Determinants Test

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Determinants Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

If $$\Delta_1=\begin{vmatrix} x & b & b\\ a & x & b\\ a & a & x\end{vmatrix}$$ and $$\Delta_2=\begin{vmatrix} x & b\\ a & x\end{vmatrix}$$ are the given determinants, then.

A
$$\Delta_1=3(\Delta_2)^2$$

B
$$(d/dx)\Delta_1=3\Delta_2$$

C
$$(d/dx)\Delta_1=3(\Delta_2)^2$$

D
$$\Delta_1=3\Delta_2^{3/2}$$

Solution
Question 2
1 / -0

If a, b, c are different and $$\begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c\\ x+b & x+c & 0\end{vmatrix}=0$$, then x is equal to.

A
a

B
b

C
c

D
$$0$$

Solution
Question 3
1 / -0

If $$\left |\begin{array}{111}6i & -3i & 1 \\4 & 3i & -1 \\20 & 3 & i \\\end {array}\right | =x+iy$$, then

A
$$x=3, y=1$$

B
$$x = 1, y=3$$

C
$$x=0, y=3$$

D
$$x=0, y=0$$

Solution

$$\begin{vmatrix} 6i & -3i & 1 \\ 4 & 3 i & -1 \\ 20 & 3 & i \end{vmatrix}$$
$$\Rightarrow 6i((3i. i) - (-3)) + 3i (4 . i + 20) + 1 (12 - 60 i)$$
$$\Rightarrow 6i (-3 + 3) + 12 (-1) + 60 i + 12 - 60 i$$
$$\Rightarrow 0$$ $$[\because i^2 = -1]$$
$$x + i y = 0 + i(0)$$
$$x = 0 , y = 0$$
Question 4
1 / -0

If A = $$\begin{bmatrix}
a & 0 & 0 \\[0.3em]
0 & a & 0 \\[0.3em]
0 & 0 & a
\end{bmatrix}$$, then the value of |A| |Adj. A|

A
$$a^3$$

B
$$a^6$$

C
$$a^9$$

D
$$a^{27}$$

Solution

As$$|A|=(a)^3$$

$$|A|(Adj. A) = |A|$$ $$\begin{bmatrix}
a & 0 & 0 \\[0.3em]
0 & a & 0 \\[0.3em]
0 & 0 & a
\end{bmatrix}$$

=$$\begin{bmatrix}
|A| & 0 & 0 \\[0.3em]
0 & |A| & 0 \\[0.3em]
0 & 0 & |A|
\end{bmatrix}$$ where $$|A|$$ = $$a^3$$

Take determinant of both sides

|A|. |Adj. A| =$$\begin{vmatrix}
|A| & 0 & 0 \\[0.3em]
0 & |A| & 0 \\[0.3em]
0 & 0 & |A|
\end{vmatrix} |A|^3 = (a^3)^3 = a^9$$
Question 5
1 / -0

If $$f(x) =$$ $$
\left |
\begin{array}{111}
1 & x & x+1 \\
2x & x(x-1) & (x+1)x \\
3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \\
\end {array}
\right |
$$
then f(100) is equal to

A
0

B
1

C
100

D
-100

Solution

$$f(x) = \begin{vmatrix} 1 & x & x + 1 \\ 2x & x(x - 1) & (x + 1)x \\ 3x(x -1) & x(x - 1) (x - 2) & (x + 1) x(x - 1) \end{vmatrix}$$
$$x \times x \times (x - 1) \, \begin{vmatrix} 1 & x & x + 1 \\ 2 & x - 1 & x + 1 \\ 3 & x - 2 & x + 1 \end{vmatrix}$$
$$x^2 (x - 1)(x + 1) \, \begin{vmatrix} 1 & x & 1 \\ 2 & x - 1 & 1 \\ 3 & x - 2 & 1 \end{vmatrix}$$
operating : $$R_1 \rightarrow R_2 - R_1$$ & $$R_2 \rightarrow R_3 - R_2$$
$$x^2 (x^2 - 1) \, \begin{vmatrix} 1 & -1 & 0 \\ 1 & -1 & 0 \\ 3 & x - 2 & 1 \end{vmatrix}$$
As two rows of the determinant is zero
$$\therefore$$ Determinant $$= 0$$
Hence, $$\boxed {f(x) = 0}$$
$$\therefore$$ for any value of $$x ;\ f(x) = 0$$
Hence, $$\boxed {f(100) = 0}$$
Question 6
1 / -0

If A = $$\begin{bmatrix}
a & 0 & 0 \\[0.3em]
0 & a & 0 \\[0.3em]
0 & 0 & a
\end{bmatrix}$$, then the value of |Adj. A| is equal to

