Matrices Test

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Matrices Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The matrix equation satisfied by $$A$$ is ______$$, $$ if $$A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}.$$

A
$$A^2-4A-5I=0$$

B
$$A^2-4A-5=0$$

C
$$A^2+4A-5I=0$$

D
$$A^2+4A-5=0$$

Solution

Consider the given matrix,

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

Consider, $$A^2-4A-5I$$

$$=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-4\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

$$=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}-\begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix}-\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$$

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

$$=0$$
Question 2
1 / -0

If $$A = \begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}$$ and $$B = \begin{bmatrix}3&1&2\\2&0&5\\1&2&0\end{bmatrix}$$, find $$AB$$.

A
$$\begin{bmatrix}5&6&-9\\4&3&-9\\-1&3&-2\end{bmatrix}$$

B
$$\begin{bmatrix}2&4&7\\1&5&3\\-1&0&-5\end{bmatrix}$$

C
$$\begin{bmatrix}7&11&-9\\4&10&-11\\-1&2&-5\end{bmatrix}$$

D
$$\begin{bmatrix}7&8&-19\\4&1&-5\\-1&12&-5\end{bmatrix}$$

Solution

$$A=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\quad B=\begin{bmatrix} 3 & 1 & 2 \\ 2 & 0 & 5 \\ 1 & 2 & 0 \end{bmatrix}\\ AB=\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\times \begin{bmatrix} 3 & 1 & 2 \\ 2 & 0 & 5 \\ 1 & 2 & 0 \end{bmatrix}\\ \quad =\begin{bmatrix} 9-6+4 & 3+0+8 & 6-15+0 \\ 6-6+4 & 2+0+8 & 4-15+0 \\ 0-2+1 & 0+0+2 & 0-5+0 \end{bmatrix}\\ \quad =\begin{bmatrix} 7 & 11 & -9 \\ 4 & 10 & -11 \\ -1 & 2 & -5 \end{bmatrix}$$
Question 3
1 / -0

$$\left[ \begin{matrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 1& -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{matrix} \right] \ $$

What will be the sum of the diagonal elements of the resultant matrix?

A
10

B
6

C
9

D
7

Solution

$$\begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}+ \begin{bmatrix} 5 & 1 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{bmatrix}= \begin{bmatrix} 6 & 1 & 0 \\ 0 & 2 & -2 \\ -2 & 0 & 2 \end{bmatrix}$$

Sum of diagonal $$=6+2+2 = 10$$
Question 4
1 / -0

If $$A=\begin{bmatrix} 1 & 0 \\ \frac { 1 }{ 2 } & 1 \end{bmatrix}$$, then $$A^{100}$$ is equal to

A
$$\begin{bmatrix} 1 & 0 \\ \left( \frac { 1 }{ 2 } \right) \times { 100 } & 1 \end{bmatrix}$$

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

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

D
$$\begin{bmatrix} 1 & 0 \\ 100 & 1 \end{bmatrix}$$

Solution

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

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

$$A^3=\begin{bmatrix} 1& 0\\ \dfrac{2}{2}& 1\end{bmatrix}\times \begin{bmatrix} 1& 0\\ \dfrac{1}{2}& 1\end{bmatrix}=\begin{bmatrix} 1& 0\\ \dfrac{3}{2}& 1\end{bmatrix}$$

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

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

So
$$A^{100}=\begin{bmatrix} 1& 0 \\ \dfrac{100}{2}&1 \end{bmatrix}=\begin{bmatrix} 1& 0\\ 50& 1\end{bmatrix}$$
Question 5
1 / -0

If$$A=\begin{bmatrix}1&3\\2&4\end{bmatrix}$$ and $$B=\begin{bmatrix}4&7\\5&6\end{bmatrix},$$ then $$AB=$$

