Determinants Test

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Determinants Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

If $$\begin{vmatrix} 1+x & 2 & 3 \\ 1 & 2+x & 3 \\ 1 & 2 & 3+x \end{vmatrix}=0$$ then $$x=$$

A
$$1$$

B
$$-1$$

C
$$-6$$

D
$$6$$

Solution

We have $$(1-x)((2+x)(3+x)-6)-2(3-2)+3(2-(2+x))=0$$
it gives $$6x^2+x^3=0$$
$$x^2(6+x)=0\implies x=0,0,-6$$
option c gives correct answer
Question 2
1 / -0

If $$A = {\left[ {\begin{array}{*{20}{c}}a\\b\\c\end{array}\begin{array}{*{20}{c}}p\\q\\r\end{array}} \right]_{3 \times 2}}$$ then determinant $$\left( {A{A^T}} \right)$$ is equal to

A
$$0$$

B
$${a^2} + {b^2} + {c^2}$$

C
$${p^2} + {q^2} + {r^2}$$

D
$${p^2} + {q^2}$$

Solution

$$A=\begin{bmatrix} a & b \\ c & p \\ q & r \end{bmatrix} \Rightarrow A^{T}= \begin{bmatrix} a & b & c \\ p & q & r \end{bmatrix}$$

$$AA^{T}= \begin{bmatrix} a & b & c \\ p & q & r \end{bmatrix} \begin{bmatrix} a & b & c \\ p & q & r \end{bmatrix}= \begin{bmatrix} a^{2}+ p^{2} & ab+pq & ac+pr \\ ab+pq & b^{2}+q^{2} & bc+qr \\ ac+pr & bc+qr & c^{2}+r^{2} \end{bmatrix}$$

$$det (AA^{T}) = (a^{2}+ p^{2}) {(b^{2}+q^{2})(c^{2}-r^{2})-(bc+qr)^{2}}$$

$$-(ab+pq){(ab+pq)(c^{2}-r^{2})-(ac+pr)(bc+qr)}$$

$$+(ac+pr) {(ab+pq)(bc+qr)- (ac+pr)(b^{2}+q^{2})}$$

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

If $${t_{1,}}{t_2}\,$$ and $${t_3}$$ distinct. and the points $$\left( {{t_1}.2a{t_1} + a{t_1}^3} \right).\left( {{t_2}.2a{t_2} + a{t_2}^3} \right),\left( {{t_3}.2a{t_3} + a{t_3}^3} \right)$$ are collinear, then $${t_1} + {t_2} + {t_3} = $$

A
$$t_1 t_2 t_3 =-1$$

B
$$t_1 +t_2 +t_3 =t_1 t_2 t_3$$

C
$$t_1 +t_2 +t_3 =0$$

D
$$t_1 +t_2 +t_3 =-1$$

Solution

Let points are $$P(t_1. 2at_1 + at_1^3)$$ $$,Q(t_1, 2at_2+at_1^3)$$ and $$R(t_3, 2at_3+at_3^3)$$

$$P,Q$$ and $$R$$ are collinear

$$\Rightarrow $$area ( $$\Delta PQR$$) $$= 0$$

$$\Rightarrow \begin{vmatrix} 1&t_1&2at_1 + at_1^3 \\1&t_2& 2at_2+at_2^3\\1& t_2&2at_3+at_3^3 \end{vmatrix}=0$$

$$\Rightarrow \begin{vmatrix} 1&t_1 &2at_1 \\ 1&t_2 &2at_2 \\1&t_3&2at_3 \end{vmatrix} + \begin{vmatrix}1&t_1 &at_1^3 \\ 1&t_2&at_2^3\\1&t_3&at_3^3 \end{vmatrix}=0$$

$$\Rightarrow 2a\begin{vmatrix}1 &t_1&t_1 \\ 1&t_2&t_2\\1&t_3&t_3 \end{vmatrix}+a\begin{vmatrix}1 &t_1&t_1^3\\1&t_2&t_2^3\\1 & t_3&t_3^3 \end{vmatrix}=0$$

