Matrices Test

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Matrices Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$ then $$A^{100}$$=..............

A
100A

B
$$2^{99}A$$

C
$$2^{100}A$$

D
99A

Solution
Question 2
1 / -0

iF $$A=\begin{bmatrix} 1& -1\\ -1& 1\end{bmatrix}$$, then the expression $$A^3-2A^2$$ is

A
a null matrix

B
an identity matrix

C
equal to A

D
equal to -A

Solution

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

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

$$=\begin{bmatrix}1+1&-1-1\\-1-1&1+1\end{bmatrix}$$

$$=\begin{bmatrix}2&-2\\-2&2\end{bmatrix}$$

$$\Rightarrow$$  $$A^2=\begin{bmatrix}2&-2\\-2&2\end{bmatrix}$$

$$\Rightarrow$$ $$A^3=$$ $$\begin{bmatrix}2&-2\\-2&2\end{bmatrix}$$ $$\begin{bmatrix} 1& -1\\ -1& 1\end{bmatrix}$$

$$=\begin{bmatrix}2+2&-2-2\\-2-2&2+2\end{bmatrix}$$

$$=\begin{bmatrix}4&-4\\-4&4\end{bmatrix}$$

$$\Rightarrow$$  $$A^3=\begin{bmatrix}4&-4\\-4&4\end{bmatrix}$$

$$\Rightarrow$$ $$A^3-2A^2=$$  $$\begin{bmatrix}4&-4\\-4&4\end{bmatrix}$$ $$-2\begin{bmatrix}2&-2\\-2&2\end{bmatrix}$$

$$=\begin{bmatrix}4&-4\\-4&4\end{bmatrix}-\begin{bmatrix}4&-4\\-4&4\end{bmatrix}$$

$$=\begin{bmatrix}4-4&-4+4\\-4+4&4-4\end{bmatrix}$$

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

$$\therefore$$ The expression $$A63-2A^2$$ is null matrix.
Question 3
1 / -0

If $$A=\begin{bmatrix} 1&2 \\ 2 &3\\3 & 4\end{bmatrix}$$ and $$B=\begin{bmatrix} 1 & 2\\ 2 & 1\end{bmatrix},$$ then which one of the following is correct?

A
Both $$AB$$ and $$BA$$ exist

B
Neither $$AB$$ nor $$BA$$ exists

C
$$AB $$ exists but $$BA$$ does not exist

D
$$AB$$ does not exists but $$BA$$ exists

Solution

$$AB$$ exists but $$BA$$ does not exist.

We know that, for matrices multiplications, with orders

$$m_1\times n_1,m_2\times n_2$$ condition is, $$n_1=m_2$$

Given, $$A_{3 \times 2}, B_{2 \times 2}$$

So, $$n_1=m_2\rightarrow 2=2$$ matrix $$AB$$ exists.

Also, for $$BA$$, we have, $$B_{2 \times 2}, A_{3 \times 2}$$

Thus, $$n_1\neq m_2\rightarrow 2\neq 3$$ matrix $$BA$$ does not exist.
Question 4
1 / -0

lf $$\left[\begin{array}{ll}
x & 1\\
1 & y
\end{array}\right]\left[\begin{array}{ll}
1 & 4\\
2 & 6
\end{array}\right] =\left[\begin{array}{ll}
4 & 14\\
7 & 22
\end{array}\right]$$, then $$(x,y)$$$$=$$

A
$$(1,-2)$$

B
$$( 2,1)$$

C
$$(3,2)$$

D
$$(2,3)$$

Solution

We have, $$\left[\begin{array}{ll}
x & 1\\
1 & y
\end{array}\right]\left[\begin{array}{ll}
1 & 4\\
2 & 6
\end{array}\right] =\left[\begin{array}{ll}
4 & 14\\
7 & 22
\end{array}\right]$$

$$\Rightarrow \left[\begin{array}{ll}
x\times1+1\times 2 & x\times4+1\times 6 \\
1\times 1+y\times 2 & 1\times 4+y\times 6
\end{array}\right]=\left[\begin{array}{ll}
4 & 14\\
7 & 22
\end{array}\right]$$

$$\Rightarrow \left[\begin{array}{ll}
x+2 & 4x+6\\
2y+1 & 6y+4
\end{array}\right]=\left[\begin{array}{ll}
4 & 14\\
7 & 22
\end{array}\right]$$

