Matrices Test

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Matrices Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=[x,y], B=\left[\begin{array}{ll}
a & h\\
h & b
\end{array}\right], C=\left[\begin{array}{l}
x\\
y
\end{array}\right]$$,
then $$\mathrm{A}\mathrm{B}\mathrm{C}=$$

A
$$(ax+hy+bxy)$$

B
$$(ax^{2}+2hxy+by^{2})$$

C
$$(ax^{2}-2hxy+by^{2})$$

D
$$(bx^{2}-2hxy+ay^{2})$$

Solution

Since, $$A = \begin{bmatrix}
x &y
\end{bmatrix}$$
$$B = \begin{bmatrix}
a &h \\
h &b
\end{bmatrix}$$
$$C =\begin{bmatrix}
x\\
y
\end{bmatrix}$$
$$ABC = \begin{bmatrix}
x &y
\end{bmatrix}\begin{bmatrix}
a &h \\
h &b
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}$$

$$=\begin{bmatrix}
xa+yh &xh+yb
\end{bmatrix}\begin{bmatrix}
x\\
y
\end{bmatrix}$$

$$=\begin{bmatrix}
x^{2}a+xyh+xyh+y^{2}b
\end{bmatrix}$$

$$=\begin{bmatrix}
x^{2}a+2xyh+y^{2}b
\end{bmatrix}$$
Question 2
1 / -0

If $$A=\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} $$, then $$ A^{3}-A^{2}=$$

A
$$2A$$

B
$$2I$$

C
$$A$$

D
$$I$$

Solution

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

$$A^{3}=A^{2}\times A=\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}$$

$$A^{3}-A^{2}=\begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}=\begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$$

$$=2\begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}=2A$$
Question 3
1 / -0

If the transpose of a matrix is equal to the additive
inverse, then matrix is called _________
matrix.

A
symmetric

B
skew symmetric

C
identity

D
inverse

Solution
Question 4
1 / -0

$$[A]_{n\times m}, [B]_{m\times m},$$ are the two matrices. If multiplication AB exist, then

A
$$n = 1, m = 1$$

B
$$n = 2, m = 1$$

C
$$n = 1, m = 3$$

D
All of the above are valid

Solution

For multiplication of two matrices to exist, number of columns in the first matrix must be equal to number of rows in the second matrix.
So for a given case, $$m = m$$ which is always true.
Therefore multiplication exists for every $$n$$ and $$m.$$
Question 5
1 / -0

If $$A=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$, then $$A^{4}=$$

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

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

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

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

Solution

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

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

$$A^{4}=A^{2}\times A^{2}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
Question 6
1 / -0

If $$I = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix},$$ then find $$I^3$$

A
$$1$$

B
$$I$$

C
$$0$$

D
does not exist

Solution

$$I^n$$ ( for any natural $$n$$ ) $$= I.$$
$$I$$ is known as the identity matrix.
Question 7
1 / -0

If $$A= \begin{bmatrix}
1 & 2 & 3\\
4 & 5 & 6
\end{bmatrix}$$ and $$B= \begin{bmatrix}
1\\
0\\

5\end{bmatrix},$$ then $$AB = $$

A
$$\begin{bmatrix}
1 & 0 & 15
\end{bmatrix}$$

B
$$\begin{bmatrix}
4 & 0 & 30
\end{bmatrix}$$

C
$$\begin{bmatrix}
16\\

34\end{bmatrix}$$

D
$$\begin{bmatrix}
16 & 34
\end{bmatrix}$$

Solution

$$AB= \begin{bmatrix}
1 &2 &3 \\
4 &5 &6
\end{bmatrix}\begin{bmatrix}
1\\
0\\
5\end{bmatrix}=\begin{bmatrix}
1+ 0+ 15\\
4+ 0+30
\end{bmatrix}=\begin{bmatrix}
16\\
34
\end{bmatrix}$$
Thus, C will be correct answer.
Question 8
1 / -0

If $$A = \begin{bmatrix}a & b\end{bmatrix},\space B = \begin{bmatrix}-b & -a \end{bmatrix}$$ and $$C = \begin{bmatrix}a \\ -a\end{bmatrix}$$, then the correct statement is

A
$$A = -B$$

B
$$A+B = A-B$$

C
$$AC = BC$$

D
$$CA = CB$$

Solution

Given $$A = \begin{bmatrix}a & b\end{bmatrix}, B = \begin{bmatrix}-b & -a \end{bmatrix}$$ and $$C = \begin{bmatrix}a \\ -a\end{bmatrix}$$
We will check by options.
Clearly , $$A \ne -B$$ as their corresponding elements are different only.
$$A+B= \begin{bmatrix}a-b & b-a\end{bmatrix}$$
$$A-B=\begin{bmatrix}a+b & b+a\end{bmatrix}$$
So, $$A+B \ne A-B$$

Now,$$AC=\begin{bmatrix}a & b\end{bmatrix}\begin{bmatrix}a \\ -a\end{bmatrix}$$
$$\Rightarrow AC=\begin{bmatrix}a^{2}-ab\end{bmatrix}$$
$$BC=\begin{bmatrix}-b & -a \end{bmatrix}\begin{bmatrix}a \\ -a\end{bmatrix}$$
$$\Rightarrow BC=\begin{bmatrix}a^{2}-ab\end{bmatrix}$$
Hence, $$AC=BC$$
Option $$C$$ is correct

