Determinants Test

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Determinants Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

$$\left| \begin{matrix} 1+i & 1-i & i \\ 1+i & i & 1+i \\ i & 1+i & 1-i \end{matrix} \right| $$ (where $$i=\sqrt {-1}$$) equals.

A
$$5i-2$$

B
$$7 4i$$

C
$$4 7i$$

D
$$48i$$

Solution

$${ \begin{vmatrix} 1+i & 1-i & i \\ 1+i & i & 1+i \\ i & 1+i & 1-i \end{vmatrix}}$$ = $$(1+i)[i(1-i) - (1+i)^{2}] - (1+i)[(1-i)^{2} - i(1+i)] + i[(1-i)(1+i) - i^{2}]$$
=$$(1+i)[i-i^{2}-1-i^{2}-2i] - (1+i)[1+i^{2}-2i-i-i^{2}] + i[1-i^{2}-i^{2}]$$
=$$(1+i)(1-i) - (1+i)(1-3i) + 3i$$
=$$1-i^{2}-1(1-3i)-i(1-3i)+3i$$
=$$1+1-1+3i-i+3i^{2}+3i$$
=$$1-3+6i$$
=$$5i-2$$
Question 2
1 / -0

If $$\displaystyle{\left| {_2^{4\,}\,\,_1^1} \right|^2} = \left| {_1^3\,\,_x^2} \right| - \left| {_{ - 2}^x\,\,_1^3} \right|,$$ then $$x$$=

A
6

B
7

C
8

D
16

Solution

$${ \begin{vmatrix} 4 & 1 \\ 2 & 1 \end{vmatrix} }^{ 2 }=\begin{vmatrix} 3 & 2 \\ 1 & x \end{vmatrix}-\begin{vmatrix} x & 3 \\ -2 & 1 \end{vmatrix}$$
$$(4-2)^2=(3x-2)-(x+6)$$
$$4=(3x-x)-8$$
$$12=2x$$
$$\Rightarrow x=6$$
Question 3
1 / -0

`If $$(8,1),(k,-4),(2,-5)$$ are collinaer, then $$k=$$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let the given point be $$A(8,1),B(K, - 4),C(2, - 5)$$
if the above point are collinear,
they will lie on same line
i.e. they will not form triangle,
Area of $$\Delta$$ABC$$=0$$
$$ \Rightarrow \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0$$
Here,
$$\begin{array}{l} { x_{ 1 } }=8,{ y_{ 1 } }=1 \\ { x_{ 2 } }=k,{ y_{ 2 } }=-4 \\ { x_{ 3 } }=2,{ y_{ 3 } }=-5 \end{array}$$
putting value,
$$\begin{array}{l} \Rightarrow \dfrac { 1 }{ 2 } \left[ { 8\left( { -4+5 } \right) +k\left( { -5-1 } \right) +2\left( { 1-4 } \right) } \right] =0 \\ \Rightarrow 8\left( { -4+5 } \right) +k\left( { -5-1 } \right) +2\left( { 1+4 } \right) =0\times 2 \\ \Rightarrow -32+40-5k-k+2+8=0 \\ \Rightarrow 8+10-6k=0 \\ \Rightarrow -6k=-18 \\ \Rightarrow 6k=18 \\ \Rightarrow k=\dfrac { { 18 } }{ 6 } \\ \Rightarrow k=3 \end{array}$$
hence,we get
$$k=3$$
so, option $$C$$ is correct.
Question 4
1 / -0

If $$A=\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right] $$ then determinant of $$[A]$$ is

A
$$1$$

B
$$-1$$

C
$$0 $$

D
$$2$$

Solution

Determinants of a $$2\times 2$$ matrix is given by:

If $$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

$$Det(A)=ad-bc$$

Hence using this above property:

If $$A=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$, then

$$Det(A)=0\times0-1\times (-1)=0+1=1$$
Question 5
1 / -0

$$\left| \begin{matrix} 1& a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{matrix} \right| =$$

A
$$(a-b)(b-c)(c-a)$$

B
$$(a+b)(c-a)$$

C
$$(a+b+c)^2$$

D
$$2(a+b+c)^2$$

Solution

$$\begin{array}{l} Let\, \, the\, \, given\, \quad determinant\, be\, \Delta .Then \\ \Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 1 & b & { { b^{ 2 } } } \\ 1 & c & { { c^{ 2 } } } \end{array} } \right| \\ =\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 0 & { b-a } & { { b^{ 2 } }-{ a^{ 2 } } } \\ 0 & { c-a } & { { c^{ 2 } }-{ a^{ 2 } } } \end{array} } \right| \, \, \left[ { applying\, \, { R_{ 2 } }\to \left( { { R_{ 2 } }-{ R_{ 1 } } } \right) and\, { R_{ 3 } }\to \left( { { R_{ 3 } }-{ R_{ 1 } } } \right) } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) .\left| { \begin{array} { *{ 20 }{ c } }1 & a & { { a^{ 2 } } } \\ 0 & 1 & { b-a } \\ 0 & 1 & { c+a } \end{array} } \right| \\ \left[ { taking\, \, \left( { b-a } \right) common\, \, from\, \, { R_{ 2 } }\, \, and\, \, \left( { c-a } \right) common\, \, from\, \, { R_{ 3 } } } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) \times 1.\left| { \begin{array} { *{ 20 }{ c } }1 & { b+a } \\ 1 & { c+a } \end{array} } \right| \, \, \left[ { anded\, \, by\, \, { C_{ 1 } } } \right] \\ =\left( { b-a } \right) \left( { c-a } \right) \left\{ { \left( { c+a } \right) -\left( { b+a } \right) } \right\} \\ =\left( { b-a } \right) \left( { c-a } \right) \left( { c-b } \right) =\left( { a-b } \right) \left( { b-c } \right) \left( { c-a } \right) . \\ Hence, \\ \Delta =\left( { a-b } \right) \left( { b-c } \right) \left( { c-a } \right) . \end{array}$$
So, option $$A$$ is correct answer.
Question 6
1 / -0

