Determinants Test

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Determinants Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

If $$A+B+C= \pi$$, then $$ \displaystyle \left| \begin{matrix} \tan { \left( A+B+C \right) } & \tan { B } & \tan { C } \\ \tan { (A+C) } & 0 & \tan { A } \\ \tan { (A+B) } & -\tan { A } & 0 \end{matrix} \right| $$ is equal to

A
$$0$$

B
$$1$$

C
$$\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{A}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{B}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{C}$$

D
$$-2$$

Solution

Let $$P=\begin{vmatrix}
tan(A+B+C) & tan B & tan C\\
tan (A+C) & 0 & tan A\\
tan(A+B) & -tan A & 0
\end{vmatrix}$$

$$=\tan(A+B+C) \tan^2A-\tan B[-\tan A \tan (A+B)]+\tan C[-\tan A \tan(A+C)]$$

$$=\tan (A+B+C) \tan^2 A+\tan A \tan B \tan(A+B)-\tan A \tan C \tan (A+C)$$

Given $$A+B+C=\pi \implies A+B=\pi-C$$ or $$A+C=\pi-B$$

$$\therefore P=\tan (\pi) \tan^2 A + \tan A \tan B(-\tan C) -\tan A \tan C(-\tan B)$$

$$=0\cdot \tan^2 A - \tan A \tan B \tan C + \tan A \tan B \tan C$$

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

$$I\mathrm{f}\mathrm{A}=\left[\begin{array}{ll}
1 & 3\\
2 & 1
\end{array}\right]$$, then the determinant $$\mathrm{A}^{2}-2\mathrm{A}$$:

A
$$5$$

B
$$25$$

C
$$-5$$

D
$$-25$$

Solution

Given $$A=\begin{pmatrix}
1 & 3\\
2 & 1
\end{pmatrix}$$
$$A^2=\begin{bmatrix}
1 & 3\\
2 & 1
\end{bmatrix}\begin{bmatrix}
1 & 3\\
2 & 1
\end{bmatrix}$$
By operation of matrix (4),
$$A^2=\begin{bmatrix}
7 & 6\\
4 & 7
\end{bmatrix}$$
So,$$A^2-2A=\begin{bmatrix}
7 & 6\\
4 & 7
\end{bmatrix}-2\begin{bmatrix}
1 & 3\\
2 & 1
\end{bmatrix}=\begin{bmatrix}
7 & 6\\
4 & 7
\end{bmatrix}-\begin{bmatrix}
2 & 6\\
4 & 2
\end{bmatrix}$$
$$=\begin{bmatrix}
5 & 0\\
0 & 5
\end{bmatrix}$$
So, $$det A^2-2A=|A^2-2A|=25$$
Question 3
1 / -0

$$\left[\begin{array}{llll}
\mathrm{c}\mathrm{o}\mathrm{s}\alpha+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\alpha & \mathrm{c}\mathrm{o}\mathrm{s}\beta+\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\beta\\
\mathrm{s}\mathrm{i}\mathrm{n}\beta+\mathrm{i}\mathrm{c}\mathrm{o}\mathrm{s}\beta\ & \mathrm{s}\mathrm{i}\mathrm{n}\alpha+\mathrm{i}\mathrm{c}\mathrm{o}\mathrm{s}\alpha &
\end{array}\right]$$ is

A
2 $$\cos\alpha$$

B
2 $$\sin\beta$$

C
$$0$$

D
$$1$$

Solution

Let $$A=\begin{bmatrix}
\cos \alpha+i \sin \alpha & \cos \beta + i sin \beta\\
\sin \beta + i \cos \beta & \sin \alpha + i \cos \alpha
\end{bmatrix}$$
$$|A|=(\cos \alpha + i \sin \alpha)(\sin \alpha+ i \cos \alpha)-(\cos \beta + i \sin \beta)(\sin \beta + i \cos \beta)$$
$$= \cos \alpha \sin \alpha + i - \cos \alpha \sin \alpha - [\cos \beta \sin \beta + i \sin \beta \cos \beta]$$
$$=0$$
Question 4
1 / -0