Now equating corresponding elements,
we get, $$(x,y)=(2,3)$$
Question 5
1 / -0

If $$D_{1}$$ and $$D_{2}$$ are two 3 x 3 diagonal matrices, then

A
$$D_{1}D_{2}$$ is a diagonal matrix

B
$$D_{1}+D_{2}$$ is a diagonal matrix

C
$$D_{1}^{2}+D_{2}^{2}$$ is a diagonal matrix

D
1, 2, 3 are correct

Solution

If $$D_{1}$$ and $$D_{2}$$ are diagonal matrix then
$$D_{1}D_{2},D_{1}+D_{2}$$ is also a diagonal matrix.
$$A=\begin{bmatrix}
a_{11} & & & \\
& a_{22} & & \\
& & . & \\
& & &a_{nn}
\end{bmatrix}$$

$$B=\begin{bmatrix}
l_{11} & & & \\
& l_{22} & & \\
& & . & \\
& & &l_{nn}
\end{bmatrix}$$

then,
$$AB=\begin{bmatrix}
a_{11}l_{11} & & & \\
& a_{22}l_{22} & & \\
& & . & \\
& & &a_{nn}l_{nn}
\end{bmatrix}$$

$$A+B=\begin{bmatrix}
a_{11}+l_{11} & & & \\
& a_{22}+l_{22} & & \\
& & . & \\
& & &a_{nn}+l_{nn}
\end{bmatrix}$$

$$A^2=\begin{bmatrix}
a_{11}^{2} & & & \\
& a_{22}^{2} & & \\
& & . & \\
& & &a_{nn}^{2}
\end{bmatrix}$$

$$\therefore D_{1}^{2}+D_{2}^{2}$$ is also a diagonal matrix.
Question 6
1 / -0

$$\left[\begin{array}{lll}
x & 0 & 0\\
y & \mathrm{z} & 0\\
l & m & n
\end{array}\right]\left[\begin{array}{lll}
a & 0 & 0\\
0 & b & 0\\
0 & 0 & c
\end{array}\right] =$$

A
$$\left[\begin{array}{lll} ax & 0 & 0\\ ay & b_{Z} & 0\\ al & mb & nc \end{array}\right]$$

B
$$\left[\begin{array}{lll} 0 & 0 & nc\\ 0 & b_{Z} & mb\\ a\mathrm{x} & ab & al \end{array}\right]$$

C
$$\left[\begin{array}{lll} a\mathrm{x} & ab & al\\ 0 & b_{Z} & mb\\ 0 & 0 & nc \end{array}\right]$$

D
None of these

Solution

We know that for the multiplication of matrices $$AB=C $$
So, $$C_{ij}=a_{ir} b_{rj}$$

$$\therefore$$ $$\begin{bmatrix}
x & 0 & 0\\
y & z & 0\\
l & m & n
\end{bmatrix}\begin{bmatrix}
a & 0& 0\\
0 & b & 0\\
0 & 0 & c
\end{bmatrix}$$

$$=\begin{bmatrix}
ax & 0 & 0\\
ay & bz& 0\\
al & mb & nc
\end{bmatrix}$$
Question 7
1 / -0

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

A
$$\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$

B
$$\begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$$

C
$$\begin{bmatrix} { 3 } & { -4 } \\ { 2 } & -1 \end{bmatrix}$$

D
None of these

Solution

Given, $$A=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\\ A^{ 2 }=\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\times \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}\\ =\begin{bmatrix} 9-4 & -12+4 \\ 3-1 & -4+1 \end{bmatrix}=\begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$$
Question 8
1 / -0

If $$\mathrm{A}= \left[\begin{array}{lll}
2 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 2
\end{array}\right]$$, then $$\mathrm{A}^{4}$$ is equal to

