Determinants Test

Result

Determinants Test - 46

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$m = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$$ and $$n = \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$$, then what is the value of the determinant of $$m \cos \theta - n \sin \theta$$?

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

$$m=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\quad n=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$
$$m\cos { \theta } -n\sin { \theta } $$
$$m\cos { \theta } =\begin{bmatrix} \cos { \theta } & 0 \\ 0 & \cos { \theta } \end{bmatrix}$$
$$n\sin { \theta } =\begin{bmatrix} 0 & \sin { \theta } \\ -\sin { \theta } & 0 \end{bmatrix}$$
$$m\cos { \theta } -n\sin { \theta } =\begin{bmatrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{bmatrix}$$
$$det=\cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } $$
$$=1$$
Option C
Question 2
1 / -0

Let $$A = \begin{pmatrix}x + 2 & 3x\\ 3 & x + 2\end{pmatrix}, B = \begin{pmatrix}x & 0\\ 5 & x + 2\end{pmatrix}$$. Then all solutions of the equation $$det (AB) = 0$$ is

A
$$1, -1, 0, 2$$

B
$$1, 4, 0, -2$$

C
$$1, -1, 4, 3$$

D
$$-1, 4, 0, 3$$

Solution

Given : $$A = \begin{pmatrix}x + 2 & 3x\\ 3 & x + 2\end{pmatrix}$$ and $$B = \begin{pmatrix}x & 0\\ 5 & x + 2\end{pmatrix}$$

$$\implies AB = \begin{bmatrix}x^{2} + 17x & 3x^{2} + 6x\\ 8x + 10 & (x + 2)^{2}\end{bmatrix} = 0$$

Now, $$det(AB) = \begin{vmatrix}x^{2} + 17x & 3x^{2} + 6x\\ 8x + 10 & (x + 2)^{2}\end{vmatrix} = 0$$
$$\implies (x^{2}+17x)(x+2)^{2}-(3x^{2}+6x)(8x+10)=0$$
$$\implies x(x+17)(x+2)^{2}-3x(x+2)(8x+10)=0$$
$$\implies x(x+2)[(x+17)(x+2)-3(8x+10)]=0$$
$$\implies x(x+2)[x^{2}+19x+34-24x-30]=0$$
$$\implies x(x+2)[x^{2}-5x+4]=0$$
$$\implies x(x + 2)(x - 4) (x - 1) = 0$$
$$\implies x = 0, -2, 1, 4$$
Question 3
1 / -0

$$\begin{vmatrix}1 & 1 & 1\\ a & b & c\\ a^2 - bc & b^2 - ca & c^2 - ab\end{vmatrix} = $$

A
$$0$$

B
$$1$$

C
$$abc$$

D
$$(a -b), \,(b-c),\, (c-a)$$

Solution

$$=\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 2 }-bc & { b }^{ 2 }-ca & { c }^{ 2 }-ab \end{vmatrix} $$

$$ =\begin{vmatrix} a & 1 & 0 \\ a-b & b & \left( c-b \right) \\ \left( a-b \right) \left( a+b \right) +c\left( a-b \right) & { b }^{ 2 }-ca & \left( c-b \right) \left( c+b \right) +a\left( c-b \right) \end{vmatrix}$$ .......... $$ { C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 2\\ }$$ and $$ { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }$$

$$ =\begin{vmatrix} 0 & 1 & 0 \\ \left( a-b \right) & b & \left( c-b \right) \\ \left( a-b \right) \left( a+b+c \right) & { b }^{ 2 }-ac & \left( c-b \right) \left( a+b+c \right) \end{vmatrix}$$

$$= \left( a-b \right) \left( c-b \right) \begin{vmatrix} 0 & 1 & 0 \\ 1 & b & 1 \\ a+b+c & { b }^{ 2 }-ac & a+b+c \end{vmatrix}$$

$$ =\left( a-b \right) \left( c-b \right) \begin{vmatrix} 0 & 1 & 0 \\ 0 & b & 1 \\ 0 & { b }^{ 2 }-ac & a+b+c \end{vmatrix}$$
$$ = 0$$
Question 4
1 / -0

If $$A = \begin{bmatrix} 1& 2\\ 3 & 4\end{bmatrix}$$, then $$A^{-1} =$$

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

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

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

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

Solution

For $$2 \times 2$$ matrix $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$, we swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant $$(ad-bc).$$
So for $$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ inverse of matrice is
$$ad-bc=1 \times 4-2 \times 3=-2$$
$$A^{-1}=$$ $$\dfrac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$$
Question 5
1 / -0

