Matrices Test

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Matrices Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle A=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix},$$ then $${A}^{50}$$ is

A
$$\begin{bmatrix} 1 & 0 \\ 0 & 50 \end{bmatrix}$$

B
$$\begin{bmatrix} 1 & 0 \\ 50 & 1 \end{bmatrix}$$

C
$$\begin{bmatrix} 1 & 25 \\ 0 & 1 \end{bmatrix}$$

D
None of these

Solution

$$\displaystyle A=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}$$
$$\displaystyle{ A }^{ 2 }=\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ \cfrac { 1 }{ 2 } & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2\left( \cfrac { 1 }{ 2 } \right) & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$
$${ A }^{ 4 }=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\\ { A }^{ 8 }=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 4 & 1 \end{bmatrix}$$
Similarly $$\displaystyle{ A }^{ n }=\begin{bmatrix} 1 & 0 \\ \cfrac { n }{ 2 } & 1 \end{bmatrix}$$ for even $$n$$
$$\Rightarrow { A }^{ 50 }=\begin{bmatrix} 1 & 0 \\ 25 & 1 \end{bmatrix}$$
Question 2
1 / -0

If $$\displaystyle \left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\: \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O$$ then $$x$$ is

A
$$2$$

B
$$-2$$

C
$$14$$

D
none of these

Solution

$$\displaystyle \left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix}\: \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O$$
$$\Rightarrow \begin{bmatrix} 16+2x & \quad 6+5x & \quad 4+x \end{bmatrix}\begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix}=O\\ \Rightarrow { x }^{ 2 }+16x+28\quad =\quad 0\\ \therefore \quad x=-2\quad or\quad x=-4$$
Question 3
1 / -0

If $$\displaystyle \begin{bmatrix} 2 & -3 \\ 1 & \lambda \end{bmatrix} \times \begin{bmatrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{bmatrix} = \begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$ then

A
$$\displaystyle \lambda = 3, \mu = 4$$

B
$$\displaystyle \lambda = 4, \mu = 3$$

C
$$\lambda = 2 , \mu = 5$$

D
none of these

Solution

Since, $$\begin{bmatrix} 2 & -3 \\ 1 & \lambda \end{bmatrix}\times \begin{bmatrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$

$$\Rightarrow \begin{bmatrix} 2-0 & 10-6 & 2\mu +9 \\ 1+0 & 5+2\lambda & \mu -3\lambda \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$

$$\Rightarrow \begin{bmatrix} 2 & 4 & 2\mu +9 \\ 1 & 5+2\lambda & \mu -3\lambda \end{bmatrix}=\begin{bmatrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{bmatrix}$$

$$\Rightarrow 2\mu +9=1$$ and $$5+2\lambda =-1$$
$$\Rightarrow \mu =-4,\lambda =-3$$
But these value does not satisfy $$\mu -3\lambda =13$$
Question 4
1 / -0

If $$\displaystyle A = \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}$$ and $$\displaystyle B = \begin{bmatrix} a^2 & ab & ac \\ ba & b^2 & bc \\ ca & cb & c^2 \end{bmatrix}$$ then $$AB$$ is equal to

A
$$[0]$$

B
$$I$$

C
$$2I$$

D
none of these

Solution

$$\displaystyle A = \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}$$ and $$\displaystyle B = \begin{bmatrix} a^2 & ab & ac \\ ba & b^2 & bc \\ ca & cb & c^2 \end{bmatrix}$$

Now, $$AB=\begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}\begin{bmatrix} a^{ 2 } & ab & ac \\ ba & b^{ 2 } & bc \\ ca & cb & c^{ 2 } \end{bmatrix}$$

$$\Rightarrow AB=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
Question 5
1 / -0

If the matrix $$\displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ then $$\displaystyle A^2$$ is

A
$$\displaystyle \begin{bmatrix} a^2 & b^2 \\ c^2 & d^2 \end{bmatrix}$$

B
$$\displaystyle \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix}$$

C
nonexistent

D
none of these

Solution

Given $$\displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

$$A^{2}=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

$$=\displaystyle \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix}$$
Question 6
1 / -0

If $$A = \begin{bmatrix} 0 & 2 & 3 \\ 3 & 5 & 7 \end{bmatrix},$$ $$B = \begin{bmatrix} 1 & 3 & 7 \\ 2 & 4 & 1 \end{bmatrix}$$ and $$A+B = \begin{bmatrix} 1 & 5 & 10 \\ 5 & k & 8 \end{bmatrix},\\ $$ then find the value of $$k .$$

A
9

B
4

C
5

D
1

Solution

Given $$A = \begin{bmatrix} 0 & 2 & 3 \\ 3 & 5 &

7 \end{bmatrix}$$ and $$B = \begin{bmatrix} 1 & 3 & 7 \\ 2 &

4 & 1 \end{bmatrix}$$

$$\therefore A+B = \mbox{sum of respective elements of A and B}=\begin{bmatrix} 1 & 5

& 10 \\ 5 & 9 & 8 \end{bmatrix} $$

Hence the value of $$k$$ is $$9$$
Question 7
1 / -0

If $$\displaystyle A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}$$ then $$\displaystyle A^2$$ is equal to

A
$$A$$

B
$$I$$

C
$$\displaystyle A^T$$

D
none of these

Solution

Given, $$\displaystyle A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}$$

$$A^{ 2 }=\begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}\begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

$$\Rightarrow A^{2}=I$$
Question 8
1 / -0

If $$A$$ be a matrix such that $$\displaystyle A \times \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$ then $$A$$ is

