Matrices Test

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Matrices Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If $$\mathrm{A}=\left[\begin{array}{ll}
0 & 1\\
1 & 0
\end{array}\right]$$, then $$\mathrm{A}^{5}=$$

A
$$I$$

B
$$\mathrm{O}$$

C
$$\mathrm{A}$$

D
$$\mathrm{A}^{2}$$

Solution

Given, A $$=$$$$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

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

$$A^5$$$$=$$$${A^2}\times{A^2}\times{A}$$

$$A^5$$$$=$$ $$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ $$=$$$$A$$
Question 2
1 / -0

If $$\mathrm{A}=\left[\begin{array}{ll}
1 & 2\\
0 & 3
\end{array}\right]$$ and $$\mathrm{B}=[3 \space-1]$$, then $$\mathrm{B}\mathrm{A}=$$

A
$$\left[\begin{array}{ll}

3 & 0\\

0 & 3

\end{array}\right]$$

B
$$[3 \ \ \ 0]$$

C
$$[3 \ \ \ 3 ]$$

D
$$[0 \ \ -3]$$

Solution

The value of $$\mathrm{BA}$$
$$=[3 \space -1]\begin{bmatrix}1 & 2\\0 & 3\end{bmatrix}$$
$$=[ 3\times 1-1\times 0 \ \ \space 3\times 2-1\times 3 ]$$
$$=[3 \ \ \ \space 3]$$
Question 3
1 / -0

$$ If \space A= \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$$, then A is

A
a nilpotent matrix

B
an involutory matrix

C
a symmetric matrix

D
an idempotent matrix

Solution

Here The transpose of the matrix A is
$$ = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$$
$$ \therefore A = A^{T}$$
And a symmetric matrix is a square matrix that is equal to its transpose.
Hence the answer is option C
Question 4
1 / -0

If A = $$\begin{bmatrix}
x & 1\\
1 & 0
\end{bmatrix}$$ and $$A^{2}$$ is identity matrix, then $$x= $$

A
$$1$$

B
$$-1$$

C
$$\pm 1$$

D
$$0$$

Solution

Since, $$A=\begin{bmatrix}
x &1 \\
1 &0
\end{bmatrix}, A^2 =I$$

$$A^{2}=\begin{bmatrix}
x &1 \\
1 &0
\end{bmatrix}\begin{bmatrix}
x &1 \\
1 &0
\end{bmatrix}$$

$$\Rightarrow \begin{bmatrix}
x^{2}+1 &x \\
x &1
\end{bmatrix}= \begin{bmatrix}
1 &0 \\
0 &1
\end{bmatrix}$$

$$\Rightarrow$$ So $$x=0$$

Option D is the correct answer.
Question 5
1 / -0

lf $$\mathrm{A}=\left[\begin{array}{ll}
2 & -1\\
3 & -2
\end{array}\right],$$ then $$\mathrm{A}^{5}=$$

A
$$I$$

B
$$A$$

C
$$-A$$

D
$$A^{2}$$

Solution

$$A^2=\begin{bmatrix}
2 & 3\\
3 & -2
\end{bmatrix}\begin{bmatrix}
2 & 3\\
3 & -2
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$$
$$A^3=AI=A$$
$$A^5=A^3(A^2)=A^3I=A(A^2I)=A$$
Question 6
1 / -0

$$L=\left[\begin{array}{lll}
2 & 3 & 5\\
4 & 1 & 2\\
1 & 2 & 1
\end{array}\right] =P+Q$$, $$P$$ is a symmetric matrix, $${Q}$$ is a skew-symmetric matrix then $${P}$$ is equal to

A
$$\left[\begin{array}{lll}

3 & 5 & 6\\

5 & 6 & 4\\

9 & 4 & 3
\end{array}\right]$$

B
$$\left[\begin{array}{lll}

2 & 3.5 & 3\\

3.5 & 1 & 2\\

3 & 2 & 1

\end{array}\right]$$

C
$$\left[\begin{array}{lll}

6 & 5 & 4\\

3 & 6 & 3\\

5 & 2 & 5

\end{array}\right]$$

D
$$\left[\begin{array}{lll}

6 & 5 & 4\\

4 & 5 & 3\\

3 & 4 & 3

\end{array}\right]$$

Solution

Given ,$$ P+Q=\begin{bmatrix}
2 & 3 & 5\\
4 & 1 & 2\\
1 & 2 & 1
\end{bmatrix}$$-----(i)

ans $$P$$ is symmetric and $$Q$$ is skew symmetric matrix. So, by property of transpose (7),

$$P=P^{T}$$ and $$Q^{T}=-Q$$

$$\therefore$$ equation (i), $$P^{T}-Q=\begin{bmatrix}
2 & 4 & 1\\
3 & 1 & 2\\
5 & 2 & 1
\end{bmatrix}$$---------(ii)

