Matrices Test

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Matrices Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\left[ \begin{matrix} 6 & 9 \\ -4 & -6 \end{matrix} \right] $$, then $$A^{2}$$=

A
$$\left[ \begin{matrix} 6 & 9 \\ -4 & 6 \end{matrix} \right]$$

B
$$\left[ \begin{matrix} 6 & 9 \\ 4 & -6 \end{matrix} \right]$$

C
$$\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$$

D
$$\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right]$$

Solution
Question 2
1 / -0

If the order of $$A$$ is $$4 \times 3$$, the order of $$B$$ is $$4 \times 5$$ and the order of $$C ,\ 7\times 3$$ then the order of $$(A^{T}B^{T})^{T}C^{T}$$ is

A
$$4 \times 5$$

B
$$3 \times 7$$

C
$$4 \times 3$$

D
$$5 \times 7$$

Solution

Order of $$A=4\times 3$$
$$\therefore $$ Order of $$A^{T}=3\times 4$$

Order of $$B=4\times 5$$
$$\therefore$$ Order of $$B^{T}=5\times 4$$

Order of $$C=7\times 3$$
$$\therefore$$Order of $$C^{T}=3\times 7$$

Order of $$(A^{T}B^{T})=4\times 5$$
Question 3
1 / -0

The inverse of a symmetric matrix is

A
symmetric

B
skew-symmetric

C
diagonal matrix

D
singular matrix

Solution

$$ A^{T}=A$$
$$A^{-1}$$ exists.
$$(A^{-1}A)^{T}=A^{T}(A^{-1})^{T}$$
$$=A(A^{-1})^{T}=I^{T}=I$$

$$A^{-1}A(A^{-1})^{T}=A^{-1}I$$
$$I(A^{-1})^{T}=(A^{-1})^{T}=A^{-1}$$

$$\therefore$$ The inverse of a symmetric matrix is also symmetric.
Question 4
1 / -0

If $$\left[ \begin{matrix} a \\ 8 \end{matrix}\begin{matrix} 5 \\ b \end{matrix} \right]$$ $$-$$ $$\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 6 \\ 2 \end{matrix} \right]$$ =$$\left[ \begin{matrix} 2 \\ 1 \end{matrix}\begin{matrix} -1 \\ 5 \end{matrix} \right],$$ then value of $$a$$ is

A
$$a=7$$

B
$$b=7$$

C
$$a=6$$

D
$$b=-7$$

Solution

Since, $$\left[ \begin{matrix} a \\ 8 \end{matrix}\begin{matrix} 5 \\ b \end{matrix} \right]$$ $$-$$ $$\left[ \begin{matrix} 4 \\ 7 \end{matrix}\begin{matrix} 6 \\ 2 \end{matrix} \right]$$ =$$\left[ \begin{matrix} 2 \\ 1 \end{matrix}\begin{matrix} -1 \\ 5 \end{matrix} \right]$$
By equating above matrices, we get
$$a-4=2$$
$$\Rightarrow a=2+4$$
$$\Rightarrow a=6$$
Question 5
1 / -0

If $$A=\left[ \begin{matrix} 1 & 0 & -1 \\ 3 & 4 & 5 \\ 0 & 6 & 7 \end{matrix} \right]$$ and $$A^{-1}=[\alpha_{ij}]_{3\times 3}$$ then $$\alpha_{23}=$$

A
$$-1/5$$

B
$$1/5$$

C
$$-2/5$$

D
$$2/5$$

Solution
Question 6
1 / -0

If $$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$$ and $$I$$ is the unit matrix of order $$2$$, then $$A^{2}$$ equals

A
$$4A-3I$$

B
$$3A-4I$$

C
$$A-I$$

D
$$A+I$$

Solution

$$\begin{array}{l} A=\left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \\ I=\left( { \begin{array} { *{ 20 }{ c } }1 & 0 \\ 0 & 1 \end{array} } \right) \\ { A^{ 2 } }=A\cdot A \\ =\left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \left( { \begin{array} { *{ 20 }{ c } }2 & { -1 } \\ { -1 } & 2 \end{array} } \right) \\ =\left( { \begin{array} { *{ 20 }{ c } }{ 2\times 2+1 } & { -2-2 } \\ { -2-2 } & { 1+4 } \end{array} } \right) \\ =\left( { \begin{array} { *{ 20 }{ c } }5 & { -4 } \\ { -4 } & 5 \end{array} } \right) \\ Now, \quad 4A-3I \\= 4\left( { \begin{array} { *{ 20 }{ c } }8 & { -4 } \\ { -4 } & 8 \end{array} } \right) -\left( { \begin{array} { *{ 20 }{ c } }3 & 0 \\ 0 & 3 \end{array} } \right) \\ = \left( { \begin{array} { *{ 20 }{ c } }{ 8-3 } & { -4-0 } \\ { -4-0 } & { 8-3 } \end{array} } \right) \\ = \left( { \begin{array} { *{ 20 }{ c } }5 & { -4 } \\ { -4 } & 5 \end{array} } \right) ={ A^{ 2 } } \end{array}$$
Question 7
1 / -0

