Determinants Test

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Determinants Test - 58

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Question 1
1 / -0

If $$\alpha, \beta$$ are the roots of $$x^2+x+1=0$$ then $$\begin{vmatrix} y+1 & \beta & \alpha\\ \beta & y+\alpha & 1\\ \alpha & 1 & y+\beta\end{vmatrix}=?$$

A
$$y^2-1$$

B
$$y(y^2-1)$$

C
$$y^2-y$$

D
$$y^3$$

Solution

$$\Rightarrow \alpha + \beta - 1$$ & $$\alpha \beta = 1$$
Now $$R_1\rightarrow R_1+R_2+R_3$$ gives
$$\begin{vmatrix} y & y & y\\ \beta & y+\alpha & 1\\ \alpha & 1 & y+\beta\end{vmatrix}$$
$$c_2\rightarrow c_2-c_1, c_3\rightarrow c_3-c_1$$ gives
$$\begin{vmatrix} y & 0 & 0\\ \beta & y+\alpha +\beta & 1-\beta\\ \alpha & 1-\alpha & y+\beta -\alpha\end{vmatrix}=y\{\{y^2-(\alpha -\beta)^2\}-(\uparrow -\alpha)(1-\beta)\}$$.
$$\Rightarrow y[y^2-((\alpha +\beta)^2-4\alpha\beta)-3]\Rightarrow y[y^2+3-3]=y^3$$.
Question 2
1 / -0

$$\begin{vmatrix} \sin ^{ 2 }{ \theta } & \cos ^{ 2 }{ \theta } \\ -\cos ^{ 2 }{ \theta } & \sin ^{ 2 }{ \theta } \end{vmatrix}=$$

A
$$\cos { 2\theta } $$

B
$$\cfrac { 1 }{ 2 } \left( 1+\cos ^{ 2 }{ 2\theta } \right) $$

C
$$\cfrac { 1 }{ 2 } \left( 1-\sin ^{ 2 }{ 2\theta } \right) $$

D
$$\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 2\theta } $$

Solution

we know that
$$(sin^2x+cos^2x)^2=sin^4x+cos^4x+2sin^2xcos^2x$$

$$\begin{vmatrix} \sin ^{ 2 }{ \theta } & \cos ^{ 2 }{ \theta } \\ -\cos ^{ 2 }{ \theta } & \sin ^{ 2 }{ \theta } \end{vmatrix}=$$

$$\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } =1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta }$$

$$ =1-\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 2\theta } $$

$$=1-\cfrac { 1 }{ 2 } \left( 1-\cos ^{ 2 }{ 2\theta } \right)$$

$$ =\cfrac { 1 }{ 2 } \left( 1+\cos ^{ 2 }{ 2\theta } \right) $$
Question 3
1 / -0

The sum of the real roots of the equation
$$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0$$ is equal to

A
$$6$$

B
$$1$$

C
$$0$$

D
$$-4$$

Solution

Given $$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0$$

By expansion, we get

$$x(-3x^2+6x)-(-6)(2x+4-3x+3)+(-1)(4x+9x)$$

$$\Rightarrow -5{ x }^{ 3 }+30x-30+5x=0\Rightarrow -5{ x }^{ 3 }+35x-30=0\Rightarrow { x }^{ 3 }-7x+6=0$$, All roots are real

So, sum of roots $$=0$$
Question 4
1 / -0

$$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix}=$$?

A
$$(a^2+b^2-c^2-d^2)$$

B
$$(a^2-b^2+c^2-d^2)$$

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

D
$$none\ of\ these$$

Solution
Question 5
1 / -0

For square matrices $$A$$ and $$B$$ of the same order, we have $$adj (AB) = ?$$

A
$$(adj\ A)(adj\ B)$$

B
$$(adj\ B)(adj\ A)$$

C
$$|AB|$$

D
None of these

Solution
Question 6
1 / -0

$$\begin{vmatrix} \cos { { 70 }^{ o } } & \sin { { 20 }^{ o } } \\ \sin { { 70 }^{ o } } & \cos { { 20 }^{ o } } \end{vmatrix}=$$?