If $$\alpha, \beta, \gamma$$ are the roots of the equation $$x^{3} + px + q = 0$$ then the value of the determinant $$\begin{vmatrix} \alpha& \beta & \gamma\\ \beta & \gamma & \alpha\\ \gamma & \alpha & \beta\end{vmatrix}$$ is

A
$$q$$

B
$$0$$

C
$$p$$

D
$$p^{2} - 2q$$

Solution

As $$ \alpha,\beta,\gamma$$ are the roots of given equation

$$\implies (\alpha +\beta +\gamma )=0$$
Now,
$$\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix}$$

$$R_1\rightarrow R_1+R_2+R_3$$

$$= \begin{vmatrix} \alpha +\beta +\gamma & \alpha +\beta +\gamma & \alpha +\beta +\gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix}$$

$$= (\alpha +\beta +\gamma )\begin{vmatrix} 1 & 1 & 1 \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix}$$

$$= (\alpha +\beta +\gamma )(\alpha \beta +\alpha \gamma +\beta \gamma -\alpha ^{ 2 }-\beta ^{ 2 }-\gamma ^{ 2 })$$

$$\implies (\alpha +\beta +\gamma )(3(\alpha \beta +\alpha \gamma +\beta \gamma )-(\alpha +\beta +\gamma )^2)=0$$ as $$(\alpha +\beta +\gamma )=0$$
Hence, option B is correct.
Question 6
1 / -0

Consider the following statements in respect of the determinant $$\begin{vmatrix}{\cos}^2\dfrac{\alpha}{2}&{\sin}^2\dfrac{\alpha}{2}\\{\sin}^2\dfrac{\beta}{2}&{\cos}^2\dfrac{\beta}{2}\end{vmatrix}$$ where $$\alpha , \beta$$ are complementary angles
1. The value of the determinant is $$\dfrac{1}{\sqrt{2}}\cos \begin{pmatrix}\dfrac{\alpha - \beta}{2}\end{pmatrix}$$.
2. The maximum value of the determinant is $$\dfrac{1}{\sqrt{2}}$$.
Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Solution:
We have, $$\triangle=\begin{bmatrix}\cos^2\cfrac{\alpha}{2}&\sin^2\cfrac{\alpha}2\\\sin^2\cfrac{\beta}2&\cos^2\cfrac{\beta}{2}\end{bmatrix}$$
$$=\cos^2\cfrac{\alpha}2\cos^2\cfrac{\beta}2-\sin^2\cfrac{\alpha}2\sin^2\cfrac{\beta}{2}$$
$$=\left(\cos\cfrac{\alpha}{2}\cos\cfrac{\beta}{2} +\sin\cfrac{\alpha}{2}\sin \cfrac{\beta}{2}\right)$$ $$\left(\cos\cfrac{\alpha}{2}\cos\cfrac{\beta}{2}-\sin\cfrac{\alpha}{2}\sin\cfrac{\beta}{2}\right)$$
$$=\cos\left(\cfrac{\alpha-\beta}{2}\right)\cos45^\circ$$ ........ $$[\alpha , \beta ]$$ are complementry angles.
$$=\cfrac1{\sqrt2}\cos\left(\cfrac{\alpha-\beta}{2}\right)$$
Maximum value of $$\cos\left(\cfrac{\alpha-\beta}{2}\right)=1$$
So, maximum value of $$\cfrac1{\sqrt2}\cos\left(\cfrac{\alpha-\beta}{2}\right)=\cfrac1{\sqrt2}$$
Hence, C is the correct option.
Question 7
1 / -0

If $$A = \begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3}\end{vmatrix}$$ and $$B = \begin{vmatrix} c_{1}& c_{2} & c_{3}\\ a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\end{vmatrix}$$ then.

