Matrices Test

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Matrices Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right] $$ and $$\displaystyle I=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] $$, then the correct statement is:

A
$$\displaystyle { A }^{ 2 }+5A-7I=O$$

B
$$\displaystyle -{ A }^{ 2 }+5A+7I=O$$

C
$$\displaystyle { A }^{ 2 }-5A+7I=O$$

D
$$\displaystyle { A }^{ 2 }+5A+7I=O$$

Solution

Now, $$\displaystyle { A }^{ 2 }=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right] $$
$$\displaystyle =\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \end{matrix} \right] $$
$$\displaystyle \therefore \quad { A }^{ 2 }-5A+7I=\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \end{matrix} \right] -\left[ \begin{matrix} 15 & 5 \\ -5 & 10 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 \\ 0 & 7 \end{matrix} \right] $$
$$\displaystyle =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] $$
Question 2
1 / -0

If AB = AC then

A
B = C

B
$$B\neq C$$

C
B need not be equal to C

D
B = -C

Solution

Value of B and C depends on the value of A. if A is non-singular, i.e non zero, then B=C.
But if A=0, then both terms AB and AC becomes zero, so B and C need not be necessarily equal.
Question 3
1 / -0

If $$\mathrm{A}^{2}=\mathrm{A},\ \mathrm{B}^{2}=\mathrm{B},\ \mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}=O$$ (Null Matrix), then $$(\mathrm{A}+\mathrm{B})^{2}=$$

A
$$\mathrm{A}-\mathrm{B}$$

B
$$\mathrm{A}+\mathrm{B}$$

C
$$\mathrm{A}^{2}-\mathrm{B}^{2}$$

D
$$0$$

Solution

$$(A + B)^{2} = A^{2} + B^{2} + AB + BA$$
$$= A + B$$
Question 4
1 / -0

$$I$$ $$A=\left[\begin{array}{ll}
0 & 1\\ 1 & 0 \end{array}\right]$$, $$A^{4}=$$
($$I$$ is an identity matrix.)

A
$$I$$

B
$$0$$

C
$$\mathrm{A}$$

D
$$4{I}$$

Solution

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

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

$$\Rightarrow A^{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\Rightarrow A^{2}=I$$
$$\Rightarrow A^{4}=I$$
Question 5
1 / -0

lf $$\mathrm{A}= \left[\begin{array}{lll}
o & c & -b\\
-c & o & a\\
b & -a & o
\end{array}\right]\mathrm{a}\mathrm{n}\mathrm{d}$$ $$ \mathrm{B}=\left[\begin{array}{lll}
a^{2} & ab & ac\\
ab & b^{2} & bc\\
ac & bc & c^{2}
\end{array}\right],$$ then $$\mathrm{A}\mathrm{B}=$$

A
$$\mathrm{A}$$

B
$$\mathrm{B}$$

C
$$I$$

D
$$\mathrm{O}$$

Solution

Multiplying the two matrices we get,

$$\begin{bmatrix} 0+abc-abc & 0+c{ b }^{ 2 }-c{ b }^{ 2 } & 0+b{ c }^{ 2 }-b{ c }^{ 2 } \\ -{ a }^{ 2 }c+0+{ a }^{ 2 }c & -abc+0+abc & -a{ c }^{ 2 }+0+a{ c }^{ 2 } \\ { a }^{ 2 }b-{ a }^{ 2 }b+0 & a{ b }^{ 2 }-a{ b }^{ 2 }+0 & abc-abc+0 \end{bmatrix}$$

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

$$= O$$

Hence the answer is option D
Question 6
1 / -0

lIf $$\mathrm{A} =\left[\begin{array}{ll}
a & 0\\
a & 0
\end{array}\right],\ \mathrm{B}=\left[\begin{array}{ll}
0 & 0\\
b & b
\end{array}\right],$$ then $$\mathrm{A}\mathrm{B}=$$

A
$$O$$

B
$$\mathrm{B}\mathrm{A}$$

C
$$\mathrm{A}\mathrm{B}$$

D
$$\mathrm{A}\mathrm{B}\mathrm{A}\mathrm{B}$$

Solution

$$\displaystyle AB = \begin{bmatrix} a&0\\a&0 \end{bmatrix} \begin{bmatrix} 0&0\\b&b \end{bmatrix}$$
$$=\begin{bmatrix} a\times 0+0\times b&a\times 0+0\times b\\a\times 0+0\times b&a\times 0+0\times b \end{bmatrix} $$
$$ =\begin{bmatrix} 0&0\\0&0 \end{bmatrix} = O$$
Question 7
1 / -0

If $$A=\left[\begin{array}{lll}
1 & -2 & 3\\
-4 & 2 & 5
\end{array}\right]$$ and $$B=\left[\begin{array}{ll}
2 & 3\\
4 & 5\\
2 & 1
\end{array}\right],$$ then

A
$$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$$ exist and equal

B
$$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$$ exist and are not equal

C
$$\mathrm{A}\mathrm{B}$$ exists and $$\mathrm{B}\mathrm{A}$$ does not exist

D
$$\mathrm{A}\mathrm{B}$$ does not exist and $$\mathrm{B}\mathrm{A}$$ exists

Solution

Since, we have $$A = 2 \times 3 , \: B = 3 \times 2$$
So, both $$AB$$ and $$BA$$ exist.
But both are not equal to each other because bith are if different order
$$\because AB \rightarrow 2 \times 2$$ and $$BA \rightarrow 3 \times 3.$$
Question 8
1 / -0

$$A=\left[\begin{array}{lll} 0 & 1 & -2\\1 & 0 & 3\\2 &-3 & 0 \end{array}\right]$$ then $$\mathrm{A}+\mathrm{A}^{\mathrm{T}}=$$