Now, $$CA=\begin{bmatrix}a \\ -a\end{bmatrix}\begin{bmatrix}a & b\end{bmatrix}$$
$$\Rightarrow CA=\begin{bmatrix}a^{2}&ab\\-a^{2}&-ab \end{bmatrix}$$
$$CB=\begin{bmatrix}a \\ -a\end{bmatrix}\begin{bmatrix}-b & -a \end{bmatrix}$$
$$\Rightarrow CB=\begin{bmatrix}-ab&-a^{2}\\ab & a^{2}\end{bmatrix}$$
Hence, $$CA\ne CB$$
Question 9
1 / -0

If $$\displaystyle A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & 3 & 2 \\ 3 & 1 & 5 \end{bmatrix}$$ and $$\displaystyle B = \begin{bmatrix} 0 & -2 & 4 \\ 1 & 3 & 2 \\ -1 & 1 & 5 \end{bmatrix}$$, then $$A + B$$ is

A
$$\displaystyle \begin{bmatrix}

1 & -2 & 4 \\

3 & 3 & 2 \\

2 & 1 & 5

\end{bmatrix}$$

B
$$\displaystyle \begin{bmatrix}

1 & -2 & 8 \\

3 & 3 & 4 \\

2 & 1 & 10

\end{bmatrix}$$

C
$$\displaystyle \begin{bmatrix}

1 & -4 & 8 \\

3 & 6 & 4 \\

2 & 2 & 10

\end{bmatrix}$$

D
none of these

Solution

Given, $$\displaystyle A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & 3 & 2 \\ 3 & 1 & 5 \end{bmatrix}$$ and $$\displaystyle B = \begin{bmatrix} 0 & -2 & 4 \\ 1 & 3 & 2 \\ -1 & 1 & 5 \end{bmatrix}$$

Therefore, $$A+B=\begin {bmatrix} 1+0 & -2-2 & 4+4 \\2+1 & 3+3 & 2+2 \\3-1 & 1+1 & 5+5\end {bmatrix}$$
$$=\begin{bmatrix} 1 & -4 & 8 \\ 3 & 6 & 4 \\ 2 & 2 & 10 \end{bmatrix}$$
Question 10
1 / -0

If $$A = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 & 8\end{bmatrix},$$ $$B = \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 & 5\end{bmatrix}$$ and $$ A+B = \begin{bmatrix}1 & a & b \\ c & 2 & 13\end{bmatrix},$$ then find the value of $$a+b+c.$$

A
$$12$$

B
$$21$$

C
$$15$$

D
$$5$$

Solution

Given, $$A = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 & 8\end{bmatrix},$$ $$B = \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 & 5\end{bmatrix}$$ and $$\quad A+B = \begin{bmatrix}1 & a & b \\ c & 2 & 13\end{bmatrix}$$ ....(1)
Therefore,
$$\quad A+B = \begin{bmatrix}2 & 3 & 4 \\ -3 & 4 &

8\end{bmatrix} + \begin{bmatrix}-1 & 4 & 7 \\ -3 & -2 &

5\end{bmatrix}=\begin{bmatrix}2-1 & 3+4 & 4+7 \\ -3-3 & 4-2 & 8+5\end{bmatrix}= \begin{bmatrix}1 & 7 & 11 \\ -6 & 2 & 13\end{bmatrix}$$ .....(2)
By equating (1) and (2), we get
$$ a+b+c=7+11-6=12$$

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Matrices Test - 18

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Matrices Test - 18

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Question 1

$$(ax+hy+bxy)$$

$$(ax^{2}+2hxy+by^{2})$$

$$(ax^{2}-2hxy+by^{2})$$

$$(bx^{2}-2hxy+ay^{2})$$

Solution

Question 2

$$2A$$

$$2I$$

$$A$$

$$I$$

Solution

Question 3

symmetric

skew symmetric

identity

inverse

Solution

Question 4

$$n = 1, m = 1$$

$$n = 2, m = 1$$

$$n = 1, m = 3$$

All of the above are valid

Solution

Question 5

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

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

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

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

Solution

Question 6

$$1$$

$$I$$

$$0$$

does not exist

Solution

Question 7

$$\begin{bmatrix}1 & 0 & 15\end{bmatrix}$$

$$\begin{bmatrix}4 & 0 & 30\end{bmatrix}$$

$$\begin{bmatrix}16\\ 34\end{bmatrix}$$

$$\begin{bmatrix}16 & 34\end{bmatrix}$$

Solution

Question 8

$$A = -B$$

$$A+B = A-B$$

$$AC = BC$$

$$CA = CB$$

Solution

Question 9

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

$$\displaystyle \begin{bmatrix}1 & -2 & 8 \\ 3 & 3 & 4 \\ 2 & 1 & 10\end{bmatrix}$$

$$\displaystyle \begin{bmatrix}1 & -4 & 8 \\ 3 & 6 & 4 \\ 2 & 2 & 10\end{bmatrix}$$

none of these

Solution

Question 10

$$12$$

$$21$$

$$15$$

$$5$$

Solution

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$$\begin{bmatrix}
1 & 0 & 15
\end{bmatrix}$$

$$\begin{bmatrix}
4 & 0 & 30
\end{bmatrix}$$

$$\begin{bmatrix}
16\\

34\end{bmatrix}$$

$$\begin{bmatrix}
16 & 34
\end{bmatrix}$$

$$\displaystyle \begin{bmatrix}

1 & -2 & 4 \\

3 & 3 & 2 \\

2 & 1 & 5

\end{bmatrix}$$

$$\displaystyle \begin{bmatrix}

1 & -2 & 8 \\

3 & 3 & 4 \\

2 & 1 & 10

\end{bmatrix}$$

$$\displaystyle \begin{bmatrix}

1 & -4 & 8 \\

3 & 6 & 4 \\

2 & 2 & 10

\end{bmatrix}$$