Add (i) and (ii), we get

$$P+P^{T}=\begin{bmatrix}
4 & 7 & 6\\
7 & 2 & 4\\
6 & 4 & 2
\end{bmatrix},
2P=\begin{bmatrix}
4 & 7 & 6\\
7 & 2 & 4\\
6 & 4 & 2
\end{bmatrix}$$

$$\Rightarrow P=\begin{bmatrix}
2 & 3.5 & 3\\
3.5 & 1 & 2\\
3 & 2 & 1
\end{bmatrix}$$
Question 7
1 / -0

If $$I=\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$$ and E =$$\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}$$, then $$\left ( 2I+3E \right )^{3}=$$

A
$$8I+ 18E$$

B
$$4I+ 36E$$

C
$$8I +36E$$

D
$$2I+ 3E$$

Solution

Consider, $$ 2I+3E$$
$$2\begin{bmatrix}
1 &0 \\
0 &1
\end{bmatrix}+3\begin{bmatrix}
0 &1 \\
0 &0
\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix}
2 &0 \\
0 &2
\end{bmatrix}+\begin{bmatrix}
0 &3 \\
0 &0
\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix}
2 &3 \\
0 &2
\end{bmatrix}$$
Now $$(2I+3E)^3$$
$$\Rightarrow \begin{bmatrix}
2 &3 \\
0 &2
\end{bmatrix}\begin{bmatrix}
2 &3 \\
0 &2
\end{bmatrix}\begin{bmatrix}
2 &3 \\
0 &2
\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix}
4 &6+6 \\
0 &4
\end{bmatrix}\begin{bmatrix}
2 &3 \\
0 &2
\end{bmatrix}$$
$$=\begin{bmatrix}
8 &12+24 \\
0 &8
\end{bmatrix}=\begin{bmatrix}
8 &36 \\
0 &8
\end{bmatrix}=8I+36E$$
$$\therefore(C)$$
Question 8
1 / -0

If in a square matrix $$A=\left[ { a }_{ ij } \right] $$, we find that $${ a }_{ ij }={ a }_{ ji }\quad \forall \quad i,j$$ , then $$A$$ is

A
Symmetric

B
Skew Symmetric

C
Idempotent

D
none of these

Solution

Given square matrix $$A=\left\{ { a }_{ ij } \right\} $$ and $${ a }_{ ij }={ a }_{ ji }$$
We know that in a square matrix if $${ a }_{ ij }={ a }_{ ji }$$ for all i and j, then the matrix is a symmetric matrix
So A is a symmetric matrix
Question 9
1 / -0

If P = $$ \begin{bmatrix}
1\\
3\\

4\end{bmatrix}$$ , Q = $$\begin{bmatrix}
2 & -1&5
\end{bmatrix}$$ then PQ =

A
$$\begin{bmatrix}2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20\end{bmatrix}$$

B
$$\begin{bmatrix}2 & -3 & 20\end{bmatrix}$$

C
$$\begin{bmatrix}2\\ -3\\ 20\end{bmatrix}$$

D
$$[19]$$

Solution

$$\text{PQ}=\begin{bmatrix} 1\\3\\4\end{bmatrix}\begin{bmatrix} 2&-1&5\end{bmatrix}$$
$$\quad = \begin{bmatrix} 1\times 2&1\times(-1)&1\times 5\\3\times2&3\times(-1)&3\times 5 \\4\times 2&4\times(-1)&4\times 5\end{bmatrix}$$
$$\quad = \begin{bmatrix} 2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20 \end{bmatrix}$$
Question 10
1 / -0

$$\mathrm{A}$$: If $$\mathrm{A}=\left\{\begin{array}{ll}
1 & -1\\
-1 & 1
\end{array}\right\} $$ and $$\mathrm{B}=\left\{\begin{array}{ll}
2 & 2\\
2 & 2
\end{array}\right\},$$ then $$\mathrm{A}\mathrm{B}=0$$
$$\mathrm{R}$$: If $$\mathrm{A}\mathrm{B}=0\Rightarrow \mathrm{A}$$ or $$\mathrm{B}$$ need not be null matrices.
The correct answer is

