Matrices Test

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Matrices Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

If $$A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$$ and $$kA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$$, then the value of $$k, a, b $$, are respectively.

A
$$-6, -12, -18$$

B
$$-6, 4, 9$$

C
$$-6, -4, -9$$

D
$$-6, 12, 18$$

Solution

If $$A=\begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$$
then,
$$kA=\begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix}$$

But it is given that
$$kA=\begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$$

So,
$$\begin{bmatrix} 0 & 2k \\ 3k & -4k \end{bmatrix}$$$$=\begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$$

$$\Rightarrow -4k=24$$
$$k=-6$$

$$3a=2k$$
$$3a=-12$$
$$a=-4$$

$$2b=3k$$
$$2b=-18$$
$$b=-9$$
Question 2
1 / -0

If $$A = \left[ {\begin{array}{*{20}{c}}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{array}} \right]$$ then $$A^6$$ = _____________.

A
$$6A$$

B
$$12A$$

C
$$16A$$

D
$$32A$$

Solution

$$A=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\\ A^{ 2 }=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\\ A^{ 4 }=A^{ 2 }A^{ 2 }=\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}=\begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix}\\ A^{ 6 }=A^{ 4 }A^{ 2 }=\begin{bmatrix} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{bmatrix}\begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}=\begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix}\\ taking\quad 32\quad out\quad we\quad get\\ A^{ 6 }=32\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}=32A$$
Question 3
1 / -0

If $$\left[ {\begin{array}{*{20}{c}}x & 4 & { - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2 & 1 & 0\\1 & 0 & 2\\0 & 2 & 4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\4\\{ - 1}\end{array}} \right] = 0$$ then $$x$$ is equal to

A
$$ - 1 + \sqrt 6 $$

B
$$ 8 \pm \sqrt 5 $$

C
$$ - 2 \pm \sqrt {10} $$

D
$$ 3 \pm \sqrt 6 $$

Solution

$$\left[ \begin{matrix} x & 4 & 1 \end{matrix} \right]_{1\times 3}\left[ \begin{matrix}2 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & 4 \end{matrix} \right]_{3\times 3}\left[ \begin{matrix} x \\ 4 \\ -1 \end{matrix} \right]_{3\times 1}=0$$
$$\left[ \begin{matrix} x & 4 & 1 \end{matrix} \right]_{1\times 3}\left[ \begin{matrix} 2x+4+0 \\ x+0-2 \\ 0+8-4 \end{matrix} \right]_{3\times 1}=0$$
$$\Rightarrow\,2{x}^{2}+4x+4x-8-4=0$$
$$\Rightarrow\,2{x}^{2}+8x-12=0$$
$$\Rightarrow\,{x}^{2}+4x-6=0$$
$$\Rightarrow\,{x}^{2}+2\times 2x+4-4-6=0$$
$$\Rightarrow\,{\left(x+2\right)}^{2}=10$$
$$\Rightarrow\,x+2=\pm\sqrt{10}$$
$$\therefore\,x=-2\pm\sqrt{10}$$
Question 4
1 / -0

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

A
$$O$$

B
$$I$$

C
$$2I$$

D
None of these

Solution

$$AB=\begin{bmatrix} 0 & c & -b \\ -c & b & a \\ b & -a & 0 \end{bmatrix}\begin{bmatrix} a^{ 2 } & ab & ac \\ ba & b^{ 2 } & bc \\ ca & cb & c^{ 2 } \end{bmatrix}\\ =\begin{bmatrix} abc-abc & b^{ 2 }c^{ 1 }-b^{ 2 }c & bc^{ 2 }-bc^{ 2 } \\ -a^{ 2 }c+a^{ 2 }c & -abc+abc & -ac^{ 2 }+ac^{ 2 } \\ a^{ 2 }b-a^{ 2 }b & ab^{ 2 }-ab^{ 2 } & abc-abc \end{bmatrix}\\ =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=0\Rightarrow (A)$$
Question 5
1 / -0

The value of $$x$$, so that $$\left\lceil 1\quad x\quad 1 \right\rceil \begin{bmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x \end{bmatrix}=0$$ is

A
$$\cfrac{-7\pm \sqrt{35}}{2}$$

B
$$\cfrac{-9\pm \sqrt{53}}{2}$$

C
$$\pm 2$$

D
$$\dfrac{-7}{10}$$

Solution

Consider, $$\begin{bmatrix} 1 & x & 1\end{bmatrix}\begin{bmatrix} 1 & 3 & 2\\ 0 & 5 & 1\\ 0 & 3 & 2\end{bmatrix}=1\begin{bmatrix} 1 & 3+5x+3 & 5\end{bmatrix}$$
$$\begin{bmatrix} 1 & 6+5x & 5\end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ x\end{bmatrix}=7+10x$$
By given, we have
$$7+10x =0$$
$$\Rightarrow x=-\dfrac{7}{10}$$.
Question 6
1 / -0