Adj $$\left ( Adj\begin{bmatrix}
2 &-3 \\
4& 6
\end{bmatrix} \right )=$$

A
$$\begin{bmatrix}

2 & -3\\

4& 6

\end{bmatrix}$$

B
$$\begin{bmatrix}

6& 3\\

-4& 2

\end{bmatrix}$$

C
$$\begin{bmatrix}

-6& 3\\

-4& -2

\end{bmatrix}$$

D
$$\begin{bmatrix}

-6& -3\\

4& -2

\end{bmatrix}$$

Solution

Adjoint of matrix is given by
$$\begin{bmatrix} M11 & -M12 \\ -M21 & M22 \end{bmatrix}$$
Here $$M11 = 6$$
$$M12 = 4$$
$$M21 = -3$$
$$M22 = 2$$
So adjoint of the matrix is
$$\begin{bmatrix} 6& -4 \\ 3 & 2 \end{bmatrix}$$
Again adjoint of this matrix is
$$\begin{bmatrix} 2&-3\\ 4&6\end{bmatrix}$$
Hence the answer is option A.
Question 5
1 / -0

lf $$\left|\begin{array}{lll}
a+x & a-x & a-x\\
a-x & a+x & a-x\\
a-x & a-x & a+x
\end{array}\right|=0$$ then the non-zero value of x=............

A
$$a$$

B
$$3a$$

C
$$2a$$

D
$$4a$$

Solution

By operation of matrix (5),
$$A=\begin{vmatrix}
a+x & a-x & a-x\\
a-x & a+x & a-x\\
a-x & a-x & a+x
\end{vmatrix}$$
$$=(a+x)[(a+x)^2-(a-x)^2]-(a-x)[a^2-x^2-(a-x)^2]+(a-x)[(a-x)^2-(a^2-x^2)]$$
$$=(a+x)(4ax)-(a-x)(a^2-x^2)+2(a-x)^3-(a-x)(a^2-x^2)$$
$$=(a+x)(4ax)+2(a-x)[a^2+x^2-2ax-a^2+x^2]$$
$$=(a+x)4ax+2(a-x)(2x^2-2ax)$$
$$A=4a^2x+4ax^2+4ax^2-4a^2x-ax^3+4ax^2$$
$$A=8ax^2+4ax^2-4x^3$$
$$A=12ax^2-4x^3$$
Given,$$ A= 12 ax^2-4x^3=0$$
$$x^2(12 a-x)=0$$
$$x=3a$$
Question 6
1 / -0

A= $$\begin{bmatrix}
3 & 0 & 0\\
0& 3 & 0\\
0& 0 & 3
\end{bmatrix}$$ ,then Adj ( A)

A
3A

B
6A

C
$$9A^{T}$$

D
$$2A^{T}$$

Solution

We have, $$A=\begin{bmatrix}
3 & 0 & 0\\
0& 3 & 0\\
0& 0 & 3
\end{bmatrix}$$
co-factor matrix of $$A$$ is,
$$B = \begin{bmatrix}
9 & 0 & 0\\
0& 9 & 0\\
0& 0 & 9
\end{bmatrix}$$
Thus
$$\therefore $$ Adj(A)$$ =B^T=\begin{bmatrix}
9 & 0 & 0\\
0& 9 & 0\\
0& 0 & 9
\end{bmatrix}=3\begin{bmatrix}
3 & 0 & 0\\
0& 3 & 0\\
0& 0 & 3
\end{bmatrix}=3A$$
Question 7
1 / -0

lf $$\left|\begin{array}{lll}
1 & 2 & x\\
4 & -1 & 7\\
2 & 4 & -6
\end{array}\right|$$ is a singular matrix, then $$x$$ is equal to

A
$$0$$

B
$$1$$

C
$$-3$$

D
$$3$$

Solution

Given $$A=\begin{vmatrix}
1 & 2 & x\\
4 & -1 & 7\\
2 & 4 & -6
\end{vmatrix}$$ is a singular matrix

So, $$detA=0$$

$$\Rightarrow 1(6-28)-2(-24-14)+x(16+2)=0$$

$$\Rightarrow -22+76+18x=0$$

$$\Rightarrow 18x=-54$$

$$\therefore x=-3$$
Question 8
1 / -0

If $$\mathrm{A}$$ is an unitary matrix then $$|A|$$ is equal to:

A
$$1$$

B
$$-1$$

C
$$\pm 1$$

D
$$2$$

Solution

Since, A is a unitary matrix
$$AA^{*}=I$$
$$det A \times det A^{*}=det I$$
$$\Rightarrow (det A)^{2}=1$$ ($$\because |A|=|\overline{A^{T}}|$$)
$$\Rightarrow det A=\pm 1$$
Question 9
1 / -0

