ c } -0\} \\ \Rightarrow \Delta =\log _{ a }{ b } { (\log _{ a }{ c) } }^{ 2 }-\log _{ a }{ b } { (\log _{ a }{ c) } }^{ 2 }=0$$
So, option A is correct.
Question 7
1 / -0

The points (2, -3), (4,3) and (5, k/2) are on the same straight line. The value(s) of k is (are):

A
$$12$$

B
$$-12$$

C
$$\pm 12$$

D
$$12$$ or $$6$$

Solution

Given (2, -3), (4, 3) and (5, $$\cfrac{k}{2}$$) are on same straight line.
$$\therefore \cfrac { 3-(-3) }{ 4-2 } =\cfrac { \cfrac { k }{ 2 } -3 }{ 5-4 } \Longrightarrow \cfrac { 6 }{ 2 } =\cfrac { \cfrac { k }{ 2 } -3 }{ 1 } \Longrightarrow 3=\cfrac { k }{ 2 } -3\\ \therefore k=12$$
Question 8
1 / -0

Let a be the square matrix of order 2 such that $$A^2 - 4A + 4I =0$$ where I is an identify matrix of order 2. .If$$ B = A ^5 - 4A^4 + 6 A^3 + 4A^2 + A $$ then Det (B) is equal to

A
162

B
$$(162)^2$$

C
256

D
$$(256)^2$$

Solution

$$A^2 - 4A + 4I = O \Rightarrow ( A- 2 I)^2 = O \Rightarrow A = 2 I$$
$$ B = A (A^4 \, + \, 4A^3\, + \, 6A^2 \, + \, 4A +I) = A(A \, + \, I)^4 = 2I (3I)^4=162 I$$
det$$(B) = (162)^2$$
Question 9
1 / -0

If $$P=\begin{bmatrix} 1 & c & 3\\ 1 & 3 & 3\\ 2 & 4 & 4\end{bmatrix}$$ is the adjoint of a $$3\times 3$$ matrix Q and det.(Q)$$=4$$, then c is equal to.

A
$$0$$

B
$$4$$

C
$$5$$

D
$$11$$

Solution

P$$=$$adj. Q
$$\Rightarrow |P|=|adj. Q|=|Q|^2$$
$$\therefore |P|=(4)^2\Rightarrow |P|=16$$
$$\Rightarrow \begin{vmatrix} 1 & c & 3\\ 1 & 3 & 3\\ 2 & 4 & 4\end{vmatrix}=16$$
$$\Rightarrow 1(0)-c(-2)+3(-2)=16\Rightarrow 2c=22$$
$$\Rightarrow c=11$$.
Question 10
1 / -0

If $$A$$ is a skew symmetric matrix, then $$\left| A \right| $$ is

A
$$1$$

B
$$-1$$

C
$$0$$

D
none

Solution

SINCE THE SKEW SYMMETRIC MATRIX CONSIST OF ELEMENTS OF OPPOSITE SIGN AT OPPOSITE SIDE OF MATRIX DIAGONAL WITH ALL THE DIAGONAL ELEMENTS AS ZERO THEREFORE THE DETERMINANT OF SKEW SYMMETRIC MATRIX IS ZERO.

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Re-Attempt Weekly Quiz Competition

Determinants Test - 22

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Determinants Test - 22

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Question 1

0

1

-1

Can be any real value.

Solution

Question 2

$$0$$

$$-6$$

$$6$$

$$1$$

Solution

Question 3

$$0$$

$$1$$

$$\log_e xyz$$

$$-\log_{e} xyz$$

Solution

Question 4

$$1$$

$$2$$

$$3$$

Infinite

Solution

Question 5

Becomes zero

Remains the same

Changes its sign

Is doubled

Solution

Question 6

$$0$$

$$1 + \log_a (a +b + c)$$

$$\log_a (ab + bc + ca)$$

$$1$$

$$\log_a(a + b + c)$$

Solution

Question 7

$$12$$

$$-12$$

$$\pm 12$$

$$12$$ or $$6$$

Solution

Question 8

162

$$(162)^2$$

256

$$(256)^2$$

Solution

Question 9

$$0$$

$$4$$

$$5$$

$$11$$

Solution

Question 10

$$1$$

$$-1$$

$$0$$

none

Solution

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