A
$$\begin{bmatrix}19&25\\28&38\end {bmatrix}$$

B
$$\begin{bmatrix}25&8\\26&14\end {bmatrix}$$

C
$$\begin{bmatrix}34&8\\58&23\end {bmatrix}$$

D
$$\begin{bmatrix}5&10\\7&10\end {bmatrix}$$

Solution

Given,
$$A=\begin{bmatrix}1&3\\2&4\end{bmatrix}$$ and $$B=\begin{bmatrix}4&7\\5&6\end{bmatrix}$$

$$AB=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}4&7\\5&6\end{bmatrix}$$

$$=\begin{bmatrix}1(4)+3(5)&1(7)+3(6)\\2(4)+4(5)& 2(7)+4(6)\end{bmatrix}$$

$$=\begin{bmatrix}19&25\\28&38\end{bmatrix}$$
Question 6
1 / -0

$$B=A+A^{2}+A^{3}+A^{4}$$
If order of $$A$$ is $$3$$ then order of $$B$$ is

A
$$3$$

B
$$6$$

C
$$2$$

D
$$9$$

Solution

The order of matrix does not change when the operation are done on it.
So, the order of $$B$$ remains same as the order of $$A$$.
Question 7
1 / -0

If $$A=\begin{bmatrix} 2 & 5 & -3 \\ -1 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix}$$ then $${A}^{2}$$ is

A
$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$

B
$$\begin{bmatrix} -13 & 22 & -7 \\ -1 & 5 & 8 \\ 15 & 25 & -7 \end{bmatrix}$$

C
$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & 7 \\ 1 & 5 & 8 \end{bmatrix}$$

D
$$\begin{bmatrix} 13 & 22 & 7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$

Solution

$$A^2=A.A=\begin{bmatrix} 2 & 5 & -3 \\ -1 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix}.\begin{bmatrix} 2 & 5 & -3 \\ -1 & 3 & 1 \\ 4 & 1 & 2 \end{bmatrix}$$

$$=\begin{bmatrix} -13\quad \quad & 22\quad \quad & 2\times -3+5\times 1+(-3)\times 2 \\ -1 & 5 & 8 \\ 15 & 25 & -7 \end{bmatrix}$$

$$\begin{bmatrix} -13 & 22 & -7 \\ -1 & 5 & 8 \\ -15 & 25 & -7 \end{bmatrix}$$
Hence, B will be correct answer.
Question 8
1 / -0

If the graph of $$y = f(x)$$ is symmetrical about the lines $$x = 1$$ and $$x = 2$$, then which of the following is true?

A
$$f(x + 1) = f(x)$$

B
$$f(x + 3) = f(x)$$

C
$$f(x + 2) = f(x)$$

D
None of these

Solution

$$f(x)$$ is symmetrical about the line $$x=1$$

So,
$$f(1-x)=f(1+x)$$ .................... (1)

$$f(x)$$ is symmetrical about the line $$x=2$$

So,
$$f(2-x)=f(2+x)$$ .................... (2)

Replace $$x$$ with $$1-x$$ in equation (1) ,

$$f(1-1+x)=f(1+1-x)$$

$$f(x)=f(2-x)$$ ..........................(3)

From (2) and (3) ,

$$f(x)=f(2+x)$$
Question 9
1 / -0

Find the inverse of the following matrix.

$$\left[ {\begin{array}{ccccccccccccccc}0&1&2\\1&2&3\\3&1&1\end{array}} \right]$$

A
$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ 4}&3&{ 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$

B
$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&-3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$

C
$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$

D
$$\left[ {\begin{array}{ccccccccccccccc}0&1&2\\1&-2&3\\3&1&-1\end{array}} \right]$$

Solution
Question 10
1 / -0

If $$A$$ and $$B$$ are two non-singular matrices and both are symmetric and commute each other, then

