Determinants Test

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Determinants Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The value of the determinant $$\displaystyle \left | \begin{matrix}
1 &\omega ^{3} &\omega ^{5} \\
\omega ^{3}&1 &\omega ^{4} \\
\omega ^{5}&\omega ^{4} &1
\end{matrix} \right |$$ , where $$\omega$$ is an imaginary cube root of unity,is

A
$$\displaystyle (1-\omega )^{2} $$

B
3

C
-3

D
none of these

Solution

$$\left| \begin{matrix} 1 & \omega ^{ 3 } & \omega ^{ 5 } \\ \omega ^{ 3 } & 1 & \omega ^{ 4 } \\ \omega ^{ 5 } & \omega ^{ 4 } & 1 \end{matrix} \right| =\left| \begin{matrix} 1 & 1 & \omega ^{ 2 } \\ 1 & 1 & \omega \\ \omega ^{ 2 } & \omega & 1 \end{matrix} \right| \\ \Rightarrow \omega ^{ 2 }-2\omega +1$$ $$(\because \omega^{3}=1\quad and\quad 1+\omega+\omega^{2}=0)$$
$$ \Rightarrow (\omega-1)^2$$
Question 2
1 / -0

Let $$\displaystyle A=\left [ \begin{matrix}1 &0 &0 \\ 2 &1 &0 \\ 3 &2 &1 \end{matrix} \right ]$$ and $$\displaystyle U_{1}, U_{2}, U_{3}$$ be column
matrices satisfying $$\displaystyle AU_{1}=\left [ \begin{matrix}1\\ 0\\ 0\end{matrix} \right ], AU_{2}=\left [ \begin{matrix}2\\ 3\\ 0\end{matrix} \right ], AU_{3}=\left [ \begin{matrix}2\\ 3\\ 1\end{matrix} \right ]$$. If U is
$$\displaystyle 3\times 3$$ matrix whose columns are $$\displaystyle U_{1}, U_{2}, U_{3}$$, then $$\displaystyle \left | U \right |=$$

A
$$\displaystyle 3$$

B
$$\displaystyle -3$$

C
$$\displaystyle\dfrac{3}{2}$$

D
$$\displaystyle 2$$

Solution

$$A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$$
Let $$U_1=\begin{bmatrix} a_{ 1 } \\ b_{ 1 } \\ c_{ 1 } \end{bmatrix}$$,$$U_2=\begin{bmatrix} a_{ 2 } \\ b_{ 2 } \\ c_{ 2 } \end{bmatrix}$$ and $$U_3=\begin{bmatrix} a_{ 3 } \\ b_{ 3 } \\ c_{ 3 } \end{bmatrix}$$
Given $$AU_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$
$$\Rightarrow a_1=1; 2a_1+b_1=0; 3a_1+2b_1+c_1=0$$
simplifying gives $$a_1=1; b_1=-2; c_1=1$$
$$AU_2=\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$$
$$\Rightarrow a_2=2; 2a_2+b_2=3; 3a_2+2b_2+c_2=0$$
simplifying gives $$a_2=2; b_2=-1; c_2=-4$$
$$AU_3=\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$$
$$\Rightarrow a_3=2; 2a_3+b_3=3; 3a_3+2b_3+c_3=1$$
simplifying gives $$a_3=2; b_3=-1; c_3=-3$$
$$\therefore U=\begin{bmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{bmatrix}$$
$$|U|=\begin{vmatrix} 1 & 2 & 2 \\ -2 & -1 & -1 \\ 1 & -4 & -3 \end{vmatrix}=3$$
Hence, option A.
Question 3
1 / -0

If the lines $$L_{1}:\lambda ^{2}x-y-1=0$$ $$L_{2}:x-\lambda ^{2}y+1=0$$ $$L_{3}:x+y-\lambda ^{2}=0$$ pass through the same point the value(s) of $$\lambda$$ equals