A
$$16\mathrm{A}$$

B
$$32 \mathrm A$$

C
$$4\mathrm{A}$$

D
$$8\mathrm{A}$$

Solution

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

$$A^{4} = 16 \begin{bmatrix}
1 &0 &0 \\
0 &1 &0 \\
0 &0 &1
\end{bmatrix}$$

$$A^{4} = 8 \times 2 \begin{bmatrix}
1 &0 &0 \\
0 &1 &0 \\
0 &0 &1
\end{bmatrix}$$

$$\therefore A^{4} = 8 \times A$$
Question 9
1 / -0

If $$A=\begin{bmatrix} a & b \\ b & a \end{bmatrix}$$ and $${ A }^{ 2 }=\begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix}$$, 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

We are given $$A=\begin{bmatrix} a & b \\ b & a \end{bmatrix}$$
Now we find $${ A }^{ 2 }=AA=\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} a & b \\ b & a \end{bmatrix}$$
$${ \Rightarrow A }^{ 2 }=\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & 2ab \\ 2ab & { a }^{ 2 }+{ b }^{ 2 } \end{bmatrix}$$
Compare it with the given $${ A }^{ 2 }$$ matrix. we get
$$\alpha ={ a }^{ 2 }+{ b }^{ 2 },\beta =2ab$$
Question 10
1 / -0

If $$A=[a_{ij}]_{3\times 3}$$ is a square matrix so that $$a_{ij}=i^{2}-j^{2}$$, then $$A$$ is a

A
unit matrix

B
symmetric marix

C
skew symmetric matrix

D
orthogonal matrix

Solution

Given:
$$A = [a_{ij}]_{(3\times3)}$$

where, $$a_{ij} = i^2-j^2$$

$$\therefore a_{ij}=0$$ if $$i=j$$

Now,
$$a_{12}=1^{2}-2^{2}=-3$$

$$a_{13}=1^{2}-3^{2}=-8$$

$$a_{21}=2^2-1^2 = 3$$

$$a_{23}=2^{2}-3^{2}=-5$$

$$a_{31}=3^2 - 1^2 = 8$$

$$a_{32}=3^2-2^2 = 5$$

$$\therefore A=\begin{bmatrix}
0 &-3 &-8 \\
3 & 0&-5 \\
8& 5 &0
\end{bmatrix}$$

Here, $$A^T=-A$$

$$\therefore\ A$$ is a skew-symmetric matrix.

Hence, option C.

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Re-Attempt Weekly Quiz Competition

Matrices Test - 24

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Matrices Test - 24

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SHARING IS CARING

Question 1

100A

$$2^{99}A$$

$$2^{100}A$$

99A

Solution

Question 2

a null matrix

an identity matrix

equal to A

equal to -A

Solution

Question 3

Both $$AB$$ and $$BA$$ exist

Neither $$AB$$ nor $$BA$$ exists

$$AB $$ exists but $$BA$$ does not exist

$$AB$$ does not exists but $$BA$$ exists

Solution

Question 4

$$(1,-2)$$

$$( 2,1)$$

$$(3,2)$$

$$(2,3)$$

Solution

Question 5

$$D_{1}D_{2}$$ is a diagonal matrix

$$D_{1}+D_{2}$$ is a diagonal matrix

$$D_{1}^{2}+D_{2}^{2}$$ is a diagonal matrix

1, 2, 3 are correct

Solution

Question 6

$$\left[\begin{array}{lll} ax & 0 & 0\\ ay & b_{Z} & 0\\ al & mb & nc \end{array}\right]$$

$$\left[\begin{array}{lll} 0 & 0 & nc\\ 0 & b_{Z} & mb\\ a\mathrm{x} & ab & al \end{array}\right]$$

$$\left[\begin{array}{lll} a\mathrm{x} & ab & al\\ 0 & b_{Z} & mb\\ 0 & 0 & nc \end{array}\right]$$

None of these

Solution

Question 7

$$\begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$

$$\begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$$

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

None of these

Solution

Question 8

$$16\mathrm{A}$$

$$32 \mathrm A$$

$$4\mathrm{A}$$

$$8\mathrm{A}$$

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

unit matrix

symmetric marix

skew symmetric matrix

orthogonal matrix

Solution

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