A
$$a^3$$

B
$$a^6$$

C
$$a^9$$

D
$$a^{27}$$

Solution

A(Adj. A) = |A| $$I_3$$ = |A| $$\begin{bmatrix}
1 & 0 & 0 \\[0.3em]
0 & 1 & 0 \\[0.3em]
0 & 0 & 1
\end{bmatrix}$$
=$$\begin{bmatrix}
|A| & 0 & 0 \\[0.3em]
0 & |A| & 0 \\[0.3em]
0 & 0 & |A|
\end{bmatrix}$$ where |A| = $$a^3$$
Take determinant of both sides
|A|. |Adj. A| =$$\begin{vmatrix}
|A| & 0 & 0 \\[0.3em]
0 & |A| & 0 \\[0.3em]
0 & 0 & |A|
\end{vmatrix} |A|^3 =|Adj. A| = |A|^{n-1} = |A|^2 = a^6$$
Question 7
1 / -0

If A = $$ \begin{bmatrix} \alpha & 2 \\ 2 & \alpha\end{bmatrix}$$ and | A$$^3$$ | = 125 then $$\alpha$$ is

A
$$\pm1$$

B
=2

C
$$\pm3$$

D
$$\pm5$$

Solution

$$A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$$. $$|A^3| = 125$$
$$|A| = \alpha^2 - 4$$
we know by properties of determinants that
$$|A^3| = |A|^3$$
$$\Rightarrow \ (\alpha^2 - 4)^3 = 125$$
$$\Rightarrow \ \alpha^2 - 4 = 5$$
$$\Rightarrow \ a^2 = 9$$
$$\Rightarrow \ \boxed {\alpha = \pm 3}$$
Hence, option $$(C)$$ is correct
Question 8
1 / -0

If the points $$(-2, -5), (2, -2), (8, a)$$ are collinear, then the value of $$a$$ is ________.

A
$$\dfrac 52$$

B
$$\dfrac 32$$

C
$$\dfrac 72$$

D
None of these

Solution

If three points are collinear, then the slope of the line joining the first two points is equal to that of the last two.

Apply the slope formula $$\dfrac{(y2-y1)}{(x2-x1).}$$

If collinear then
$$\dfrac { -2+5 }{ 2+2 } =\dfrac { a+2 }{ 6 } \\ \dfrac { 9 }{ 2 } =a+2\\ a=\dfrac { 5 }{ 2 } $$
Question 9
1 / -0

If $$ \begin{vmatrix}a & a & x \\ m & m & m \\b & x & b\end{vmatrix}=0$$ then $$x$$ is:

A
$$a$$

B
$$b$$

C
$$a$$ or $$b$$

D
$$0$$

Solution

Given $$\begin{vmatrix} a & a & x \\ m & m & m \\ b & x & b \end{vmatrix}=0$$

Expanding the determinant we get

$$a(mb-mx)-a(mb-mb)+x(mx-mb)=0\\ \Rightarrow a(mb-mx)-x(mb-mx)=0\\ \Rightarrow (a-x)(mb-mx)=0\\ \Rightarrow (a-x)(b-x)=0$$

The solution for this quadratic equation is either $$x=a$$ or $$x=b$$
Question 10
1 / -0

The points $$(-a, -b), (0, 0), (a, b)$$ $$(a^2,ab)$$ are

A
collinear

B
vertices of rectangle

C
vertices of parallelgram

D
none of these

Solution

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Re-Attempt Weekly Quiz Competition

Determinants Test - 50

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Determinants Test - 50

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SHARING IS CARING

Question 1

$$\Delta_1=3(\Delta_2)^2$$

$$(d/dx)\Delta_1=3\Delta_2$$

$$(d/dx)\Delta_1=3(\Delta_2)^2$$

$$\Delta_1=3\Delta_2^{3/2}$$

Solution

Question 2

a

b

c

$$0$$

Solution

Question 3

$$x=3, y=1$$

$$x = 1, y=3$$

$$x=0, y=3$$

$$x=0, y=0$$

Solution

Question 4

$$a^3$$

$$a^6$$

$$a^9$$

$$a^{27}$$

Solution

Question 5

0

1

100

-100

Solution

Question 6

$$a^3$$

$$a^6$$

$$a^9$$

$$a^{27}$$

Solution

Question 7

$$\pm1$$

=2

$$\pm3$$

$$\pm5$$

Solution

Question 8

$$\dfrac 52$$

$$\dfrac 32$$

$$\dfrac 72$$

None of these

Solution

Question 9

$$a$$

$$b$$

$$a$$ or $$b$$

$$0$$

Solution

Question 10

collinear

vertices of rectangle

vertices of parallelgram

none of these

Solution

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