Determinants Test

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Determinants Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The value of the determinant $$\begin{vmatrix}a & b & 0\\ 0 & a & b\\ b & 0 &a \end{vmatrix}$$ is equal to

A
$$a^3-b^3$$

B
$$a^3+b^3$$

C
0

D
none of these

Solution

$$\begin{vmatrix} a & b & 0 \\ 0 & a & b \\ b & 0 & a \end{vmatrix}=a\left( { a }^{ 2 }-0 \right) -b\left( 0-{ b }^{ 2 } \right) +0={ a }^{ 3 }+{ b }^{ 3 }$$
Question 2
1 / -0

If $$A$$ is any skew-symmetric matrix of odd order then $$\left| A \right| $$ equals

A
$$-1$$

B
$$0$$

C
$$1$$

D
none of these

Solution

if $$A$$ is skew symmetric matrix
then $$A=-A^T$$
Therefore, $$|A|=-|A^T|=-|A|$$
$$\Rightarrow 2|A|=0$$
$$\Rightarrow|A|=0$$

Ans: B
Question 3
1 / -0

If $$\begin{vmatrix} 6i & -3i & 1\\ 4 & 3i & -1 \\ 20 & 3 & i\end{vmatrix}=x+iy$$, then

A
$$x = 3, y = 1$$

B
$$x = 1, y = 3$$

C
$$x = 0, y = 3$$

D
$$x = 0, y = 0$$

Solution

$$\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \end{vmatrix}=x+iy$$
$$ \Rightarrow 6i\left( 3{ i }^{ 2 }+3 \right) +3i\left( 4i+20 \right) +1\left( 12-60i \right) =x+iy$$ $$\Rightarrow 0=x+iy$$
Therefore, $$x=y=0$$

Ans: D
Question 4
1 / -0

If $$\begin{vmatrix} x & y\\ 4 & 2 \end{vmatrix}=7$$ and $$\begin{vmatrix} 2 & 3\\ y & x \end{vmatrix}=4$$ then

A
$$x=-3, y=-\dfrac {5}{2}$$

B
$$x=-\dfrac {5}{2}, y=-3$$

C
$$x=-3, y=\dfrac {5}{2}$$

D
$$x=-\dfrac {5}{2}, y=3$$

Solution

$$\begin{vmatrix} x & y\\ 4 & 2 \end{vmatrix}=7$$
$$\Rightarrow 2x-4y=7$$
$$\begin{vmatrix} 2 & 3\\ y & x \end{vmatrix}=4$$
$$\Rightarrow 2x-3y=4$$
Therefore, $$x=-5/2,y=-3$$

Ans: B
Question 5
1 / -0

If $$\begin{vmatrix}a & -b & -c\\-a & b & -c \\ -a & -b & -c\end{vmatrix}+\lambda abc=0$$, then $$\lambda$$ is equal to

A
$$2$$

B
$$4$$

C
$$-2$$

D
$$-4$$

Solution

$$\begin{vmatrix} a & -b & -c \\ -a & b & -c \\ -a & -b & -c \end{vmatrix}+\lambda abc=0$$

Substitute $$a=b=c=1$$

$$\begin{vmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & -1 \end{vmatrix}+\lambda =0$$

$$(-2-2)+\lambda=0$$

$$\Rightarrow \lambda =4$$
Question 6
1 / -0

The cofactors of elements in second row of the determinant $$\begin{vmatrix} 2 & -1 & 4 \\ 4 & 2 & -3 \\1 & 1 & 2 \end{vmatrix}$$ are

A
$$5, 6, 4$$

B
$$6, 0, -3$$

C
$$5, 1, 8$$

D
$$6, 0, 3$$

Solution

$$\begin{vmatrix} 2 & -1 & 4 \\ 4 & 2 & -3 \\ 1 & 1 & 2 \end{vmatrix}\\ { C }_{ 21 }=-\begin{vmatrix} -1 & 4 \\ 1 & 2 \end{vmatrix}=6\\ { C }_{ 22 }=\begin{vmatrix} 2 & 4 \\ 1 & 2 \end{vmatrix}=0\\ { C }_{ 23 }=-\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}=-3$$
Question 7
1 / -0

