Determinants Test

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Determinants Test - 36

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Question 1
1 / -0

If $$\Delta =\begin{vmatrix}
a & 5-i & 7+i\\
5+i & b & 3+i\\
7-i & 3-i & c
\end{vmatrix}$$, then $$\Delta $$ is always

A
real

B
imaginary

C
$$0$$

D
None of these

Solution

$$\Delta =\begin{vmatrix} a & 5-i & 7+i \\ 5+i & b & 3+i \\ 7-i & 3-i & c \end{vmatrix}$$

$$=abc-a(9-i^{ 2 })+(i-5)(5c+ic-(7-i)(3+i))+(7+i)[(5+i)(3-i)-b(7-i)]$$

$$=abc-10a+5ic-c-25c-5ic-(i-5)(21+4i+1)+(7+i)(16-2i)-50b$$

$$=abc-10a-26c-2i+114+114+2i-50b$$

$$\Delta=abc-10a-26c+228-50b$$
Hence, $$\Delta $$ is real.
Question 2
1 / -0

If the lines $$\displaystyle y-x=5,3x+4y=1$$ and $$\displaystyle y=mx+3$$ are concurrent then the value of m is

A
$$\displaystyle \frac{19}{5}$$

B
$$\displaystyle 1$$

C
$$\displaystyle \frac{5}{19}$$

D
none of these

Solution

Given lines $$\displaystyle y-x=5,3x+4y=1$$ and $$\displaystyle y=mx+3$$
For concurrency,
$$\begin{vmatrix} -1 & 1 & 5 \\ 3 & 4 & 1 \\ -m & 1 & 3 \end{vmatrix}=0$$
$$\Rightarrow -5+19m=0$$
$$\Rightarrow \displaystyle m =\frac{5}{19}$$
Question 3
1 / -0

If $$\Delta =\begin{vmatrix}
\cos \theta /2 & 1 & 1\\
1 & \cos \theta /2 & -\cos \theta /2\\
-\cos \theta /2 & 1 & -1
\end{vmatrix}$$, If the minimun of $$\Delta $$ is $$m_{1}$$ and maximum of $$\Delta $$ is $$m_{2}$$, then $$\left [ m_{1}, m_{2} \right ]$$ are related

A
[-4, -2]

B
[2, 4]

C
[-4, 0]

D
[0, 2]

Solution

$$\Delta =\begin{vmatrix} \cos \theta /2 & 1 & 1 \\ 1 & \cos \theta /2 & -\cos \theta /2 \\ -\cos \theta /2 & 1 & -1 \end{vmatrix}$$

$$=-1(-1-\cos ^{ 2 }{ \theta /2 } )+1(1+\cos ^{ 2 }{ \theta /2 } )$$

$$=2+2\cos ^{ 2 }{ \theta /2 } $$

$$=3+\cos\theta$$

Now, $$\because -1\le \cos\theta \le 1$$
$$\Rightarrow 2\le 3+\cos\theta \le 4$$
Hence, $$m_{1}=2,m_{2}=4$$
Question 4
1 / -0

If $$A=\begin{pmatrix}
1 & 2 & 1\\
-1 & 0 & 3\\
2 & -1 & 1
\end{pmatrix}$$ then characteristic equation is given by

A
$$-\lambda ^{3}+2\lambda ^{2}-4\lambda +18=0$$

B
$$\lambda ^{3}+2\lambda ^{2}+4\lambda +18=0$$

C
$$2\lambda ^{3}-\lambda ^{2}+6\lambda -2=0$$

D
None of these

Solution

The characteristic equation of A is given by

$$|A-\lambda I|=0$$

$$\begin{vmatrix} 1-\lambda & 2 & 1 \\ -1 & -\lambda & 3 \\ 2 & -1 & 1-\lambda \end{vmatrix}=0$$

$$(1-\lambda)[-\lambda(1-\lambda)+3]-2(-1(1-\lambda)-6))+1(1+2\lambda)=0$$

$$\Rightarrow -\lambda ^{3}+2\lambda ^{2}-4\lambda +18=0$$
Question 5
1 / -0

Let $$ \displaystyle \begin{vmatrix}x^{2}+x &2x-1 &x+3 \\3x+1 &2+x^{2} &x^{3} \\x-3 &x^{2}+4 &2x\end{vmatrix} =px^{2}+qx^{6}+rx^{5}+sx^{4}+tx^{3}+ux^{2}+vx+w$$
then which of the following is not true?