If A be square matrix of order n and k is a scalar, then adj (KA) is:

A
$$K^{n}(adjA)$$

B
K (adj A)

C
$$K^{n-1}(adjA)$$

D
$$K^{n+1}(adjA)$$

Solution
Question 7
1 / -0

If $$A = \left[ \begin{array} { c c } { a b } & { b ^ { 2 } } \\ { - a ^ { 2 } } & { - a b } \end{array} \right]$$ and $$A ^ { n } = 0$$ then the minimum value of $$n$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$A=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}$$

$$A^2=\begin{bmatrix}ab&b^2\\-a^2&-ab\end{bmatrix}\begin{bmatrix}ab&b^2\\-a^z&-ab\end{bmatrix}=\begin{bmatrix}a^2b^2-a^2b^2&ab^3-ab^3\\-a^3b+a^3b&-a^2b^2+a^2b^2\end{bmatrix}$$

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

$$A^2=0$$

$$\therefore n=2$$.
Question 8
1 / -0

If $$A$$ is matrix of order $$m\times n$$ and $$B$$ is a matrix such that $$AB'$$ and $$B'A$$ are both defined, then order of matrix $$B$$ is

A
$$m\times m$$

B
$$n\times n$$

C
$$n\times m$$

D
$$m\times n$$

Solution

Let $$B$$ is $$p\times q$$ matrix
$$\Rightarrow B'$$ has $$q\times p$$ order.
In order to define $$AB',$$ we must have
$$\{(m\times n)\times (q\times p)\}\Rightarrow n=q$$
Also, for $$B'A$$ to be defined,
$$\{(q\times p)\times (m\times n)\}\Rightarrow p=m$$
$$\Rightarrow B$$ is $$m\times n$$ matrix
Hence, D will be correct answer.
Question 9
1 / -0

If $$A=\left[ \begin{matrix} a & b \\ b & a \end{matrix} \right] $$ and $$A^{2}=\left[ \begin{matrix} \alpha & \beta \\ \beta & \alpha \end{matrix} \right] $$ then

A
$$\alpha=a^{2}+b^{2},\ \beta=2ab$$

B
$$\alpha=a^{2}+b^{2},\ \beta=a^{2}b^{2}$$

C
$$\alpha=2ab,\ \beta=a^{2}+b^{2}$$

D
$$\alpha=a^{2}+b^{2},\ \beta=ab$$

Solution
Question 10
1 / -0

If $$A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$$, then $$8A^{-4}$$ is equal to

A
$$145A^{-1}+27I$$

B
$$145A^{-1}-27I$$

C
$$27I - 145A^{-1}$$

D
$$29A^{-1} +9I$$

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Matrices Test - 40

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Matrices Test - 40

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Question 1

$$-6, -12, -18$$

$$-6, 4, 9$$

$$-6, -4, -9$$

$$-6, 12, 18$$

Solution

Question 2

$$6A$$

$$12A$$

$$16A$$

$$32A$$

Solution

Question 3

$$ - 1 + \sqrt 6 $$

$$ 8 \pm \sqrt 5 $$

$$ - 2 \pm \sqrt {10} $$

$$ 3 \pm \sqrt 6 $$

Solution

Question 4

$$O$$

$$I$$

$$2I$$

None of these

Solution

Question 5

$$\cfrac{-7\pm \sqrt{35}}{2}$$

$$\cfrac{-9\pm \sqrt{53}}{2}$$

$$\pm 2$$

$$\dfrac{-7}{10}$$

Solution

Question 6

$$K^{n}(adjA)$$

K (adj A)

$$K^{n-1}(adjA)$$

$$K^{n+1}(adjA)$$

Solution

Question 7

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 8

$$m\times m$$

$$n\times n$$

$$n\times m$$

$$m\times n$$

Solution

Question 9

$$\alpha=a^{2}+b^{2},\ \beta=2ab$$

$$\alpha=a^{2}+b^{2},\ \beta=a^{2}b^{2}$$

$$\alpha=2ab,\ \beta=a^{2}+b^{2}$$

$$\alpha=a^{2}+b^{2},\ \beta=ab$$

Solution

Question 10

$$145A^{-1}+27I$$

$$145A^{-1}-27I$$

$$27I - 145A^{-1}$$

$$29A^{-1} +9I$$

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