Determinants Test

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Determinants Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

$$D\mathrm{e}\mathrm{t} \left\{\begin{array}{lll}
2 & 45 & 55\\
1 & 29 & 32\\
3 & 68 & 87
\end{array}\right\}=\ldots.$$ .

A
45

B
64

C
54

D
32

Solution

$$Det\begin{bmatrix}
2 & 45 & 55\\
1 & 29 & 32\\
3 & 68 & 87
\end{bmatrix}=2(29 \times 87 - 68 \times 32)-45(87 - 3 \times 32)+55(68-3 \times 29)$$

$$= 2\times 347 - 45(-9)+55(-19)$$

$$=54$$
Question 2
1 / -0

Find x if it is given that:
$$\det \left[\begin{array}{lll}
2 & 0 & 0\\
4 & 3 & 0\\
4 & 6 & x
\end{array}\right]=42$$

A
$$8$$

B
$$7$$

C
$$6$$

D
$$21/4$$

Solution

Given $$det\begin{pmatrix}
2 & 0 & 0\\
4 & 3 & 0\\
4 & 6 & x
\end{pmatrix}=42$$
By operation of matrix (5)
$$x(6)=42$$
$$\Rightarrow x=7$$
Question 3
1 / -0

$$\mathrm{If}$$ $$\left|\begin{array}{lll}
1 & 0 & 0\\
2 & 3 & 4\\
5 & -6 & x
\end{array}\right|$$ $$= 45$$ $$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$$ $$\mathrm{x}=$$

A
$$4$$

B
$$7$$

C
$$- 5$$

D
$$-7$$

Solution

Given $$\begin{vmatrix}
1 & 0 & 0\\
2 & 3 & 4\\
5 & -6 & x
\end{vmatrix}=45$$
By operation of matrix (5),
$$1(3x+24)=45$$
$$3x=21$$
$$\Rightarrow x=7$$
Question 4
1 / -0

$$\begin{vmatrix} 4\sin^{2}\theta & \cos^{2}\theta \\ 3\sec^{2}\theta & \text{cosec}^{2}\theta \end{vmatrix}=$$

A
$$8\sin^{2}\theta \cos^{2}\theta $$

B
$$4\sin 2\theta \cos 2\theta $$

C
$$1$$

D
$$4\cos^{3}\theta -3\cos \theta $$

Solution

The value of $$\begin{vmatrix} 4\sin^{2}\theta & \cos^{2}\theta \\ 3\sec^{2}\theta & \text{cosec}^{2}\theta \end{vmatrix}$$ is
$$=4\sin^{2}\theta .\text{cosec}^{2}\theta -3\sec^{2}\theta .\cos^{2}\theta$$
$$=4-3=1$$
Question 5
1 / -0

$$Adj \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}=\begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}\Rightarrow \left [ a\ b\ \right ]=$$ ____

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

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

C
$$\begin{bmatrix} 4 & 1 \end{bmatrix}$$

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

Solution

Given, Adj$$\left[ \begin{matrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{matrix} \right] =\left[ \begin{matrix} 5 & 9 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 6 \end{matrix} \right] $$
Adjoint of matrix = Transpose of cofactor matrix
$$C_{11}=\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix}=5$$

$${ C }_{ 12 }=\begin{vmatrix} -1 & -2 \\ 0 & 1 \end{vmatrix}=1$$

$${ C }_{ 13 }=\begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix}=-2$$

$${ C }_{ 21 }=-\begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix}=4$$

$${ C }_{ 22 }=\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix}=1$$

$${ C }_{ 23 }=-\begin{vmatrix} 1 & 0 \\ 0 & 2 \end{vmatrix}=-2$$

$${ C }_{ 31 }=\begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix}=-4$$

$${ C }_{ 32 }=-\begin{vmatrix} 1 & 2 \\ -1 & -2 \end{vmatrix}=0$$

$${ C }_{ 33 }=\begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix}=1$$
Cofactor matrix $$C=\left[ \begin{matrix} 5 & 1 & -2 \\ 4 & 1 & -2 \\ -4 & 0 & 1 \end{matrix} \right] $$
Adj matrix$$=\left[ \begin{matrix} 5 & 4 & -4 \\ 4 & 1 & 0 \\ -2 & -2 & 1 \end{matrix} \right] =\left[ \begin{matrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{matrix} \right] $$
$$\left[ \begin{matrix} a & b \end{matrix} \right] =\left[ \begin{matrix} 4 & 1 \end{matrix} \right] $$
Question 6
1 / -0

If $$A$$ is a $$3\times 3$$ matrix and $$\text{det} (3A)=k(\text{det} A)$$, then $$k=$$

A
$$9$$

B
$$6$$

C
$$1$$

D
$$27$$

Solution

The non zero determinant of a scalar multiple of a n×n matrix is given by the following property.
$$|kA|=kn|A|$$
⟹$$|3A|=3^3|A|=27|A|$$
$$k=27$$
Question 7
1 / -0