A determinant of second order is made with the elements $$0$$ and $$1.$$ The number of determinants with non-negative values is:

A
3

B
10

C
11

D
13

Solution

There are only three determinants of second order with negative value, $$\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix},\begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix},\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}$$
Number of possible determinants with elements $$0$$ and $$1$$ are $${ 2 }^{ 4 }=16$$
therefore, number of determinants with non-negative values is $$13.$$
Question 10
1 / -0

If $$A=\displaystyle \int_{1}^{sin\theta}\frac{t}{1+t^{2}}dt$$ and
$$B=\displaystyle \int_{1}^{cosec\theta}\frac{1}{t(1+t^{2})}dt$$, then the value of determinant $$\begin{vmatrix}
A & A^{2}& B\\
e^{A+B}& B^{2} &-1 \\
1& A^{2}+B^{2} & -1
\end{vmatrix}$$ is

A
$$\sin\theta$$

B
$$cosec \theta$$

C
$$0$$

D
$$1$$

Solution

$$A=\displaystyle \int _{ 1 }^{ sin\theta } \frac { t }{ 1+t^{ 2 } } dt$$
$$\displaystyle B=\int _{ 1 }^{ cosec\theta } \frac { 1 }{ t(1+t^{ 2 }) } dt$$

Put $$\displaystyle z=\frac { 1 }{ t } \quad $$
$$\displaystyle dz=-\frac { 1 }{ { t }^{ 2 } } dt$$

$$\displaystyle B=\int _{ 1 }^{ \sin { \theta } } \frac { -z }{ (z^{ 2 }+1) } dz$$
$$\displaystyle =\int _{ 1 }^{ \sin { \theta } } \frac { -t }{ (t^{ 2 }+1) } dt$$
$$\Rightarrow B=-A$$

Now, $$\begin{vmatrix} A & A^{ 2 } & B \\ e^{ A+B } & B^{ 2 } & -1 \\ 1 & A^{ 2 }+B^{ 2 } & -1 \end{vmatrix}$$

$$=\begin{vmatrix} A & A^{ 2 } & -A \\ e^{ 0 } & A^{ 2 } & -1 \\ 1 & 2A^{ 2 } & -1 \end{vmatrix}$$

$$=-\begin{vmatrix} A & A^{ 2 } & A \\ 1 & A^{ 2 } & 1 \\ 1 & 2A^{ 2 } & 1 \end{vmatrix}$$

$$=0$$ (On expanding the determinant)

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Determinants Test - 26

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Determinants Test - 26

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Question 1

$$0$$

$$1$$

$$\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{A}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{B}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{C}$$

$$-2$$

Solution

Question 2

$$5$$

$$25$$

$$-5$$

$$-25$$

Solution

Question 3

2 $$\cos\alpha$$

2 $$\sin\beta$$

$$0$$

$$1$$

Solution

Question 4

$$\begin{bmatrix}2 & -3\\ 4& 6\end{bmatrix}$$

$$\begin{bmatrix}6& 3\\ -4& 2\end{bmatrix}$$

$$\begin{bmatrix}-6& 3\\ -4& -2\end{bmatrix}$$

$$\begin{bmatrix}-6& -3\\ 4& -2\end{bmatrix}$$

Solution

Question 5

$$a$$

$$3a$$

$$2a$$

$$4a$$

Solution

Question 6

3A

6A

$$9A^{T}$$

$$2A^{T}$$

Solution

Question 7

$$0$$

$$1$$

$$-3$$

$$3$$

Solution

Question 8

$$1$$

$$-1$$

$$\pm 1$$

$$2$$

Solution

Question 9

3

10

11

13

Solution

Question 10

$$\sin\theta$$

$$cosec \theta$$

$$0$$

$$1$$

Solution

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$$\begin{bmatrix}

2 & -3\\

4& 6

\end{bmatrix}$$

$$\begin{bmatrix}

6& 3\\

-4& 2

\end{bmatrix}$$

$$\begin{bmatrix}

-6& 3\\

-4& -2

\end{bmatrix}$$

$$\begin{bmatrix}

-6& -3\\

4& -2

\end{bmatrix}$$