A
$$\displaystyle w=21,v=75$$

B
$$\displaystyle q=0,s=-4 $$

C
$$\displaystyle p=0,t=-8$$

D
$$\displaystyle p=q=-1$$

Solution

Given, $$\begin{vmatrix} x^{ 2 }+x & 2x-1 & x+3 \\ 3x+1 & 2+x^{ 2 } & x^{ 3 } \\ x-3 & x^{ 2 }+4 & 2x \end{vmatrix}=px^{ 2 }+qx^{ 6 }+rx^{ 5 }+sx^{ 4 }+tx^{ 3 }+ux^{ 2 }+vx+w$$

$$(x^{ 2 }+x)[2x(2+x^{ 2 })-x^{ 3 }(4+x^{ 2 })]-(2x-1)[2x(3x+1)-x^{ 3 }(x-3)]+(x+3)[(3x+1)(x^{ 2 }+4)-(x-3)(x^{ 2 }+2)]=px^{ 2 }+qx^{ 6 }+rx^{ 5 }+sx^{ 4 }+tx^{ 3 }+ux^{ 2 }+vx+w$$

$$-x^{ 7 }-x^{ 6 }+0x^{ 5 }+4x^{ 4 }+8x^{ 3 }+34x^{ 2 }+75x+21=px^{ 7 }+qx^{ 6 }+rx^{ 5 }+sx^{ 4 }+tx^{ 3 }+ux^{ 2 }+vx+w$$

On comparing both sides we get $$p=q=-1w=21,v=75p=-1,t=8,q=0,t=-8$$
Question 6
1 / -0

If $$A=\begin{bmatrix}
-1 & -3 & -3\\
3 & 1 & -3\\
3 & -3 & 1
\end{bmatrix}$$ then adj (A) is

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

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

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

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

Solution

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

$$ C=\begin{bmatrix} -8 & -12 & -12 \\ 12 & 8 & -12 \\ 12 & -12 & 8 \end{bmatrix}$$

$$adj A=C^{T}=\begin{bmatrix} -8 & 12 & 12 \\ -12 & 8 & -12 \\ -12 & -12 & 8 \end{bmatrix}$$

$$=4\begin{bmatrix} -2 & 3 & 3 \\ -3 & 2 & -3 \\ -3 & -3 & 2 \end{bmatrix}$$
Question 7
1 / -0

$$\displaystyle A_{1},B_{1},C_{1}$$ are respectively the co-factors of $$\displaystyle a_{1},b_{1},c_{1}$$ of the determinant $$\displaystyle \Delta = \begin{vmatrix}a_{1} &b_{1} &c_{1} \\a_{2} &b_{2} &c_{2} \\a_{3} &b_{3} &c_{3}\end{vmatrix}$$ then $$\displaystyle \begin{vmatrix}B_{2} &C_{2} \\B_{3} &C_{3}\end{vmatrix}$$ equals

A
$$\displaystyle a_{1}a_{3}\Delta $$

B
$$\displaystyle (a_{1}-b_{1})\Delta $$

C
$$\displaystyle a_{1} \Delta $$

D
None of these

Solution

For $$\Delta =\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}$$
Let $$R=\begin{vmatrix} { A }_{ 1 } & { B }_{ 1 } & { C }_{ 1 } \\ { A }_{ 2 } & { B }_{ 2 } & { C }_{ 2 } \\ { A }_{ 3 } & { B }_{ 2 } & { C }_{ 3 } \end{vmatrix}$$ is the matrix of cofactor
Then $${ a }_{ 1 }\Delta =\begin{vmatrix} { B }_{ 2 } & { C }_{ 2 } \\ { B }_{ 2 } & { C }_{ 3 } \end{vmatrix}$$
Question 8
1 / -0