A
$$A = -B$$

B
$$A = B$$

C
$$B = 0$$

D
$$B = A^{2}$$

Solution

Let determinant of matrix, $$\begin{vmatrix} c_{ 1 } & c_{ 2 } & c_{ 3 } \\ { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \end{vmatrix}$$ be $$y$$

$$\begin{vmatrix} c_{ 1 } & c_{ 2 } & c_{ 3 } \\ { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \end{vmatrix}$$
exchanging row $$1$$ and row $$2$$, the determinant becomes $$-y$$
$$ \begin{vmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ c_{ 1 } & c_{ 2 } & c_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \end{vmatrix}$$
exchanging row $$2$$ and row $$3$$ the determinant becomes $$y$$
$$ \begin{vmatrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ c_{ 1 } & c_{ 2 } & c_{ 3 } \end{vmatrix}$$
Now taking transpose the determinant will remain $$y$$
$$ \begin{vmatrix} { a }_{ 1 } & b_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & b_{ 3 } & c_{ 3 } \end{vmatrix}\\ \\ $$
So, $$A=B$$
(b)
Question 8
1 / -0

If $$\begin{vmatrix}1 & sin \theta &1 \\ -sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix}$$ then,

A
$$\Delta =0$$

B
$$\Delta \in (0, \infty )$$

C
$$\Delta \in [-1, 2]$$

D
$$\Delta \in [2, 4]$$

Solution

Let $$\Delta = \begin{vmatrix}1 & sin \theta &1 \\ -sin \theta & 1 & sin \theta\\ -1 & -sin \theta & 1\end{vmatrix}$$
$$=1(1+sin^2 \theta)-sin \theta (0)+1(sin^2 \theta +1)$$
$$= 2 (1+sin^2 \theta)$$
$$\because 0 \leq sin^2 \theta \leq 1$$
$$\Rightarrow 1\leq 1+sin^2 \theta\leq 2$$
$$\Rightarrow 2 \leq 2 (1+sin^2 \theta)\leq 4$$
$$\Rightarrow 2 \leq \Delta \leq 4$$
$$\Rightarrow \Delta \in [2,4]$$
Question 9
1 / -0

If $$ C = 2 \cos \theta $$ , then the value of the determinant $$ \triangle = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix} $$ is :

A
$$ \dfrac { \sin 4 \theta } { \sin \theta} $$

B
$$ \dfrac { 2 \sin^2 \theta } { \sin \theta} $$

C
$$ 4 \cos^2 \theta (2 \cos \theta - 1) $$

D
None of these above

Solution

Let $$ \triangle = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix} = C (C^2-1) -1(C-6) = C^3 - 2C + 6 $$
Put $$ C = 2 \cos \theta $$ we get
$$ \triangle = (2 \cos \theta)^3 - 2(2 \cos \theta) + 6 $$
$$ = 8 \cos^3 \theta - 4 \cos \theta + 6 $$
Question 10
1 / -0

If $$\begin{vmatrix} x & 2 & 8 \\ 2 & 8 & x \\ 8 & x & 2 \end{vmatrix}=\begin{vmatrix} 3 & x & 7 \\ x & 7 & 3 \\ 7 & 3 & x \end{vmatrix}=\begin{vmatrix} 5 & 5 & x \\ 5 & x & 5 \\ x & 5 & 5 \end{vmatrix}=0$$ then $$x$$ is equal to

A
$$0$$

B
$$-10$$

C
$$3$$

D
None of these

Solution

On applying $${ R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$ in each determinant, we can take out $$\left( x+10 \right) $$ common.
Then, we get $$x+10=0$$
$$\Rightarrow x=-10$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Determinants Test - 46

Result

Determinants Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 2

$$1, -1, 0, 2$$

$$1, 4, 0, -2$$

$$1, -1, 4, 3$$

$$-1, 4, 0, 3$$

Solution

Question 3

$$0$$

$$1$$

$$abc$$

$$(a -b), \,(b-c),\, (c-a)$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$$q$$

$$0$$

$$p$$

$$p^{2} - 2q$$

Solution

Question 6

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 7

$$A = -B$$

$$A = B$$

$$B = 0$$

$$B = A^{2}$$

Solution

Question 8

$$\Delta =0$$

$$\Delta \in (0, \infty )$$

$$\Delta \in [-1, 2]$$

$$\Delta \in [2, 4]$$

Solution

Question 9

$$ \dfrac { \sin 4 \theta } { \sin \theta} $$

$$ \dfrac { 2 \sin^2 \theta } { \sin \theta} $$

$$ 4 \cos^2 \theta (2 \cos \theta - 1) $$

None of these above

Solution

Question 10

$$0$$

$$-10$$

$$3$$

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.