Determinants Test

Result

Determinants Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{bmatrix} 2 & 1 & -1 \\ 0 & 1 & 4 \\ 0 & 0 & 3 \end{bmatrix}$$, then $$tr(adj(adj\ A))$$ is equal to

A
$$18$$

B
$$24$$

C
$$36$$

D
$$48$$

Solution
Question 2
1 / -0

Let $$A =[a_{ij}]$$ be a $$3 \times 3 $$ matrix whose determinant is $$5$$. Then the determinant of the matrix $$B = [ 2^{i-j} a_{ij} ]$$ is

A
$$5$$

B
$$10$$

C
$$20$$

D
$$40$$

Solution
Question 3
1 / -0

$$A=\begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}$$ and A (adj A)=KI, then the value of 'K' is ...

A
2

B
-2

C
10

D
-10

Solution
Question 4
1 / -0

If $$\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{matrix} \right]$$ then $$|adj\ (adj\ A)|$$ is equal to

A
$$18^{3}$$

B
$$18^{2}$$

C
$$18^{4}$$

D
$$18^{6}$$

Solution
Question 5
1 / -0

If $$A=\begin{bmatrix} 1 & -2 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$$, then $$A.adj(a)=$$

A
$$\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$

B
$$\begin{bmatrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{bmatrix}$$

C
$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

D
$$\begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix}$$

Solution
Question 6
1 / -0

If $$A$$ is a square matrix of order $$n$$ and $$|A|=D$$ and $$|adj A|=D^{\prime}$$, then

A
$$DD^{\prime}=D^{2}$$

B
$$DD^{-1}=D^{-1}$$

C
$$DD^{\prime}=D^{n-1}$$

D
$$DD^{\prime}=D^{n}$$

Solution
Question 7
1 / -0

If the points (k, 2 - 2k) (1 - k, 2k) and (-k -4, 6 -2x) be collinear the possible values of k are

A
-$$\dfrac{1}{2}$$

B
$$\dfrac{1}{2}$$

C
1

D
- 1

Solution

Given three points
$$(K,2-2K)$$
$$(1-K,2K)$$
$$(-K-4,6-2K)$$
Noe three points $$\left( { x }_{ 1 },{ y }_{ 1 } \right) \left( { x }_{ 2 },{ y }_{ 2 } \right) \left( { x }_{ 3 },{ y }_{ 3 } \right) $$ will be collinear
if
$$\Rightarrow \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] =0\quad \longrightarrow \left( 1 \right) $$
Putting $${ x }_{ 1 }=K;{ y }_{ 1 }=2-2K$$
$${ x }_{ 2 }=1-K;{ y }_{ 2 }=2K$$
$${ x }_{ 3 }=-K-4;{ y }_{ 3 }=6-2K$$
in the equation $$(1)$$ we get
$$\Rightarrow K\left( 2K-6+2K \right) +\left( 1-K \right) \left( 6-2K-2+2K \right) +\left( K-4 \right) \left( 2-2K-2K \right) =0$$
$$\Rightarrow K\left( 4K-6 \right) +\left( 1-K \right) \left( 4 \right) +\left( -K-4 \right) \left( 2-4K \right) =0$$
$$\Rightarrow { 4K }^{ 2 }-6K+4-4K-\left( 2K-{ 4K }^{ 2 }+8-16K \right) =0$$
$$\Rightarrow { 4K }^{ 2 }-6K+4-4K-2K+{ 4K }^{ 2 }-8+16K=0$$
$$\Rightarrow { 8K }^{ 2 }+4K-4=0$$
$$\Rightarrow { 2K }^{ 2 }+K-1=0$$
$$\Rightarrow { 2K }^{ 2 }+2K-K-1=0$$
$$\Rightarrow 2K\left( K+1 \right) -1\left( K+1 \right) =0$$
$$\Rightarrow \left( 2K-1 \right) \left( K+1 \right) =0$$
$$\therefore$$ $$K=1/2,-1$$
[option B, option D]
Question 8
1 / -0

If $$A = \begin{bmatrix} 4 & 2 \\ 3 & 4 \end{bmatrix}$$ then |adj A| is equal to

A
16

B
10

C
6

D
none of these

Solution
Question 9
1 / -0

If $$A = \begin{bmatrix}1 & 2 & -1\\ -1 & 2 & 2\\ 2 & -1 & 1\end{bmatrix}$$, then $$:|adj (adj.A)|=$$

A
$$(17)^1$$

B
$$(17)^2$$

C
$$(17)^3$$

D
$$(17)^4$$

Solution
Question 10
1 / -0

If $$A=\begin{bmatrix} a & c & b\\ b & a & c\\ c & b & a\end{bmatrix}$$ then the cofactor of $$a_{32}$$ in $$A+A^T$$ is?

A
$$-(2a(b+c)-(b+c)^2)$$

B
$$ac-b^2$$

C
$$a^2-bc$$

D
$$2a(a+c)-(a+c)^2$$

Solution