Determinant is zero if two rows of a determinant are same

$$\Rightarrow 0+a \begin{vmatrix} 1&t_1&t_1^3 \\1&t_2 &t_2^3\\ 1&t_3&t_3^3 \end{vmatrix}=0$$

$$R_2 \to R_2 - R_1$$ $$R_3\to R_3-R_1$$

$$\Rightarrow \begin{vmatrix} 1&t_1&t_1^3 \\ 1&t_2-t_1&t_2^3-t_1^3\\0&t_2-t_1&t_3^3-t_1^3 \end{vmatrix}=0$$

$$\Rightarrow (t_2-t_1)(t_3-t_1) \begin{vmatrix} 1&t_1&t_1^3 \\0&1&t_1^2+t_1t_2+t_2^2\\0&1&t_1^2+t_1t_3+t_3^2\end{vmatrix}=0$$

$$\Rightarrow (t_2-t_1)(t_2-t_1)[t_1^2 + t_!t_2+t_3^2 - t_1^2 - t_1t_2-t_2^2]=0$$

$$\Rightarrow (t_2-t_1)(t_3-t_1)[t_2^3 + t_1t_3-t_1t_2-t_2^2] = 0$$

$$\Rightarrow (t_2-t_1)(t_3-t_1)(t_3-t_2)(t_1+t_2+t_3)=0$$

as $$t_1\neq t_2 \neq t_3$$

$$\Rightarrow t_2-t_1$$ or $$t_3-t_1$$ or $$t_3-t_2$$ can't be zero

$$\Rightarrow t_1+t_2+t_3 = 0$$
Question 4
1 / -0

If $$\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}} \right|$$ for $$x \ne 0,\,y \ne 0$$ then $$\Delta $$ is

A
Divisible by neither $$x$$ nor $$y$$

B
Divisible by both $$x$$ and $$y$$

C
Divisible by $$x$$ but not $$y$$

D
Divisible by $$y$$ but not $$x$$

Solution

$$\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{matrix} \right|$$
$$ \Delta =\left( 1+x \right) \left( 1+y \right) -1-1\left( 1+y-1 \right) +1\left( 1-1-x \right)$$
$$ \Delta =\left( 1+x \right) \left( 1+y \right) -1-y-x$$
$$ \Delta =xy+1+x+y-1-y-x$$
$$ \Delta =xy$$
It is divisible by both $$x$$ and $$y$$
$$B$$ is correct.
Question 5
1 / -0

If $$\alpha, \beta, \gamma, $$ are the roots of $$x^3+ax^2+b=0$$, then the value of $$\left| \begin{matrix} \alpha & { \beta } & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{matrix} \right| $$ is

A
$$-a^3$$

B
$$a^3-3b$$

C
$$a^3$$

D
$$a^2-3b$$

Solution
Question 6
1 / -0

Find the value of the determinant $$\begin{vmatrix} 1 & 0 & 0 \\ 2 & \cos { x } & \sin { x } \\ 3 & \sin { x } & \cos { x } \end{vmatrix}$$.

A
$$\cos{2x}$$

B
$$1$$

C
$$0$$

D
$$\sin{2x}$$

Solution

Given
$$\begin{vmatrix} 1 & 0 & 0 \\ 2 & cosx & sinx \\ 3 & sinx & cosx \end{vmatrix}$$
$$=1(\cos x\times\cos x-\sin x\times\sin x)$$
$$=\cos^x-\sin^2x$$
We know that
$$\cos 2x=\cos^2x-\sin^2x$$
$$=\cos 2x$$
Question 7
1 / -0

The value of $$x$$ for which the matrix $$A=\begin{bmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{bmatrix}$$
Is non-singular

A
$$x\in R-\{-1,\ 2\}$$

B
$$x\in R-\{1,\ 2\}$$

C
$$x\in R-\{-2,\ 2\}$$

D
$$None\ of\ these$$

Solution

$$A=\begin{bmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{bmatrix}$$
matrix $$'A'$$ is non-singular

$$\therefore |A|\neq 0$$
$$=\begin{vmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix}\neq 0$$
$$\Rightarrow (x-1)[x^2+1-2x-1]-1(x-2)+1(1-x+1)\neq 0$$
$$\Rightarrow (x-1).x(x-2)+2(2-x)\neq 0$$
$$\Rightarrow (x-1)x(x-2)-2(x-2)\neq 0$$
$$\Rightarrow (x-2)(x^2-2x+x-2)\neq 0$$
$$\Rightarrow (x-2)(x(x-2)+1(x-2))\neq 0$$
$$\Rightarrow (x-2)^2(x+1)\neq 0$$
$$\therefore x\neq 2,\ x\neq -1$$
$$\therefore x\in R-\{-1,\ 2\}$$
Question 8
1 / -0

If the points $$(k, 2-2k)$$, $$(1-k, 2k)$$ and $$(-k-4, 6-2k)$$ be collinear, the number of possible values of $$k$$ are

