Matrices Test

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Matrices Test - 34

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Question 1
1 / -0

If $$A=[x\, y\, z], B = \begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}$$ and $$C=\begin{bmatrix}x\\y\\z\end{bmatrix}$$
Then, $$ABC = 0$$, if

A
$$[ax^2+by^2+cz^2+2gxy+2fyz+2czx]=0$$

B
$$[ax^2+cy^2+bz^2+xy+yz+zx]=0$$

C
$$[ax^2+by^2+cz^2+2hxy+2by+2cz]=0$$

D
$$[ax^2+by^2+cz^2+2gzx+2fyz+2hxy]=0$$

Solution

$$ A = [x\,y\,z] B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} c = \begin{bmatrix} x \\ y \\ z \end{bmatrix} $$
$$ A.B = [x\,y\,z] \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix} $$
$$ = [ax+hy+gz\,\,\,\,hx+by+fz\,\,\,\,gx+fy+cz] $$
$$ A.B.C = (AB)\times \begin{bmatrix} x \\ y \\ z\end{bmatrix} = [ax^{2}+hyx+gyx+hxy+by^{2}+fzy+gxz+fyz+cz^{2}] $$
$$ ABC = [ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy] $$
$$ ABC = 0 $$
when,
$$ ax^{2}+by^{2}+cz^{2}+2hxy+2gzx+2fzy = 0 $$
Question 2
1 / -0

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

A
Null matrix

B
Unit matrix

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

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

Solution

$$A=\begin{bmatrix} -2&4\\-1&2 \end{bmatrix}$$
$$A.A.=A^{2}= \begin{bmatrix} -2&4\\-1&2 \end{bmatrix} \begin{bmatrix} -2&4\\-1&2 \end{bmatrix} = \begin{bmatrix} 4-4\,\,-2\times 4+4\times 2\\-1x-2+2x-1\,\,-1\times 4+2\times 2 \end{bmatrix} $$
$$ A^{2}=\begin{bmatrix} 0&0\\0&0 \end{bmatrix} = O =$$ null matrix.
Question 3
1 / -0

If $$P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$$, then $${P}^{5}$$ is equal to

A
$$P$$

B
$$2P$$

C
$$-P$$

D
$$-2P$$

Solution

Given $$P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$$

$${P}^{2}=P.P=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}$$

$$=\begin{pmatrix} 4+2-4 & -4-6+8 & -8-8+12 \\ -2-3+4 & 2+9-8 & 4+12-12 \\ 2+2-3 & -2-6+6 & -4-8+9 \end{pmatrix}$$

$$=\begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{pmatrix}=P$$

$$\therefore$$ $${P}^{4}={P}^{2}=P$$

$$\Rightarrow$$ $${P}^{5}={P}^{2}=P$$
Question 4
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

$$A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ { A }^{ 2 }=A\times A=\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\\ \quad \quad =\begin{bmatrix} 4+1 & -2-2 \\ -2-2 & 1+4 \end{bmatrix}=\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}$$
From the given option to get first entry i.e $$5$$ of $$A^2$$, only option $$(A)$$ is satisfied. Or we can check as below
$$4A-3I=\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-3\times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 8 & -4 \\ -4 & 8 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\\ \quad \quad \quad \quad =\begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}={ A }^{ 2 }$$
Question 5
1 / -0

If $$A$$ is a matrix such that $$\begin{pmatrix}2&1 \\3&2 \end{pmatrix} A(1 \space \,1)=\begin{pmatrix}1&1 \\0&0 \end{pmatrix}$$ then $$A= $$

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

B
$$(2 \,1)$$

C
$$\begin{pmatrix}1&0 \\-1&1 \end{pmatrix}$$

D
$$\begin{pmatrix}2 \\-3 \end{pmatrix}$$

Solution

$$\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}A \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$
To maintain the rules of matrix multiplication, $$A$$ has to be of the order $$2 \times 1$$
$$\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$

$$\Rightarrow \begin{pmatrix} 2x + y \\ 3x + 2y \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$

$$\Rightarrow \begin{pmatrix} 2x + y & 2x + y \\ 3x + 2y & 3x + 2y \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$

$$\Rightarrow 2x + y = 1$$ and $$3x + 2y = 0$$
$$\Rightarrow 4x + 2y - (3x + 2y) = 2 - 0$$
$$\Rightarrow x = 2$$ and $$y = -3$$
Question 6
1 / -0

If $$A$$ and $$B$$ are any two matrices, then

A
$$AB=BA$$

B
$$AB=I$$

C
$$AB=0$$

D
$$AB$$ may or may not be defined

Solution

$$\textbf{Step 1: Finding the order}$$
$$\text{Matrix multiplication is only defined for two matrices whose order is of the form}$$
$$a\times b \text{ and }b\times c \text{ respectively}$$
$$\text{Since we cant say anything about the order of the given matrices it may or may not be defined}$$
$$\textbf{Hence, AB may or may not be defined}$$
Question 7
1 / -0

For the matrices $$A=\begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix}$$ $$B=\begin{bmatrix} 2 & 2 \\ -1 & 4 \end{bmatrix}$$
What is $$AB$$?