Directions For Questions

Based on this information answer the questions given below.
Two $$n\times n$$ square matrices A and B are said to be similar if there exists a non-singular matrix P such that
$$P^{-1}A P=B$$

...view full instructions

If A and B are similar matrices such that $$det (AB)=0, $$ then

A
$$det (A)=0 $$ and $$ det (B)=0$$

B
$$det (A)=0$$ and $$ det (B) \neq 0$$

C
$$A=0$$ and $$B=0$$

D
$$A=0 $$ or $$B=0$$

Solution

Given. two n$$\times$$n square matrices A and B
Given, $$B=P^{-1}AP$$
$$det\left( B \right) =det\left( { P }^{ -1 }AP \right) $$
$$det\left( B \right) =det\left( { P }^{ -1 }AP \right) $$
$$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right) $$
$$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right) $$
$$det\left( B \right) =det\left( A \right) $$
$$det\left( AB \right) =\left( det\left( A \right) \right) \left( det\left( B \right) \right) $$
If$$\left( det\left( A \right) \right) =0$$ then $$\left( det\left( B \right) \right) $$ also zero.
Question 8
1 / -0

$$\displaystyle \begin{vmatrix}cos\: C &tan\: A &0 \\ sin\: B &0 &-tan\: A \\ 0 &sin\: B &cos\: C \end{vmatrix}$$ has the value

A
$$0$$

B
$$1$$

C
$$\sin A \sin B \cos B$$

D
none of these

Solution

$$\begin{vmatrix} cos\: C & tan\: A & 0 \\ sin\: B & 0 & -tan\: A \\ 0 & sin\: B & cos\: C \end{vmatrix}\\$$

$$ =cosC\left( tanAsinB \right) -tanA\left( sinBcosC \right) \\ $$

$$=0$$

Hence, option 'A' is correct.
Question 9
1 / -0

Two $$n\times n$$ square matrices A and B are said to be similar if there exists a non-singular matrix P such that $$P^{-1}A P=B$$.
If A and B are similar matrices such that $$det (A)=1$$, then

A
$$det (B)=1$$

B
$$det (A)=-det (B)$$

C
$$det (B)=-1$$

D
none of these

Solution

Given, two n$$\times$$n square matrices A and B
Given, $$B=P^{-1}AP$$
$$det\left( B \right) =det\left( { P }^{ -1 }AP \right) $$
$$det\left( B \right) =\left( det\left( { P }^{ -1 } \right) \right) \left( det\left( A \right) \right) \left( det\left( P \right) \right) $$
$$det\left( B \right) =\left( det\left( { P } \right) \right) ^{ -1 }\left( det\left( A \right) \right) \left( det\left( P \right) \right) $$ $$[\because \quad|A||A^{-1}|=|AA^{-1}|=|I|=1]$$
$$det\left( B \right) =det\left( A \right) =1$$
Question 10
1 / -0

If$$\displaystyle \left | \begin{matrix}-12 &0 &\lambda \\ 0& 2& -1\\ 2& 1 &15 \end{matrix} \right |=-360$$, then the value of $$\lambda$$,is

A
$$-1$$

B
$$-2$$

C
$$-3$$

D
$$4$$

Solution

$$\left| \begin{matrix} -12 & 0 & \lambda \\ 0 & 2 & -1 \\ 2 & 1 & 15 \end{matrix} \right| =-360$$

Expanding along first column

$$\Rightarrow (-12)(31)-4\lambda =-360$$

$$\Rightarrow \lambda =-3$$

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Re-Attempt Weekly Quiz Competition

Determinants Test - 19

Result

Determinants Test - 19

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SHARING IS CARING

Question 1

45

64

54

32

Solution

Question 2

$$8$$

$$7$$

$$6$$

$$21/4$$

Solution

Question 3

$$4$$

$$7$$

$$- 5$$

$$-7$$

Solution

Question 4

$$8\sin^{2}\theta \cos^{2}\theta $$

$$4\sin 2\theta \cos 2\theta $$

$$1$$

$$4\cos^{3}\theta -3\cos \theta $$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$9$$

$$6$$

$$1$$

$$27$$

Solution

Question 7

$$det (A)=0 $$ and $$ det (B)=0$$

$$det (A)=0$$ and $$ det (B) \neq 0$$

$$A=0$$ and $$B=0$$

$$A=0 $$ or $$B=0$$

Solution

Question 8

$$0$$

$$1$$

$$\sin A \sin B \cos B$$

none of these

Solution

Question 9

$$det (B)=1$$

$$det (A)=-det (B)$$

$$det (B)=-1$$

none of these

Solution

Question 10

$$-1$$

$$-2$$

$$-3$$

$$4$$

Solution

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