If the points $$(a, 1), (2, -1)$$ and $$\left(\dfrac{1}{2}, 2\right)$$ are collinear, then $$a$$ is equal to:

A
$$1$$

B
$$0$$

C
$$2$$

D
$$\dfrac{1}{4}$$

Solution

Slope of $$AB=$$ slope of $$BC$$
$$ \Rightarrow \dfrac{{ - 1 - 1}}{{2 - a}} = \dfrac{{2 - \left( { - 1} \right)}}{{\dfrac{1}{2} - 2}}$$
$$ \Rightarrow \dfrac{{ - 2}}{{2 - a}} = \dfrac{{3 \times 2}}{{ - 3}}$$
$$ \Rightarrow 6 = 12 - 6a$$
$$ \Rightarrow 6a = 6$$
$$a=1$$

Hence, option $$A$$ is correct answer.
Question 7
1 / -0

If $$A=\begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix}$$, the determinant of matrix $$A$$ is

A
$$13$$

B
$$12$$

C
$$17$$

D
$$-13$$

Solution

Determinants of a $$2\times 2$$ matrix is given by:

If $$A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

$$Det(A)=ad-bc$$

Hence using this above property:

If $$A=\begin{bmatrix} 5 & 1 \\ 2 & 3 \end{bmatrix}$$, then

$$Det(A)=5\times3-1\times 2=15-2=13$$
Question 8
1 / -0

Find the determinant:
$$\begin{vmatrix} 1 & 2 & 1 \\ 2 & 2 & 2 \\ 3 & 1 & 4 \end{vmatrix}$$

A
$$-2$$

B
$$0$$

C
$$3$$

D
$$1$$

Solution

$$\begin{array}{l} Let\, \, the\, \, er\min ant\, \, be\, \Delta \\ Now, \\ \Delta =\left| { \begin{array} { *{ 20 }{ c } }1 & 2 & 1 \\ 2 & 2 & 2 \\ 3 & 1 & 4 \end{array} } \right| \\ \Delta =1\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 1 & 4 \end{array} } \right| -2\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 3 & 4 \end{array} } \right| +1\left| { \begin{array} { *{ 20 }{ c } }2 & 2 \\ 3 & 1 \end{array} } \right| \\ \Delta =1\left( { 8-2 } \right) -2\left( { 8-6 } \right) +1\left( { 2-6 } \right) \\ \Delta =6-4-4 \\ \Delta =-2 \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$$
Question 9
1 / -0

What is the value of the determinant
$$\begin{vmatrix} 1!& 2! & 3!\\ 2! & 3! & 4! \\ 3!& 4!& 5!\end{vmatrix}$$$$?$$

A
$$0$$

B
$$12$$

C
$$24$$

D
$$36$$

Solution

Given,

$$\begin{pmatrix}1!&2!&3!\\ 2!&3!&4!\\ 3!&4!&5!\end{pmatrix}$$

$$=1!\cdot \det \begin{pmatrix}3!&4!\\ 4!&5!\end{pmatrix}-2!\cdot \det \begin{pmatrix}2!&4!\\ 3!&5!\end{pmatrix}+3!\cdot \det \begin{pmatrix}2!&3!\\ 3!&4!\end{pmatrix}$$

$$=1!\cdot \:144-2!\cdot \:96+3!\cdot \:12$$

$$=144-192+72$$

$$=24$$
Question 10
1 / -0

lf $$\mathrm{A}=\left[\begin{array}{lll}
1 & 5 & -6\\
-8 & 0 & 4\\
3 & -7 & 2
\end{array}\right]$$, then the cofactor of -7=......

A
44

B
43

C
40

D
39

Solution

By property of cofactor of matrix,
$$A=\begin{bmatrix}
1 & 5 & -6\\
-8 & 0 & 4\\
3 & -7 & 2
\end{bmatrix}$$
minor of $$-7=\begin{vmatrix}
1 & -6\\
-8 & 4
\end{vmatrix}=4-48=-44$$
So, cofactor of (-7)=minor of $$(-7) \times (-1)^{3+2}=-[minor \;of \; (-7)]$$
$$=-(-44)$$
$$=44$$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 24

Result

Determinants Test - 24

Score

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SHARING IS CARING

Question 1

$$5i-2$$

$$7 4i$$

$$4 7i$$

$$48i$$

Solution

Question 2

6

7

8

16

Solution

Question 3

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

$$1$$

$$-1$$

$$0 $$

$$2$$

Solution

Question 5

$$(a-b)(b-c)(c-a)$$

$$(a+b)(c-a)$$

$$(a+b+c)^2$$

$$2(a+b+c)^2$$

Solution

Question 6

$$1$$

$$0$$

$$2$$

$$\dfrac{1}{4}$$

Solution

Question 7

$$13$$

$$12$$

$$17$$

$$-13$$

Solution

Question 8

$$-2$$

$$0$$

$$3$$

$$1$$

Solution

Question 9

$$0$$

$$12$$

$$24$$

$$36$$

Solution

Question 10

44

43

40

39

Solution

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