If $$\displaystyle A= \begin{bmatrix}2 &14 &17 \\0 &\sin 2x &\cos 2x \\0 &\cos 2x &\sin 2x \end{bmatrix}$$ then $$\displaystyle \left | A \right |$$ equals

A
$$\displaystyle \cos 2x $$

B
$$-2$$

C
$$\displaystyle -2 \cos 4x $$

D
$$\displaystyle \sin 4x $$

Solution

$$\displaystyle A= \begin{bmatrix}2 &14 &17 \\0 &\sin 2x &\cos 2x \\0 &\cos 2x &\sin 2x \end{bmatrix}$$

$$\displaystyle \left | A \right |=2(\sin^{2}2x-\cos^{2}2x)=-2\cos 4x$$
Question 9
1 / -0

If $$x$$ is a non-real cube root of $$-2$$, then the value of
$$\begin{vmatrix}
1 & 2x & 1\\
x^{2} & 1 & 3x^{2}\\
2 & 2x & 1
\end{vmatrix}$$ equals to

A
$$-7$$

B
$$-13$$

C
$$0$$

D
$$-12$$

Solution

Given, $$x=\sqrt [ 3 ]{ -2 } $$

$$\Rightarrow x^{3}=-2$$

$$\begin{vmatrix} 1 & 2x & 1 \\ x^{ 2 } & 1 & 3x^{ 2 } \\ 2 & 2x & 1 \end{vmatrix}$$

$$=(1-6x^{3})-2x(-5x^{2})+1(2x^{3}-2)$$

$$=13-20-6$$

$$=-13$$
Question 10
1 / -0

If a. b, c are negative and different real numbers then $$\displaystyle \Delta=\begin{vmatrix}a &b &c \\ b &c &a \\c &a &b \end{vmatrix}$$ is

A
$$\displaystyle < 0 $$

B
0

C
$$\displaystyle > 0 $$

D
$$\displaystyle \leq 0 $$

Solution

$$\displaystyle \Delta=\begin{vmatrix}a &b &c \\ b &c &a \\c &a &b \end{vmatrix}$$

$$\displaystyle \Delta =(a+b+c) \begin{vmatrix} 1 & b & c\\ 1 & c & a\\ 1 & a & b\end{vmatrix}$$

$$ =-(a+b+c) [a^{2}+b^{2}+c^{2}-ab-bc-ca] \Delta =-\left(\dfrac{a+b+c}{2}\right) [(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$$

$$\displaystyle \Delta > 0 $$ as $$\displaystyle a \neq b \neq c,a,b,c < 0$$

$$\displaystyle \therefore a+b+c <0 $$

$$\displaystyle \therefore \frac{-(a+b+c)}{2}>0 $$

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Determinants Test - 36

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Determinants Test - 36

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Question 1

real

imaginary

$$0$$

None of these

Solution

Question 2

$$\displaystyle \frac{19}{5}$$

$$\displaystyle 1$$

$$\displaystyle \frac{5}{19}$$

none of these

Solution

Question 3

[-4, -2]

[2, 4]

[-4, 0]

[0, 2]

Solution

Question 4

$$-\lambda ^{3}+2\lambda ^{2}-4\lambda +18=0$$

$$\lambda ^{3}+2\lambda ^{2}+4\lambda +18=0$$

$$2\lambda ^{3}-\lambda ^{2}+6\lambda -2=0$$

None of these

Solution

Question 5

$$\displaystyle w=21,v=75$$

$$\displaystyle q=0,s=-4 $$

$$\displaystyle p=0,t=-8$$

$$\displaystyle p=q=-1$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$\displaystyle a_{1}a_{3}\Delta $$

$$\displaystyle (a_{1}-b_{1})\Delta $$

$$\displaystyle a_{1} \Delta $$

None of these

Solution

Question 8

$$\displaystyle \cos 2x $$

$$-2$$

$$\displaystyle -2 \cos 4x $$

$$\displaystyle \sin 4x $$

Solution

Question 9

$$-7$$

$$-13$$

$$0$$

$$-12$$

Solution

Question 10

$$\displaystyle < 0 $$

0

$$\displaystyle > 0 $$

$$\displaystyle \leq 0 $$

Solution

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