A
Both $$A^{-1}B$$ and $$A^{-1}B^{-1}$$ are symmetric

B
$$A^{-1}B$$ is symmetric but $$A^{-1}B^{-1}$$ is not symmetric

C
$$A^{-1}B^{-1}$$ is symmetric but $$A^{-1}B$$ is not symmetric

D
Neither $$A^{-1}B$$ nor $$A^{-1}B^{-1}$$ are symmetric

Solution

$$\begin{array}{l} \left| A \right| \ne 0\, \, \& \, \, \left| B \right| \ne 0 \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { { B^{ -1 } } } \right) ^{ T } }{ \left( { { A^{ -1 } } } \right) ^{ T } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { { B^{ T } } } \right) ^{ -1 } }{ \left( { { A^{ T } } } \right) ^{ -1 } }\, \, \, \, \left[ \begin{array}{l} { A^{ T } }=A \\ { B^{ T } }=B \end{array} \right] \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ B^{ -1 } }{ A^{ -1 } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { AB } \right) ^{ -1 } }\, \, \, \, \, \, \left[ { AB=BA } \right] \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ \left( { BA } \right) ^{ -1 } } \\ { \left( { { A^{ -1 } }{ B^{ -1 } } } \right) ^{ T } }={ A^{ -1 } }{ B^{ -1 } } \\ { A^{ -1 } }{ B^{ -1 } }\, is\, \, a\, symmetric\, \, matrix \end{array}$$
similarly $$A^{-1}B$$ is also symmetric matrix.

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Re-Attempt Weekly Quiz Competition

Matrices Test - 39

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Matrices Test - 39

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Question 1

$$A^2-4A-5I=0$$

$$A^2-4A-5=0$$

$$A^2+4A-5I=0$$

$$A^2+4A-5=0$$

Solution

Question 2

$$\begin{bmatrix}5&6&-9\\4&3&-9\\-1&3&-2\end{bmatrix}$$

$$\begin{bmatrix}2&4&7\\1&5&3\\-1&0&-5\end{bmatrix}$$

$$\begin{bmatrix}7&11&-9\\4&10&-11\\-1&2&-5\end{bmatrix}$$

$$\begin{bmatrix}7&8&-19\\4&1&-5\\-1&12&-5\end{bmatrix}$$

Solution

Question 3

10

6

9

7

Solution

Question 4

$$\begin{bmatrix} 1 & 0 \\ \left( \frac { 1 }{ 2 } \right) \times { 100 } & 1 \end{bmatrix}$$

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

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

$$\begin{bmatrix} 1 & 0 \\ 100 & 1 \end{bmatrix}$$

Solution

Question 5

$$\begin{bmatrix}19&25\\28&38\end {bmatrix}$$

$$\begin{bmatrix}25&8\\26&14\end {bmatrix}$$

$$\begin{bmatrix}34&8\\58&23\end {bmatrix}$$

$$\begin{bmatrix}5&10\\7&10\end {bmatrix}$$

Solution

Question 6

$$3$$

$$6$$

$$2$$

$$9$$

Solution

Question 7

$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$

$$\begin{bmatrix} -13 & 22 & -7 \\ -1 & 5 & 8 \\ 15 & 25 & -7 \end{bmatrix}$$

$$\begin{bmatrix} -13 & 22 & -7 \\ 15 & 25 & 7 \\ 1 & 5 & 8 \end{bmatrix}$$

$$\begin{bmatrix} 13 & 22 & 7 \\ 15 & 25 & -7 \\ -1 & 5 & 8 \end{bmatrix}$$

Solution

Question 8

$$f(x + 1) = f(x)$$

$$f(x + 3) = f(x)$$

$$f(x + 2) = f(x)$$

None of these

Solution

Question 9

$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ 4}&3&{ 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$

$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&-3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$

$$\displaystyle \left[ {\begin{array}{ccccccccccccccc}{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\{ - 4}&3&{ - 1}\\{ - \frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}\end{array}} \right]$$

$$\left[ {\begin{array}{ccccccccccccccc}0&1&2\\1&-2&3\\3&1&-1\end{array}} \right]$$

Solution

Question 10

Both $$A^{-1}B$$ and $$A^{-1}B^{-1}$$ are symmetric

$$A^{-1}B$$ is symmetric but $$A^{-1}B^{-1}$$ is not symmetric

$$A^{-1}B^{-1}$$ is symmetric but $$A^{-1}B$$ is not symmetric

Neither $$A^{-1}B$$ nor $$A^{-1}B^{-1}$$ are symmetric

Solution

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