If $$\begin{bmatrix} 3 & 2 & -1 \\ 4 & 9 & 2 \\ 5 & 0 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 7 \\ 2 \end{bmatrix}$$, then $$(x,\ y,\ z)=$$

A
$$(1,\ -1,\ 1)$$

B
$$(2,\ -1,\ -4)$$

C
$$(3,\ 0,\ 6)$$

D
$$(2,\ -1,\ 4)$$

Solution

$$\begin{bmatrix} 3 & 2 & -1\\ 4 & 9 & 2\\ 5 & 0 & -2\end{bmatrix}\begin{bmatrix} x \\ y\\ z\end{bmatrix}=\begin{bmatrix} 0\\ 7\\ 2\end{bmatrix}$$
Thus, multiplying
$$3x+2y-z=0$$ …………..$$(1)$$
$$4x+9y+2z=7$$ ………..$$(2)$$
$$5x-2z=2$$ …………$$(3)$$
Thus $$z=\dfrac{5x-2}{2}$$
$$\therefore 3x+2y=\dfrac{5x-2}{2}$$
$$\therefore 6x+4y=5x-2$$
$$\therefore x+4y=-2$$ …………$$(4)$$
Also,
$$4x+9y+2\left(\dfrac{5x-2}{2}\right)=7$$
$$9x+9y=9$$
$$\therefore x+y=1$$ ………..$$(5)$$
Solving $$(4)$$ and $$(5)$$
$$3y=-3$$
$$y=-1$$
$$\therefore x=1-y$$
$$=1-(-1)=2$$
$$\therefore x=2$$
$$\therefore z=\dfrac{5x-2}{2}=\dfrac{5(2)-2}{2}$$
$$\therefore z=4$$
$$=4$$
$$(x, y, z)=(2, -1, 4)$$.
Question 8
1 / -0

If A is a 2 X 2 matrix such that $$A^{2009} + A^{2008}$$= I, then : $$(A^{2008})^{-1}$$=

A
$$A^{2008} + I$$

B
$$A^{2009} + 1$$

C
A + I

D
A

Solution
Question 9
1 / -0

If $$\left[ \begin{matrix} 2 & -3 \\ 1 & \lambda \end{matrix} \right] \times \left[ \begin{matrix} 1 & 5 & \mu \\ 0 & 2 & -3 \end{matrix} \right] =\left[ \begin{matrix} 2 & 4 & 1 \\ 1 & -1 & 13 \end{matrix} \right],$$ then

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

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

C
no real values of $$\lambda, \mu$$ are possible

D
none of these

Solution

Given,
$$\pmatrix{2&-3\\1&\lambda}\times \pmatrix{1&5& \mu\\0&2&-3}$$=$$\pmatrix{2&4&1\\1&-1&13}$$

$$\pmatrix{2&4&2\mu-9\\1&5+2\lambda&\mu-3\lambda}=\pmatrix{2&4&1\\1&-1&13}$$

Comparing both sides we get
$$2\mu+9=1\implies 2\mu=-8\implies \mu=-4$$

Also $$5+2\lambda=-1\implies 2\lambda=6\implies \lambda=3$$

Hence, we get $$\mu=-4,\lambda=3.$$
Question 10
1 / -0

Let p be a non-singular matrix, $$1+p+p^{2}+....+p^{n}=0$$ (0 denotes the null matrix) then $$p^{-1}=$$

A
$$p^{n}$$

B
-$$p^{n}$$

C
-(1+p+...+$$p^{n}$$)

D
none

Solution

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Matrices Test - 41

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Matrices Test - 41

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Question 1

$$\left[ \begin{matrix} 6 & 9 \\ -4 & 6 \end{matrix} \right]$$

$$\left[ \begin{matrix} 6 & 9 \\ 4 & -6 \end{matrix} \right]$$

$$\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$$

$$\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right]$$

Solution

Question 2

$$4 \times 5$$

$$3 \times 7$$

$$4 \times 3$$

$$5 \times 7$$

Solution

Question 3

symmetric

skew-symmetric

diagonal matrix

singular matrix

Solution

Question 4

$$a=7$$

$$b=7$$

$$a=6$$

$$b=-7$$

Solution

Question 5

$$-1/5$$

$$1/5$$

$$-2/5$$

$$2/5$$

Solution

Question 6

$$4A-3I$$

$$3A-4I$$

$$A-I$$

$$A+I$$

Solution

Question 7

$$(1,\ -1,\ 1)$$

$$(2,\ -1,\ -4)$$

$$(3,\ 0,\ 6)$$

$$(2,\ -1,\ 4)$$

Solution

Question 8

$$A^{2008} + I$$

$$A^{2009} + 1$$

A + I

A

Solution

Question 9

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

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

no real values of $$\lambda, \mu$$ are possible

none of these

Solution

Question 10

$$p^{n}$$

-$$p^{n}$$

-(1+p+...+$$p^{n}$$)

none

Solution

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