A
$$1$$

B
$$0$$

C
$$\cos 50^o$$

D
$$\sin 50^o$$

Solution

Let us consider $$\Delta=\begin{vmatrix} \cos { { 70 }^{ o } } & \sin { { 20 }^{ o } } \\ \sin { { 70 }^{ o } } & \cos { { 20 }^{ o } } \end{vmatrix}$$

$$\Delta =\cos 70^0\cos 20^0-\sin 70^0\sin 20^0$$

$$=\cos(70^0+20^0)$$

$$=\cos 90^0$$

$$\Delta =0$$

Option $$B$$.
Question 7
1 / -0

Evaluate : $$\begin{vmatrix} \sin { { 23 }^{ o } } & -\sin { { 7 }^{ o } } \\ \cos { { 23 }^{ o } } & \cos { { 7 }^{ o } } \end{vmatrix}$$

A
$$\dfrac {\sqrt 3}{2}$$

B
$$\dfrac {1}{2}$$

C
$$\sin 16^o$$

D
$$\cos 16^o$$

Solution
Question 8
1 / -0

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

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

B
$$\begin{bmatrix} -d& b\\ c & -a\end{bmatrix}$$

C
$$\begin{bmatrix} d& -b\\ -c & a\end{bmatrix}$$

D
$$\begin{bmatrix} -d& -b\\ c & a\end{bmatrix}$$

Solution
Question 9
1 / -0

If $$A$$ is a $$3-rowed$$ square matrix and $$|A| = 5$$ then $$|adj\ A| = ?$$

A
$$5$$

B
$$25$$

C
$$125$$

D
None of these

Solution

$$|adj\ A| = |A|^{(n - 1)} = |A|^{2} = 5^{2} = 25$$.
Question 10
1 / -0

If $$|A| = 3$$ and $$A^{-1} = \begin{bmatrix}3 & -1\\ \dfrac {-5}{3} & \dfrac {2}{3}\end{bmatrix}$$ then $$adj\ A = ?$$

A
$$\begin{bmatrix}
9 & 3\\
-5 & -2
\end{bmatrix}$$

B
$$\begin{bmatrix}
9 & -3\\
-5 & 2
\end{bmatrix}$$

C
$$\begin{bmatrix}
-9 & 3\\
5 & -2
\end{bmatrix}$$

D
$$\begin{bmatrix}
9 & -3\\
5 & -2
\end{bmatrix}$$

Solution

$$A^{-1} = \dfrac {1}{|A|}\cdot adj\ A \rightarrow adj\ A = |A| \cdot A^{-1} = 3A^{-1} = \begin{bmatrix} 9& -3\\ -5 & 2\end{bmatrix}$$.

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Re-Attempt Weekly Quiz Competition

Determinants Test - 58

Result

Determinants Test - 58

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SHARING IS CARING

Question 1

$$y^2-1$$

$$y(y^2-1)$$

$$y^2-y$$

$$y^3$$

Solution

Question 2

$$\cos { 2\theta } $$

$$\cfrac { 1 }{ 2 } \left( 1+\cos ^{ 2 }{ 2\theta } \right) $$

$$\cfrac { 1 }{ 2 } \left( 1-\sin ^{ 2 }{ 2\theta } \right) $$

$$\cfrac { 1 }{ 2 } \sin ^{ 2 }{ 2\theta } $$

Solution

Question 3

$$6$$

$$1$$

$$0$$

$$-4$$

Solution

Question 4

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

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

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

$$none\ of\ these$$

Solution

Question 5

$$(adj\ A)(adj\ B)$$

$$(adj\ B)(adj\ A)$$

$$|AB|$$

None of these

Solution

Question 6

$$1$$

$$0$$

$$\cos 50^o$$

$$\sin 50^o$$

Solution

Question 7

$$\dfrac {\sqrt 3}{2}$$

$$\dfrac {1}{2}$$

$$\sin 16^o$$

$$\cos 16^o$$

Solution

Question 8

$$\begin{bmatrix} d& -c\\ -b & a\end{bmatrix}$$

$$\begin{bmatrix} -d& b\\ c & -a\end{bmatrix}$$

$$\begin{bmatrix} d& -b\\ -c & a\end{bmatrix}$$

$$\begin{bmatrix} -d& -b\\ c & a\end{bmatrix}$$

Solution

Question 9

$$5$$

$$25$$

$$125$$

None of these

Solution

Question 10

$$\begin{bmatrix}9 & 3\\ -5 & -2\end{bmatrix}$$

$$\begin{bmatrix}9 & -3\\ -5 & 2\end{bmatrix}$$

$$\begin{bmatrix}-9 & 3\\ 5 & -2\end{bmatrix}$$

$$\begin{bmatrix}9 & -3\\ 5 & -2\end{bmatrix}$$

Solution

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$$\begin{bmatrix}
9 & 3\\
-5 & -2
\end{bmatrix}$$

$$\begin{bmatrix}
9 & -3\\
-5 & 2
\end{bmatrix}$$

$$\begin{bmatrix}
-9 & 3\\
5 & -2
\end{bmatrix}$$

$$\begin{bmatrix}
9 & -3\\
5 & -2
\end{bmatrix}$$