A
$$1$$

B
$$\sqrt{2}$$

C
$$2$$

D
$$0$$

Solution

The given equations passes through same point.So, they are concurrent lines

$$\Rightarrow \left| \begin{matrix} { \lambda }^{ 2 } & -1 & -1 \\ 1 & -{ \lambda }^{ 2 } & 1 \\ 1 & 1 & -{ \lambda }^{ 2 } \end{matrix} \right| =0$$

$$\Rightarrow { \lambda }^{ 6 }-3{ \lambda }^{ 2 }-2=0$$
By doing synthetic division we get,
$$(\lambda^4-2\lambda^2+1)(\lambda^2-2)=0$$
$$(\lambda^2-1)^2(\lambda^2-2)=0$$
$$\lambda=\pm \sqrt 2 or \lambda=\pm1$$
But here $$\lambda=\pm1\ does\ not\ satisfies\ ,hence\ \lambda=\sqrt2$$
Option B satisfies above equation
Question 4
1 / -0

If A is a square matrix such that $$\displaystyle \left | \begin{matrix}4 &0 &0 \\0 &4 &0 \\0 &0 &4\end{matrix} \right |$$=

A
$$4$$

B
$$16$$

C
$$64$$

D
$$256$$

Solution

We know that the determinant of a matrix $$A=\left[ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right] $$ is:

$$\left| { A } \right| =a(ei-fh)-b(di-fg)+c(dh-eg)$$

Here, the given matrix is $$A=\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right]$$, so, let us find the determinant of matrix $$A$$ as shown below:

$$\left| { A } \right| =4[(4\times 4)-(0\times 0)]-0[(0\times 4)-(0\times 0)]+0[(0\times 0)-(0\times 4)]=4(16-0)-0(0-0)+0(0-0)$$
$$=4\times 16=64$$

Hence, the determinant of the matrix $$\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right]$$ is $$64$$.
Question 5
1 / -0

If $$\displaystyle \omega$$ is an imaginary cube root of unity,then the value of
$$\left | \begin{matrix}
a &b\omega ^{2} & a\omega \\
b\omega & c &b\omega ^{2} \\
c\omega ^{2}&a\omega &c
\end{matrix} \right |$$,is ?

A
$$\displaystyle a^{3}+b^{3}+c^{3}$$

B
$$\displaystyle a^{2}b-b^{2}c$$

C
0

D
$$\displaystyle a^{3}b+b^{3}+3abc $$

Solution

$$\triangle =\begin{vmatrix} a & b{ \omega }^{ 2 } & a\omega \\ b\omega & c & b{ \omega }^{ 2 } \\ c{ \omega }^{ 2 } & a{ \omega } & c \end{vmatrix}\quad \quad 1+\omega +{ \omega }^{ 2 }=0;{ \omega }^{ 3 }=1;{ \omega }^{ 4 }=\omega $$

$$\Rightarrow a\left( { c }^{ 2 }-ab{ \omega }^{ 3 } \right) -b{ \omega }^{ 2 }\left( bc{ \omega }-bc\omega^4 \right) +a\omega \left( ab{ \omega }^{ 2 }-{ c }^{ 2 }{ \omega }^{ 2 } \right) $$

$$\Rightarrow a\left( { c }^{ 2 }-ab{ \omega }^{ 3 } \right) -b{ \omega }^{ 2 }\left( bc{ \omega }-bc\omega \right) +a\left( ab-{ c }^{ 2 } \right) $$

$$=a\left( { c }^{ 2 }-ab \right) -a\left( c^2-ab \right) =0$$
$$\therefore \triangle =0$$
Option C
Question 6
1 / -0

If $$\displaystyle a\neq b\neq c,$$ are value of x which satisfies the equation $$\displaystyle \left | \begin{matrix}0 &x-a &x-b \\ x+a &0 &x-c \\ x+b &x+c &0 \end{matrix} \right |=0$$ is given by

A
$$x=0$$

B
$$x=c$$

C
$$x=b$$

D
$$x=a$$

Solution

$$\displaystyle \left | \begin{matrix}0 &x-a &x-b \\ x+a &0 &x-c \\ x+b &x+c &0 \end{matrix} \right |=0$$

$$\Rightarrow (x-a)(x+b)(x-c)+(x-b)(x+a)(x+c)=0$$

We can now check by options
$$for \,\,option\,\, A, $$ put $$x=0$$ in above equation we get,

$$\Rightarrow (0-a)(0+b)(0-c)+(0-b)(0+a)(0+c)=0$$

$$\Rightarrow abc-abc=0$$

So, $$x=0$$ satisfies the equation.
Question 7
1 / -0

If $$A\displaystyle= \left | \begin{matrix}a &b &c \\ x &y &z \\ p &q &r \end{matrix} \right |$$ and $$B=\left | \begin{matrix}q &-b &y \\ -p&a &-x \\ r&-c &z \end{matrix} \right |$$, then