A
$$\displaystyle \begin{bmatrix}

2 & 4 \\

1 & -1

\end{bmatrix}$$

B
$$\displaystyle \begin{bmatrix}

-1 & 1 \\

4 & 2

\end{bmatrix}$$

C
$$\displaystyle \begin{bmatrix}

4 & 2 \\

-1 & 1

\end{bmatrix}$$

D
none of these

Solution

$$\displaystyle A \times \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$

We will check using options
Option A
Consider, $$A\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$
   $$=\begin{bmatrix} 2 & 4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$

  $$=\begin{bmatrix} 6 & 12 \\ 0 & -6 \end{bmatrix}$$
$$\ne RHS$$

Hence, option $$A$$ cannot be matrix $$A$$.

Option B,
Consider, $$A\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$
  $$\begin{bmatrix} -1 & 1 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$

  $$=\begin{bmatrix} 0 & 6 \\ 6 & 0 \end{bmatrix}$$

  $$\ne RHS$$
Hence, option $$B$$ cannot be matrix $$A$$.

Option C,
Consider, $$A\begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix}$$

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

$$=\begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$

  $$= RHS$$
Hence, option C is correct for $$A$$.
Question 9
1 / -0

If $$\displaystyle \begin{bmatrix} x+y & y \\ 2x & x-y \end{bmatrix} \: \begin{bmatrix} 2 \\ -2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$ then $$x-y$$ is equal to

A
$$2$$

B
$$-2$$

C
$$4$$

D
$$6$$

Solution

Comparing LHS and RHS gives us
$$2x+2y-2y=3$$ ...(i)
And
$$4x-2x+2y=2$$ ...(ii)
from i
$$2x=3$$ and from ii
$$2x+2y=2$$
Therefore
$$3+2y=3$$
$$y=\dfrac{-1}{2}$$ and $$x=\dfrac{3}{2}$$
Hence
$$x-y=\dfrac{3}{2}-(\dfrac{-1}{2})$$
$$=\dfrac{4}{2}$$
$$=2$$.
Question 10
1 / -0

If $$A = \begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$$

If $$A^2 - 4A =pI $$ where $$I$$ and $$O$$ are the unit matrix and the null matrix of order $$3$$ respectively. Find the value of $$p$$

A
$$p=2$$

B
$$p=3$$

C
$$p=4$$

D
$$p=5$$

Solution

Given

$$\quad A = \begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$$

$$\therefore

A^2 = A.A = \begin{pmatrix}1& 2 & 2 \\ 2 & 1 & 2 \\ 2

& 2 & 1\end{pmatrix}\times\begin{pmatrix}1& 2 & 2 \\ 2

& 1 & 2 \\ 2 & 2 & 1\end{pmatrix}$$

$$\quad

= \begin{pmatrix}1+4+4 & 2+2+4 &

2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 &

4+4+1 \end{pmatrix}$$

$$\quad

= \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8

& 9\end{pmatrix}$$

$$\therefore \quad A^2 - 4A

= \begin{pmatrix}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8

& 9\end{pmatrix} - 4\begin{pmatrix}1& 2 & 2 \\ 2 & 1

& 2 \\ 2 & 2 & 1\end{pmatrix}$$

$$\quad

= \begin{pmatrix}9&8&8 \\ 8&9&8

\\ 8&8&9\end{pmatrix} - \begin{pmatrix}4&8&8 \\

8&4&4 \\ 4&4&8\end{pmatrix} =

\begin{pmatrix}5&0&0 \\ 0&5&0 \\

0&0&5\end{pmatrix}=5I$$

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Re-Attempt Weekly Quiz Competition

Matrices Test - 30

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Matrices Test - 30

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Question 1

$$\begin{bmatrix} 1 & 0 \\ 0 & 50 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 \\ 50 & 1 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 25 \\ 0 & 1 \end{bmatrix}$$

None of these

Solution

Question 2

$$2$$

$$-2$$

$$14$$

none of these

Solution

Question 3

$$\displaystyle \lambda = 3, \mu = 4$$

$$\displaystyle \lambda = 4, \mu = 3$$

$$\lambda = 2 , \mu = 5$$

none of these

Solution

Question 4

$$[0]$$

$$I$$

$$2I$$

none of these

Solution

Question 5

$$\displaystyle \begin{bmatrix} a^2 & b^2 \\ c^2 & d^2 \end{bmatrix}$$

$$\displaystyle \begin{bmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{bmatrix}$$

nonexistent

none of these

Solution

Question 6

9

4

5

1

Solution

Question 7

$$A$$

$$I$$

$$\displaystyle A^T$$

none of these

Solution

Question 8

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

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

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

none of these

Solution

Question 9

$$2$$

$$-2$$

$$4$$

$$6$$

Solution

Question 10

$$p=2$$

$$p=3$$

$$p=4$$

$$p=5$$

Solution

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$$\displaystyle \begin{bmatrix}

2 & 4 \\

1 & -1

\end{bmatrix}$$

$$\displaystyle \begin{bmatrix}

-1 & 1 \\

4 & 2

\end{bmatrix}$$

$$\displaystyle \begin{bmatrix}

4 & 2 \\

-1 & 1

\end{bmatrix}$$