A set of points which do not lie on the same line are called as

A
collinear

B
non-collinear

C
concurrent

D
square

Solution

A set of points which do not lie on the same line are called as non collinear points
Question 8
1 / -0

$$A$$ and $$B$$ are two points and $$C$$ is any point collinear with $$A$$ and $$B$$. IF $$AB=10$$, $$BC=5$$, then $$AC$$ is equal to:

A
either $$15$$ or $$5$$

B
necessarily $$5$$

C
necessarily $$16$$

D
none of these

Solution

Since $$C$$ is collinear with $$A$$ and $$B, C$$ lies either
(i) to the left of point B or
(ii) to the right of point B
$$\therefore$$ In case (i) $$AC=AB-BC=10-5=5$$
In case (ii) $$AC=AB+BC=10+5=15$$
Question 9
1 / -0

If $$A=\begin{bmatrix} 3 & -5 \\ -1 & 0 \end{bmatrix}$$, then adj. $$A$$ is equal to

A
$$\begin{bmatrix} 3 & 5 \\ 1 & 0 \end{bmatrix}$$

B
$$\begin{bmatrix} 0 & -5 \\ -1 & 3 \end{bmatrix}$$

C
$$\begin{bmatrix} 0 & 5 \\ 1 & 3 \end{bmatrix}$$

D
$$\begin{bmatrix} 0 & -1 \\ -5 & 3 \end{bmatrix}$$

Solution

co-factor of $$A_{11}=0$$; co-factor of $$A_{12}=5$$; co-factor of $$A_{21}=1$$; co-factor of $$A_{22}=3$$
$$adjA=\begin{bmatrix} 0 & 5 \\ 1 & 3 \end{bmatrix}$$

Ans: C
Question 10
1 / -0

The value of the determinant $$\begin{vmatrix} 1 & 2 & 3\\ 3 & 5 & 7\\ 8 & 14 & 20\end{vmatrix}$$ is equal to

A
$$20$$

B
$$10$$

C
$$0$$

D
$$45$$

Solution

The value of the determinant is $$\begin{vmatrix} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{vmatrix}\\ =1\left( 100-98 \right) -2\left( 60-56 \right) +3\left( 42-40 \right)$$
$$ =2-8+6=0$$

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Determinants Test - 20

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Determinants Test - 20

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Question 1

$$a^3-b^3$$

$$a^3+b^3$$

0

none of these

Solution

Question 2

$$-1$$

$$0$$

$$1$$

none of these

Solution

Question 3

$$x = 3, y = 1$$

$$x = 1, y = 3$$

$$x = 0, y = 3$$

$$x = 0, y = 0$$

Solution

Question 4

$$x=-3, y=-\dfrac {5}{2}$$

$$x=-\dfrac {5}{2}, y=-3$$

$$x=-3, y=\dfrac {5}{2}$$

$$x=-\dfrac {5}{2}, y=3$$

Solution

Question 5

$$2$$

$$4$$

$$-2$$

$$-4$$

Solution

Question 6

$$5, 6, 4$$

$$6, 0, -3$$

$$5, 1, 8$$

$$6, 0, 3$$

Solution

Question 7

collinear

non-collinear

concurrent

square

Solution

Question 8

either $$15$$ or $$5$$

necessarily $$5$$

necessarily $$16$$

none of these

Solution

Question 9

$$\begin{bmatrix} 3 & 5 \\ 1 & 0 \end{bmatrix}$$

$$\begin{bmatrix} 0 & -5 \\ -1 & 3 \end{bmatrix}$$

$$\begin{bmatrix} 0 & 5 \\ 1 & 3 \end{bmatrix}$$

$$\begin{bmatrix} 0 & -1 \\ -5 & 3 \end{bmatrix}$$

Solution

Question 10

$$20$$

$$10$$

$$0$$

$$45$$

Solution

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