Determinants Test

Result

Determinants Test - 25

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The vectorial angle of a point $$P$$ on the line joining the points $$(r_{1}, \theta _{1})$$ and $$(r_{2},\theta _{2})$$ is $$\dfrac{\theta_{1} +\theta _{2}}{2}$$ then the length of radius vector of $$P$$ is

A
$$\displaystyle \frac{r_{1} - r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$

B
$$\displaystyle \frac{r_{1} + r_{2}}{r_{1} r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$

C
$$\displaystyle \frac{r_{1} r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$

D
None of these

Solution

Given points are $$A(r_{1},\theta_{1}),B(r_{2},\theta_{2})$$
Eq of line passing through two given points is given by:
$$y-y_{1}=\cfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$$
$$\therefore y-\theta_{1}=\cfrac{\theta_{2}-\theta_{1}}{r_{2}-r_{1}}(x-r_{1})$$
Now, angle between point P and line is $$\cfrac{\theta_{1}+\theta_{2}}{2}$$
Let point P meet at point Q on the line.
The radius will be $$R=PQ\cos\theta$$
$$\therefore R=PQ\cos(\cfrac{\theta_{1}+\theta_{2}}{2})$$
Question 2
1 / -0

$$\mathrm{D}\mathrm{e}\mathrm{t} \left\{\begin{array}{lll}
1^{2} & 2^{2} & 3^{2}\\
2^{2} & 3^{2} & 4^{2}\\
3^{2} & 4^{2} & 5^{2}
\end{array}\right\}=\ldots$$.

A
$$-8$$

B
$$-7$$

C
$$-6$$

D
$$-21/4$$

Solution

$$Det\begin{pmatrix}
1^2 & 2^2 & 3^2\\
2^2 & 3^2 & 4^2\\
3^2 & 4^2 & 5^2
\end{pmatrix}=det\begin{bmatrix}
1 & 4 & 9\\
4 & 9 & 16\\
9 & 16 & 25
\end{bmatrix}$$

$$= 1(25 \times 9 - 16^2)-4[4 \times 25 - 9 \times 16]+9[4 \times 16 - 9^2]$$

$$=225 - 256- 400 + 576 + 576-729$$

$$=-8$$
Question 3
1 / -0

$$\det \left[\begin{array}{lll}
18 & 40 & 89\\
40 & 89 & 198\\
89 & 198 & 440
\end{array}\right]=$$

A
$$-8$$

B
$$-6$$

C
$$-1$$

D
$$0$$

Solution

Det $$\displaystyle \begin{bmatrix} 18 & 40 & 89 \\ 40 & 89 & 198 \\ 89 & 198 & 440 \end{bmatrix}=\begin{vmatrix} 18 & 40 & 89 \\ 40 & 89 & 198 \\ 89 & 198 & 440 \end{vmatrix}$$

$$\displaystyle =18\left( 89\times 440-189\times 198 \right) $$ $$\displaystyle -40\left( 40\times 440-89\times 198 \right) $$ $$\displaystyle + 89\left( 40\times 198-89\times 89 \right) $$

$$\displaystyle =18\times 89\times 440-18\times 198\times 198$$ $$\displaystyle -{ \left( 40 \right) }^{ 2 }\times 440+40\times 89\times 198$$ $$\displaystyle +89\times 40\times 198-{ \left( 89 \right) }^{ 3 }=-1$$
Question 4
1 / -0

If the entries in a $$3\times 3$$ determinant are either $$0$$ or 1, then the greatest value of their determinats is:

A
1

B
2

C
3

D
9

Solution

Case I: If all the entries in one row are 0 , then the determinant is 0.
Case II: If all the entries in one row are 1 , then the maximum value of determinant is 1.
$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$$

Case II:If exactly one 1 is in one row, then the maximum value is 1.
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

Case III:If exactly two 1 is in one row, then the maximum value is 2.
$$\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}$$

Hence, the largest value of the determinant with entries either 0 or 1 is 2.
Question 5
1 / -0

The value of $$\left|\begin{array}{ll}
2+i & 2-i\\
1+i & 1-i
\end{array}\right|$$ is:

A
$$\mathrm{A}$$ complex quantity

B
real quantity

C
$$0$$

D
cannot be determined

Solution

Let $$A=\begin{vmatrix}
2+i & 2-i\\
1+i & 1-i
\end{vmatrix}$$

$$det A=(2+i)(1-i)-(2-i)(1+i)$$

$$=2-2i+i-i^2-(2+2i-i-i^2)$$

$$=3-i-(3+i)=-2i$$

$$\Rightarrow det A= -2i\Rightarrow A $$ complex quantity.
Question 6
1 / -0

$$\left|\begin{array}{lll}
1 & \omega & \omega^{2}\\
\omega & \omega^{2} & 1\\
\omega^{2} & 1 & \omega
\end{array}\right|=\ldots$$....(where $$\omega$$ is the cube root of unity)

A
$$0$$

B
$$1$$

C
2

D
$$-1$$

Solution

$$\Rightarrow w$$ is cube root of unity
$$\Rightarrow w^3-1=0$$

By operation of matrix, (5),
$$\begin{vmatrix}
1 & w & w^2\\
w & w^2 & 1\\
w^2 & 1 & w
\end{vmatrix}=1(w^3-1)-w(w^2-w^2)+w^2(w-w^4)$$ $$=2w^3-1-w^6=2-1-1=0$$
Question 7
1 / -0