A
$$A= 2B$$

B
$$A= B$$

C
$$A= -B$$

D
none of these

Solution

$$B=\begin{vmatrix}q &-b &y \\ -p&a &-x \\ r&-c &z \end{vmatrix}$$

$$=-\begin{vmatrix}q &b &y \\ -p&-a &-x \\ r&c &z \end{vmatrix}$$$$=\begin{vmatrix}q &b &y \\ p&a &x \\ r&c &z \end{vmatrix}$$

$$=-\begin{vmatrix}p &a &x \\ q&b &y \\ r&c &z \end{vmatrix}$$$$=-\begin{vmatrix}a &x &p \\ b&y &q \\ c&z &r \end{vmatrix}$$

$$=-\begin{vmatrix}a &b &c \\ x &y &z \\ p &q &r \end{vmatrix}=-A$$ ($$\because |A|=|A^{T}|$$)

Hence, option C.
Question 8
1 / -0

Number of values of $$a$$ for which the lines $$2x+y-1=0, ax+3y-3=0, 3x+2y-2=0$$ are concurrent is

A
0

B
1

C
2

D
$$\infty$$

Solution

Here coefficient matrix, $$\Delta = \begin{vmatrix} 2 & 1 & -1 \\ a & 3 & -3 \\ 3 & 2 & -2 \end{vmatrix}$$
Using $$C_2 \to C_2+C_3$$
$$\Delta = \begin{vmatrix} 2 & 0 & -1 \\ a & 0 & 3 \\ 3 & 0 & -2 \end{vmatrix}=0$$
Clearly $$\Delta=0$$, Hence given lines are concurrent for all values of $$a$$
Question 9
1 / -0

The value of $$\displaystyle \left | \begin{matrix}
11 & 12 &13 \\
12&13 &14 \\
13&14 &15
\end{matrix} \right |$$,is

A
1

B
0

C
-1

D
67

Solution

$$\left| \begin{matrix} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{matrix} \right| $$
$${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }$$
$$=\left| \begin{matrix} 11 & 12 & 13 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{matrix} \right| $$
$$=2\left| \begin{matrix} 11 & 12 & 13 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right| $$
$$=0$$
Question 10
1 / -0

Let $$\begin{vmatrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}=A{ x }^{ 4 }+B{ x }^{ 3 }+C{ x }^{ 2 }+Dx+E$$. Then the value of $$5A+4B+3C+2D+E$$ is equal to

A
zero

B
-16

C
16

D
-11

Solution

$$\begin{vmatrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}=A{ x }^{ 4 }+B{ x }^{ 3 }+C{ x }^{ 2 }+Dx+E$$

Consider $$LHS=\begin{vmatrix} x & 2 & x \\ { x }^{ 2 } & x & 6 \\ x & x & 6 \end{vmatrix}$$

$$=x(6x-6x)-2(6x^2-6x)+x(x^3-x^2)$$

$$={ x }^{ 4 }-{ x }^{ 3 }-12{ x }^{ 2 }+12x$$

Comparing with RHS, we get
$$A=1,B=-1, C=-12, D=12, E=0$$

So, $$5A+4B+3C+2D+E=-11$$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 32

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Determinants Test - 32

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SHARING IS CARING

Question 1

$$\displaystyle (1-\omega )^{2} $$

3

-3

none of these

Solution

Question 2

$$\displaystyle 3$$

$$\displaystyle -3$$

$$\displaystyle\dfrac{3}{2}$$

$$\displaystyle 2$$

Solution

Question 3

$$1$$

$$\sqrt{2}$$

$$2$$

$$0$$

Solution

Question 4

$$4$$

$$16$$

$$64$$

$$256$$

Solution

Question 5

$$\displaystyle a^{3}+b^{3}+c^{3}$$

$$\displaystyle a^{2}b-b^{2}c$$

0

$$\displaystyle a^{3}b+b^{3}+3abc $$

Solution

Question 6

$$x=0$$

$$x=c$$

$$x=b$$

$$x=a$$

Solution

Question 7

$$A= 2B$$

$$A= B$$

$$A= -B$$

none of these

Solution

Question 8

0

1

2

$$\infty$$

Solution

Question 9

1

0

-1

67

Solution

Question 10

zero

-16

16

-11

Solution

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