A
$$4$$

B
$$2$$

C
$$1$$

D
$$3$$

Solution

$$(k, 2-2k), (1-k, 2k)$$ & $$(-k-4, 6-2k)$$
if collinear the $$m_1=m_2$$
$$\dfrac{2k-2+2k}{1-k-k}=\dfrac{6-2k-2k}{-k-4-1+k}$$
$$\dfrac{4k-2}{1-2k}=\dfrac{6-4k}{-5}$$
$$\dfrac{2k-1}{1-2k}=\dfrac{3-2k}{-5}$$
$$5=3-2k$$
$$2k=-2$$
$$k=-1$$
Only $$1$$ value possible.
Question 9
1 / -0

If $$f(x)=\begin{bmatrix} \cos { x } & -\sin { x } & 0 \\ \sin { x } & \cos { x } & 0 \\ 0 & 0 & 1 \end{bmatrix}$$, then $$f(x+y)$$ is equal to:

A
$$f(x)+f(y)$$

B
$$f(x)-f(y)$$

C
$$f(x).f(y)$$

D
None of these

Solution

Expanfing along $$ c_{3}, f(n) = cos^{2}x+sin^{2}x = 1 $$
$$ \Rightarrow f(x+y) = 1 ,f(y) = 1 $$
$$ \Rightarrow f(x+y) = f(n)f(y) $$ is true only
$$ \Rightarrow (c) $$
Question 10
1 / -0

If $$A=\quad \begin{bmatrix} 1 & -1 & 1 \\ 0 & 2 & -3 \\ 2 & 1 & 0 \end{bmatrix}, B=(adj\quad A)$$ and $$C=5A$$, then $$\cfrac { \left| adj\quad B \right| }{ \left| C \right| } $$ is equal to

A
$$5$$

B
$$25$$

C
$$-1$$

D
$$1$$

Solution

We have
$$\begin{matrix} A=\left[ { \begin{array} { *{ 20 }{ c } }1 & { -1 } & 1 \\ 0 & 2 & { -3 } \\ 2 & 1 & 0 \end{array} } \right] \\ B=\left( { adj\, A } \right) \\ And, \\ C=5A \\ Now \\ \frac { { \left| { adj\, B } \right| } }{ { \left| C \right| } } =? \\ \left| A \right| =1\left( { 0+3 } \right) +1\left( 6 \right) +1\left( { -4 } \right) \\ \left| A \right| =5 \\ \frac { { \left| { adj\, \left( { adj\, A } \right) } \right| } }{ { \left| { 5A } \right| } } =\frac { { { { \left| A \right| }^{ { { \left( { 3-1 } \right) }^{ 2 } } } } } }{ { 125\left| A \right| } } \\ =\frac { { { { \left| A \right| }^{ 4 } } } }{ { 125|A| } } \\ =\frac { { |A| } }{ { 125 } } =1 \\ \end{matrix}$$
Hence,
The option $$(A)$$ is the correct answer.

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Re-Attempt Weekly Quiz Competition

Determinants Test - 54

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Determinants Test - 54

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-

Rank

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SHARING IS CARING

Question 1

$$1$$

$$-1$$

$$-6$$

$$6$$

Solution

Question 2

$$0$$

$${a^2} + {b^2} + {c^2}$$

$${p^2} + {q^2} + {r^2}$$

$${p^2} + {q^2}$$

Solution

Question 3

$$t_1 t_2 t_3 =-1$$

$$t_1 +t_2 +t_3 =t_1 t_2 t_3$$

$$t_1 +t_2 +t_3 =0$$

$$t_1 +t_2 +t_3 =-1$$

Solution

Question 4

Divisible by neither $$x$$ nor $$y$$

Divisible by both $$x$$ and $$y$$

Divisible by $$x$$ but not $$y$$

Divisible by $$y$$ but not $$x$$

Solution

Question 5

$$-a^3$$

$$a^3-3b$$

$$a^3$$

$$a^2-3b$$

Solution

Question 6

$$\cos{2x}$$

$$1$$

$$0$$

$$\sin{2x}$$

Solution

Question 7

$$x\in R-\{-1,\ 2\}$$

$$x\in R-\{1,\ 2\}$$

$$x\in R-\{-2,\ 2\}$$

$$None\ of\ these$$

Solution

Question 8

$$4$$

$$2$$

$$1$$

$$3$$

Solution

Question 9

$$f(x)+f(y)$$

$$f(x)-f(y)$$

$$f(x).f(y)$$

None of these

Solution

Question 10

$$5$$

$$25$$

$$-1$$

$$1$$

Solution

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