If$$ A=\left\{\begin{array}{ll}
1 & 4\\
2 & 8
\end{array}\right\}$$ and $$B=\left\{\begin{array}{ll}
x & y\\
y & x
\end{array}\right\}$$ then the cofactor of $$\mathrm{a}_{21}$$ in $$\mathrm{A}\mathrm{B}$$ is:

A
$$-y-4x$$

B
$$y+4x$$

C
$$2x+8y$$

D
$$-2x-8y$$

Solution

Given $$A=\begin{pmatrix}
1 & 4\\
2 & 8
\end{pmatrix} $$and $$B=\begin{pmatrix}
x & y\\
y & x
\end{pmatrix}$$

So, By operation of multiplication,

$$AB=\begin{pmatrix}
x+4y & y+4x\\
2x+8y & 2y+8x
\end{pmatrix}$$

So, By property of minor (2),

$$M_{21}=$$minor of $$a_{21}=|2x + 8y|$$

So, cofactor of $$a_{21}=M_{21}(-1)^{2+1}=-M_{21}=-2x-8y$$
Question 8
1 / -0

lf $$A=\left\{\begin{array}{lll}
a & c & b\\
b & a & c\\
c & b & a
\end{array}\right\}$$ then the cofactor of $$a_{32}$$ in $$\mathrm{A}+\mathrm{A}^{\mathrm{T}}$$ is

A
$$-(2a(b+c)-(b+c)^{2})$$

B
$$ac-b^{2}$$

C
$$a^{2}-bc$$

D
$$2a(a+c)-(a+c)^{2}$$

Solution

By property of transpose, we get
$$A^{T}=\begin{bmatrix}
a & b & c\\
c & a & b\\
b & c & a
\end{bmatrix}$$
So, $$A+A^{T}=\begin{bmatrix}
2a & b+c & b+c\\
b+c & 2a & b+c\\
c+b & b+c & 2a
\end{bmatrix}$$
So, $$ M_{32}=$$ minor of $$a_{32}=\begin{vmatrix}
2a & b+c\\
b+c & c+c
\end{vmatrix}=2a(b+c)-(b+c)^2$$
$$=2ab+2ac-b^2-c^2-2bc$$
So, cofactor of $$a_{32}=M_{32}(-1)^{3+2}$$
$$=-M_{32}$$
$$=-[2a(b+c)-(b+c)^2]$$
Question 9
1 / -0

If $$A=\left[\begin{array}{lll}
1^{2} & 2^{2} & 3^{2}\\
2^{2} & 3^{2} & 4^{2}\\
3^{2} & 4^{2} & 5^{2}
\end{array}\right]$$, then the minor of $$\mathrm{a}_{22}$$ is

A
$$-56$$

B
$$51$$

C
$$-43$$

D
$$41$$

Solution

Given, $$A=\begin{bmatrix}
1^2 & 2^2 & 3^2\\
2^2 & 3^2 & 4^2\\
3^2 & 4^2 & 5^2
\end{bmatrix}$$
So, By property of minor (1),
minor of $$a_{22}=M_{22}(-1)^{2+2}=M_{22}$$
$$=\begin{vmatrix}
1^2 & 3^2\\
3^2 & 5^2
\end{vmatrix}$$
$$=25 - 9 \times 9$$
$$=-56$$
Question 10
1 / -0

$$\left|\begin{array}{lll}
\mathrm{a}+\mathrm{b} & \mathrm{a} & \mathrm{b}\\
\mathrm{a} & \mathrm{a}+\mathrm{c} & \mathrm{c}\\
\mathrm{b} & \mathrm{c} & \mathrm{b}+\mathrm{c}
\end{array}\right|=$$

A
4 abc

B
abc

C
$$2\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$$

D
$$4\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$$

Solution

$$\begin{vmatrix}
a+b & a & b\\
a & a+c & c\\
b & c & b+c
\end{vmatrix}$$
$$=(a+b)[(a+c)(b+c)-c^2]-a[ab+ac-bc]+b[ac-ab-bc]$$
$$=(a+b)[ab+ac+cb]-a^2b-a^2c+abc+abc-ab^2-b^2c$$
$$=a^2b+a^2c+abc+ab^2+abc+cb^2+abc+cb^2-a^2b-a^2c+abc+abc-ab^2-b^2c$$
$$=4abc$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Determinants Test - 25

Result

Determinants Test - 25

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle \frac{r_{1} - r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$

$$\displaystyle \frac{r_{1} + r_{2}}{r_{1} r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$

$$\displaystyle \frac{r_{1} r_{2}}{r_{1} + r_{2}} \cos (\frac{\theta _{1} - \theta _{2}}{2})$$

None of these

Solution

Question 2

$$-8$$

$$-7$$

$$-6$$

$$-21/4$$

Solution

Question 3

$$-8$$

$$-6$$

$$-1$$

$$0$$

Solution

Question 4

1

2

3

9

Solution

Question 5

$$\mathrm{A}$$ complex quantity

real quantity

$$0$$

cannot be determined

Solution

Question 6

$$0$$

$$1$$

2

$$-1$$

Solution

Question 7

$$-y-4x$$

$$y+4x$$

$$2x+8y$$

$$-2x-8y$$

Solution

Question 8

$$-(2a(b+c)-(b+c)^{2})$$

$$ac-b^{2}$$

$$a^{2}-bc$$

$$2a(a+c)-(a+c)^{2}$$

Solution

Question 9

$$-56$$

$$51$$

$$-43$$

$$41$$

Solution

Question 10

4 abc

abc

$$2\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$$

$$4\mathrm{a}^{2}\mathrm{b}^{2}\mathrm{c}^{2}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.