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

0 & 2 & 0\\

2 & 0 & 0\\

0 & 0 & 0

\end{array}\right]$$

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

1 & 0 & 0\\

0 & 3 & 0\\

0 & 0 & 4

\end{array}\right]$$

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

2 & 0 & 0\\

0 & 2 & 0\\

0 & 0 & 2

\end{array}\right]$$

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

2 & 0 & 2\\

0 & 2 & 0\\

0 & 0 & 2

\end{array}\right]$$

Solution

Given, $$A=\begin{bmatrix}
0 & 1 & -2\\
1 & 0& 3\\
2 & -3 & 0
\end{bmatrix}$$
So, By property of transpose (1),
$$A^{T}=\begin{bmatrix}
0 & 1 & 2\\
1 & 0 & -3\\
-2 & 3 & 0
\end{bmatrix}$$
So, By operation of matrixes (2),
$$A+A^{T}=\begin{bmatrix}
0 & 2 & 0\\
2 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$
Question 9
1 / -0

$$\left[\begin{array}{ll}
x & 0\\
0 & y
\end{array}\right]\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=$$

A
$$\left[\begin{array}{ll} ax & b_{X}\\ yc & dy \end{array}\right]$$

B
$$\left[\begin{array}{ll} ax & 0\\ 0 & dy \end{array}\right]$$

C
$$\left[\begin{array}{ll} ay & cy\\ bx & dy \end{array}\right]$$

D
$$\left[\begin{array}{ll} 0& ax\\ dy & 0 \end{array}\right]$$

Solution

The value of$$ \begin{bmatrix} x&0\\0&y \end{bmatrix} \begin{bmatrix} a&b\\c&d \end{bmatrix}$$
$$=\begin{bmatrix} x\times a+0\times c&x\times b+0\times d\\ 0\times a+y\times c&0\times b+y\times d \end{bmatrix} $$
$$ =\begin{bmatrix} ax&bx\\cy&dy \end{bmatrix} $$
Question 10
1 / -0

If $$\mathrm{A}=\left[\begin{array}{lll}
1 & -3 & -4\\
-1 & 3 & 4\\
1 & -3 & -4
\end{array}\right]$$, then $$\mathrm{A}^{2}=$$

A
$$A$$

B
$$- A$$

C
Null matrix

D
$$2A$$

Solution

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

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Re-Attempt Weekly Quiz Competition

Matrices Test - 17

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Matrices Test - 17

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Question 1

$$\displaystyle { A }^{ 2 }+5A-7I=O$$

$$\displaystyle -{ A }^{ 2 }+5A+7I=O$$

$$\displaystyle { A }^{ 2 }-5A+7I=O$$

$$\displaystyle { A }^{ 2 }+5A+7I=O$$

Solution

Question 2

B = C

$$B\neq C$$

B need not be equal to C

B = -C

Solution

Question 3

$$\mathrm{A}-\mathrm{B}$$

$$\mathrm{A}+\mathrm{B}$$

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

$$0$$

Solution

Question 4

$$I$$

$$0$$

$$\mathrm{A}$$

$$4{I}$$

Solution

Question 5

$$\mathrm{A}$$

$$\mathrm{B}$$

$$I$$

$$\mathrm{O}$$

Solution

Question 6

$$O$$

$$\mathrm{B}\mathrm{A}$$

$$\mathrm{A}\mathrm{B}$$

$$\mathrm{A}\mathrm{B}\mathrm{A}\mathrm{B}$$

Solution

Question 7

$$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$$ exist and equal

$$\mathrm{A}\mathrm{B},\ \mathrm{B}\mathrm{A}$$ exist and are not equal

$$\mathrm{A}\mathrm{B}$$ exists and $$\mathrm{B}\mathrm{A}$$ does not exist

$$\mathrm{A}\mathrm{B}$$ does not exist and $$\mathrm{B}\mathrm{A}$$ exists

Solution

Question 8

$$\left[\begin{array}{lll}0 & 2 & 0\\2 & 0 & 0\\0 & 0 & 0\end{array}\right]$$

$$\left[\begin{array}{lll}1 & 0 & 0\\0 & 3 & 0\\0 & 0 & 4\end{array}\right]$$

$$\left[\begin{array}{lll}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2\end{array}\right]$$

$$\left[\begin{array}{lll}2 & 0 & 2\\0 & 2 & 0\\0 & 0 & 2\end{array}\right]$$

Solution

Question 9

$$\left[\begin{array}{ll} ax & b_{X}\\ yc & dy \end{array}\right]$$

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

$$\left[\begin{array}{ll} ay & cy\\ bx & dy \end{array}\right]$$

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

Solution

Question 10

$$A$$

$$- A$$

Null matrix

$$2A$$

Solution

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

0 & 2 & 0\\

2 & 0 & 0\\

0 & 0 & 0

\end{array}\right]$$

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

1 & 0 & 0\\

0 & 3 & 0\\

0 & 0 & 4

\end{array}\right]$$

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

2 & 0 & 0\\

0 & 2 & 0\\

0 & 0 & 2

\end{array}\right]$$

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

2 & 0 & 2\\

0 & 2 & 0\\

0 & 0 & 2

\end{array}\right]$$