Determinants Test

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Determinants Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$\begin{bmatrix}3 & -5\\ -1 & 2\end{bmatrix}$$

B
$$\begin{bmatrix}3 & -1\\ -5 & 2\end{bmatrix}$$

C
$$\begin{bmatrix}-1 & 2\\ 3 & -5\end{bmatrix}$$

D
None of these

Solution
Question 2
1 / -0

$$\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=$$?

A
$$1$$

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

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

D
$$none\ of\ these$$

Solution

Let us consider $$\Delta=\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=$$

$$\Delta=\cos^215^0-\sin^2 15^0$$

$$=\cos 30^0$$.........$$(\because \cos 2\theta=\cos^2 \theta-\sin^2 \theta)$$

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

Hence $$\begin{vmatrix} \cos { { 15 }^{ o } } & \sin { { 15 }^{ o } } \\ \sin { { 15 }^{ o } } & \cos { { 15 }^{ o } } \end{vmatrix}=\dfrac{\sqrt{3}}{2}$$

Option $$C$$.
Question 3
1 / -0

When the determinant $$\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}$$ is expanded in powers of $$\sin x$$, then the constant term in that expression is

A
$$1$$

B
$$0$$

C
$$-1$$

D
$$2$$

Solution

$$f(x)=\begin{vmatrix} \cos { 2x } & \sin ^{ 2 }{ x } & \cos { 4x } \\ \sin ^{ 2 }{ x } & \cos { 2x } & \cos ^{ 2 }{ x } \\ \cos { 4x } & \cos ^{ 2 }{ x } & \cos { 2x } \end{vmatrix}$$
the requried constant term is
$$\quad f(0)=\begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix}=\quad \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix}=1(0-1)=-1$$
Question 4
1 / -0

If the determinant $$\begin{vmatrix} \cos 2x & \sin ^{ 2 } x & \cos 4x \\ \sin ^{ 2 } x & \cos 2x & \cos ^{ 2 } x \\ \cos 4x & \cos ^{ 2 } x & \cos 2x \end{vmatrix}$$ is expanded in powers of $$\sin x$$ then the constant term in the expansion is

A
$$1$$

B
$$2$$

C
$$-1$$

D
$$-2$$

Solution

Let $$\begin{vmatrix}\cos 2x & \sin^2 x &\cos 4x \\ \sin^2 x &\cos 2x &\cos^2x \\\cos 4x &\cos^2x &\cos 2x\end{vmatrix}= a_{1} +a_{2}\sin ^2 x +a_{3}\sin ^3 x + ....$$ . . . . . . . . . . . $$(i)$$
The constant term in the expansion is clearly the term $$a_{1}$$.
Putting $$x=0$$ on both sides of $$(i)$$ we get-
$$\begin{vmatrix}1 & 0 & 1\\ 0 & 1 &1 \\ 1 & 1 &1\end{vmatrix} = a_{1}$$
$$\Rightarrow a_{1}=\begin{vmatrix}1 & 0 & 1\\ 0 & 1 &1 \\ 0 & 1 &0\end{vmatrix}$$ ($$R_{3} \rightarrow R_{3} - R_{1}$$)
$$\Rightarrow a_{1} = 1(0 - 1) = -1$$ (Expanding along $$C_{1}$$)
Thus, the constant term is $$-1$$.

The correct answer is option C.
Question 5
1 / -0

If $$\begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c \end{vmatrix}=0$$, then the line $$ax+by+c=0$$ passes through the fixed point whcih is

A
$$(1,2)$$

B
$$(1,1)$$

C
$$(-2,1)$$

D
$$(1,0)$$

Solution

applying $${C}_{1}\rightarrow a{C}_{!}$$ and then $${C}_{1}\rightarrow {C}_{1}+b{C}_{2}+c{C}_{3}$$ and taking $$\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) $$ common from $${C}_{1}$$ we get
$$\Delta =\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } \begin{vmatrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{vmatrix}=\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 0 & a+c & -b \end{vmatrix}$$
$$({R}_{2}\rightarrow {R}_{2}-{R}_{1},{R}_{3}\rightarrow {R}_{3}-{R}_{1})$$
$$=\cfrac { \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }{ a } (-bc+{ a }^{ 2 }+ab+ac+bc)=\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) (a+b+c)$$
hence $$\Delta =0\Rightarrow a+b+c=0$$
therefore line $$ax+by+c=0$$ passes through the fixed point $$(1,1)$$
Question 6
1 / -0