A
Both $$A$$ and $$R$$ are true, $$R$$ is correct explanation to $$ A$$

B
Both $$A$$ and $$R$$ are true but $$R$$ is not correct explanation to $$A$$

C
$$A$$ is true, $$R$$ is false

D
$$A$$ is false, $$R$$ is true

Solution

If $$\mathrm{A}=\left\{\begin{array}{ll}
1 & -1\\
-1 & 1
\end{array}\right\} ;\mathrm{B}=\left\{\begin{array}{ll}
2 & 2\\
2 & 2
\end{array}\right\}$$
Then, $$\mathrm{AB}=\left\{\begin{array}{ll}
1\times 2-1\times 2 & 1\times 2-1\times 2\\
-1\times 2+1\times 2 & -1\times 2+1\times 2
\end{array}\right\} =O$$
Hence both statement are correct and Reason is correct explanation of Assertion.

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Matrices Test - 25

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Matrices Test - 25

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Question 1

$$I$$

$$\mathrm{O}$$

$$\mathrm{A}$$

$$\mathrm{A}^{2}$$

Solution

Question 2

$$\left[\begin{array}{ll}3 & 0\\0 & 3\end{array}\right]$$

$$[3 \ \ \ 0]$$

$$[3 \ \ \ 3 ]$$

$$[0 \ \ -3]$$

Solution

Question 3

a nilpotent matrix

an involutory matrix

a symmetric matrix

an idempotent matrix

Solution

Question 4

$$1$$

$$-1$$

$$\pm 1$$

$$0$$

Solution

Question 5

$$I$$

$$A$$

$$-A$$

$$A^{2}$$

Solution

Question 6

$$\left[\begin{array}{lll}3 & 5 & 6\\5 & 6 & 4\\9 & 4 & 3\end{array}\right]$$

$$\left[\begin{array}{lll}2 & 3.5 & 3\\3.5 & 1 & 2\\3 & 2 & 1\end{array}\right]$$

$$\left[\begin{array}{lll}6 & 5 & 4\\3 & 6 & 3\\5 & 2 & 5\end{array}\right]$$

$$\left[\begin{array}{lll}6 & 5 & 4\\4 & 5 & 3\\3 & 4 & 3\end{array}\right]$$

Solution

Question 7

$$8I+ 18E$$

$$4I+ 36E$$

$$8I +36E$$

$$2I+ 3E$$

Solution

Question 8

Symmetric

Skew Symmetric

Idempotent

none of these

Solution

Question 9

$$\begin{bmatrix}2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20\end{bmatrix}$$

$$\begin{bmatrix}2 & -3 & 20\end{bmatrix}$$

$$\begin{bmatrix}2\\ -3\\ 20\end{bmatrix}$$

$$[19]$$

Solution

Question 10

Both $$A$$ and $$R$$ are true, $$R$$ is correct explanation to $$ A$$

Both $$A$$ and $$R$$ are true but $$R$$ is not correct explanation to $$A$$

$$A$$ is true, $$R$$ is false

$$A$$ is false, $$R$$ is true

Solution

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$$\left[\begin{array}{ll}

3 & 0\\

0 & 3

\end{array}\right]$$

$$\left[\begin{array}{lll}

3 & 5 & 6\\

5 & 6 & 4\\

9 & 4 & 3
\end{array}\right]$$

$$\left[\begin{array}{lll}

2 & 3.5 & 3\\

3.5 & 1 & 2\\

3 & 2 & 1

\end{array}\right]$$

$$\left[\begin{array}{lll}

6 & 5 & 4\\

3 & 6 & 3\\

5 & 2 & 5

\end{array}\right]$$

$$\left[\begin{array}{lll}

6 & 5 & 4\\

4 & 5 & 3\\

3 & 4 & 3

\end{array}\right]$$