A
$$\begin{bmatrix} 3 & 5 \\ -1 & 3 \end{bmatrix}$$

B
$$\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}$$

C
$$\begin{bmatrix} -1 & 14 \\ 1 & -4 \end{bmatrix}$$

D
$$\begin{bmatrix} 2 & 4 \\ -1 & -7 \end{bmatrix}$$

E
none of the above

Solution

$$A$$ and $$B$$ are given by $$A=\left[ \begin{matrix} 1 & 3 \\ 0 & -1 \end{matrix} \right] , B=\left[ \begin{matrix} 2 & 2 \\ -1 & 4 \end{matrix} \right]$$
Thus, $$AB=\left[ \begin{matrix} (1\times 2)+(3\times -1) & (1\times 2)+(3\times 4) \\ (0\times 2)+(-1\times -1) & (0\times 2)+(-1\times 4) \end{matrix} \right] $$
$$=\left[ \begin{matrix} 2-3 & 2+12 \\ 0+1 & 0-4 \end{matrix} \right] $$
$$=\left[ \begin{matrix} -1 & 14 \\ 1 & -4 \end{matrix} \right]$$
Question 8
1 / -0

If $$A = \begin{bmatrix} 0& 1\\ 1 & 0\end{bmatrix}$$, then $$A^{2}$$ is equal to ______

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

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

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

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

Solution

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

Hence $$A^2=AA=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix} 0&1\\1&0\end{bmatrix}$$
$$\quad = \begin{bmatrix}0\times 0+1\times 1&0\times 1+1\times 0\\1\times 0+0\times 1 &1\times 1+0\times 0\end{bmatrix}$$
$$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$$
Question 9
1 / -0

$$\begin{bmatrix} 3& 2\\ -1 & -2\\ 4 & 5\end{bmatrix} \begin{bmatrix} 0& a\\ b & c \end{bmatrix} = \begin{bmatrix}-4 &9 \\ 4 & -7\\ -10 & 19\end{bmatrix}$$
Find $$a, b$$ and $$c$$.

A
$$2, -1, 3$$

B
$$1, -2, 3$$

C
$$3, -2, 1$$

D
$$1, 2, 3$$

Solution

The product is $$\begin{bmatrix} 3 & 2 \\ -1 & -2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 0 & a \\ b & c \end{bmatrix}=$$ $$\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix}$$
By given, we have

$$\begin{bmatrix} 2b & 3a+2c \\ -2b & -2c-a \\ 5b & 5c+4a \end{bmatrix}$$ $$=\begin{bmatrix} -4 & 9 \\ 4 & -7 \\ -10 & 19 \end{bmatrix}$$

By comparing, we get
$$b=-2 , a = 1 , c = 3$$
Question 10
1 / -0

If $$P=\begin{pmatrix}2&-2&-4 \\-1&3&4\\1&-2&-3\end{pmatrix}$$ then $$P^5$$ equals

A
$$P$$

B
$$2P$$

C
$$-P$$

D
$$-2P$$

Solution

$$P=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$$
$${ P }^{ 2 }=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$$
$$=\begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}=P$$
Now,
$${ P }^{ 2 }=P\quad { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 3 }\Rightarrow { P }^{ 5 }={ P }^{ 4 }$$
$$\left( \because { P }^{ 2 }=P \right) \Rightarrow { P }^{ 5 }={ P }^{ 2 }\times { P }^{ 2 }$$
$${ P }^{ 2 }=P\times P$$
$$\Rightarrow { P }^{ 5 }={ P }^{ 2 }=P$$
$$\therefore { P }^{ 5 }=P$$
$$\therefore $$Option A is correct

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Re-Attempt Weekly Quiz Competition

Matrices Test - 34

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Matrices Test - 34

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SHARING IS CARING

Question 1

$$[ax^2+by^2+cz^2+2gxy+2fyz+2czx]=0$$

$$[ax^2+cy^2+bz^2+xy+yz+zx]=0$$

$$[ax^2+by^2+cz^2+2hxy+2by+2cz]=0$$

$$[ax^2+by^2+cz^2+2gzx+2fyz+2hxy]=0$$

Solution

Question 2

Null matrix

Unit matrix

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

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

Solution

Question 3

$$P$$

$$2P$$

$$-P$$

$$-2P$$

Solution

Question 4

$$4A - 3I$$

$$3A - 4I$$

$$A - I$$

$$A + I$$

Solution

Question 5

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

$$(2 \,1)$$

$$\begin{pmatrix}1&0 \\-1&1 \end{pmatrix}$$

$$\begin{pmatrix}2 \\-3 \end{pmatrix}$$

Solution

Question 6

$$AB=BA$$

$$AB=I$$

$$AB=0$$

$$AB$$ may or may not be defined

Solution

Question 7

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

$$\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}$$

$$\begin{bmatrix} -1 & 14 \\ 1 & -4 \end{bmatrix}$$

$$\begin{bmatrix} 2 & 4 \\ -1 & -7 \end{bmatrix}$$

none of the above

Solution

Question 8

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

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

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

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

Solution

Question 9

$$2, -1, 3$$

$$1, -2, 3$$

$$3, -2, 1$$

$$1, 2, 3$$

Solution

Question 10

$$P$$

$$2P$$

$$-P$$

$$-2P$$

Solution

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