If A is singular matrix , then adj A is

A
singular

B
non-singular

C
symmetric

D
not defined

Solution

$$ A adj A = |A| I $$

$$ \Rightarrow |A adj A | = |A|^n $$ [ if A is of order $$ n \times n ] $$

$$ \Rightarrow |A| |adj A| = |A|^n $$

$$ \Rightarrow |adj A| = |A|^{n-1} $$

Now, A is singular .

$$ \therefore |A| = 0 $$

$$ \Rightarrow |adj A | = 0 $$

Hence adj A is singular.
Question 7
1 / -0

There are two values of a which makes determinant $$ \Delta =\left| \begin{matrix} 1\quad & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{matrix} \right| =86 $$ then sum of these number is

A
$$ 4 $$

B
$$ 5 $$

C
$$ - 4 $$

D
$$ 9 $$

Solution

we have $$ \Delta =\left| \begin{matrix} 1\quad & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{matrix} \right| =86 $$

$$ \Rightarrow 1(2a^2 +4) -2(-4a -20) + 0 = 86 $$ [expanding along first column ]

$$ \Rightarrow 2a^2 +4 +8a +40 = 86 $$

$$ \Rightarrow 2a^2 + 8a +44 -86 = 0 $$

$$ \Rightarrow a^2 +4a -21 = 0 $$

$$ \Rightarrow a^2 +7a -3a-21 =0 $$

$$ \Rightarrow (a +7)(a-3) = 0 $$

$$ a = -7 $$ and $$ 3 $$

$$ \therefore $$ Required sum = $$ -7 +3 = -4 $$
Question 8
1 / -0

If $$ \begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}$$ then value of $$ x $$ is

A
$$ 3 $$

B
$$ \pm 3 $$

C
$$ \pm 6 $$

D
$$ 6 $$

Solution

$$ \because \begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix} $$
$$ \Rightarrow 2x^2 -40 = 18 +14 $$
$$ \Rightarrow 2x^2 = 32 +40 $$
$$ \Rightarrow x^2 = \frac {72}{2} = 36 $$
$$ \therefore x = \pm 6 $$
Question 9
1 / -0

Choose the correct answer from the given alternatives in the following question:
If $$ A = \begin{bmatrix} 2 & -4 \\ 3 & 1 \end{bmatrix} $$, then the adjoint of matrix $$A$$ is

A
$$ \begin{bmatrix} -1 & 3 \\ -4 & 1 \end{bmatrix} $$

B
$$ \begin{bmatrix} 1 & 4 \\ -3 & 2 \end{bmatrix} $$

C
$$ \begin{bmatrix} 1 & 3 \\ 4 & -2 \end{bmatrix} $$

D
$$ \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} $$

Solution
Question 10
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Choose the correct answer from the given alternatives in the following question:
If $$ A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} $$ and $$ A ({\text{adj}} A) = KI $$ , then the value of $$k$$ is

A
$$1$$

B
$$-1$$

C
$$0$$

D
$$-3$$

Solution

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Re-Attempt Weekly Quiz Competition

Determinants Test - 59

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Determinants Test - 59

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SHARING IS CARING

Question 1

$$\begin{bmatrix}3 & -5\\ -1 & 2\end{bmatrix}$$

$$\begin{bmatrix}3 & -1\\ -5 & 2\end{bmatrix}$$

$$\begin{bmatrix}-1 & 2\\ 3 & -5\end{bmatrix}$$

None of these

Solution

Question 2

$$1$$

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

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

$$none\ of\ these$$

Solution

Question 3

$$1$$

$$0$$

$$-1$$

$$2$$

Solution

Question 4

$$1$$

$$2$$

$$-1$$

$$-2$$

Solution

Question 5

$$(1,2)$$

$$(1,1)$$

$$(-2,1)$$

$$(1,0)$$

Solution

Question 6

singular

non-singular

symmetric

not defined

Solution

Question 7

$$ 4 $$

$$ 5 $$

$$ - 4 $$

$$ 9 $$

Solution

Question 8

$$ 3 $$

$$ \pm 3 $$

$$ \pm 6 $$

$$ 6 $$

Solution

Question 9

$$ \begin{bmatrix} -1 & 3 \\ -4 & 1 \end{bmatrix} $$

$$ \begin{bmatrix} 1 & 4 \\ -3 & 2 \end{bmatrix} $$

$$ \begin{bmatrix} 1 & 3 \\ 4 & -2 \end{bmatrix} $$

$$ \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} $$

Solution

Question 10

$$1$$

$$-1$$

$$0$$

$$-3$$

Solution

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