Matrices Test

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Matrices Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

If $$A = \begin{vmatrix} 5 & x-2 \\ 2x+3 & x+1 \end{vmatrix}$$ is symmetric, then $$x = $$_____

A
$$4$$

B
$$5$$

C
$$-5$$

D
$$-4$$

Solution

A=$$ \begin {vmatrix} 5&x-2\\2x-3&x+1\end{vmatrix}$$
$$\implies A^T=$$$$ \begin {vmatrix} 5&2x-3\\x-2&x+1\end{vmatrix}$$
Since $$A$$ is symmetric, $$\implies A=A^T$$
$$\implies x-2=2x+3$$
$$\implies x=-5$$
Question 2
1 / -0

If $$A=\begin{bmatrix}x&y&z\end{bmatrix},$$ $$ B=\begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}, C=\begin{bmatrix}\alpha & \beta & \gamma \end{bmatrix}^{T}$$ then $$ABC$$ is

A
Not defined

B
a $$1\times1$$ matrix

C
a $$3\times3$$ matrix

D
a $$3\times2$$ matrix

Solution

$$A=\begin{bmatrix} x & y & z \end{bmatrix}\quad \quad B=\begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix},\quad \quad C=\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}\\ \quad \quad \quad 1\times 3\qquad \qquad \quad \quad \quad \ 3\times 3\qquad \qquad \qquad \ 3\times 1$$

So $$ABC$$ is a $$1 \times1$$ matrix
Question 3
1 / -0

If $$A=\begin{bmatrix} 3 & 3 & 3\\ 3 & 3 & 3\\ 3 & 3 & 3\end{bmatrix}$$, then $$A^3=$$ ___________.

A
$$243$$A

B
$$81$$A

C
$$27$$A

D
$$729$$A

Solution

$$\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}=\begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}=A$$

$$\implies A =\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}=\begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}=3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$

$$A^2=\begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}\times \begin{bmatrix}3 & 3 & 3 \\ 3 & 3 & 3 \\3 & 3 & 3 \\\end{bmatrix}$$

$$=3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$

$$=9\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}\times\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$

$$=27\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$

$$A^3=A^2.A$$
$$=27\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix} \times 3\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$

$$=81\begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix} \times \begin{bmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\1 & 1 & 1 \\\end{bmatrix}$$

$$=81A$$

The answer is option (B).
Question 4
1 / -0

If $$A =\begin{pmatrix} -2& 2\\ 2 & -2\end{pmatrix}$$, then which one of the following is correct?

A
$$A^{2} = -2A$$

B
$$A^{2} = -4A$$

C
$$A^{2} = -3A$$

D
$$A^{2} = 4A$$

Solution

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

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

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

$$=\begin{bmatrix} 8 & -8 \\ -8 & 8 \end{bmatrix}=-4\begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}=-4A$$
Question 5
1 / -0

If A = $$ \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix}$$ , B = $$ \begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$$ whenever $$A^2 \, = \, B$$
then values of $$\alpha$$ is

A
$$1$$

B
$$-1$$

C
$$4$$

D
no real value of $$\alpha$$

Solution

$$A^2$$ = B$$\Rightarrow \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix} \begin{bmatrix} \alpha & 0 \\ 1 & 1\end{bmatrix} \, = \, \begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} \alpha^2 & 0 \\ \alpha+1 & 1\end{bmatrix} \, = \, \begin{bmatrix} 1 & 0 \\ 5 & 1\end{bmatrix}$$
$$\Rightarrow \alpha^2$$ = 1, $$\alpha$$ + 1 = 5
$$\therefore \alpha \, = \, \pm 1\, , \, \alpha \, = \, 4$$
There is no real value of $$\alpha$$ which satisfies both.
Question 6
1 / -0

Let A = $$\begin{bmatrix}
0 & 0 & -1 \\[0.3em]
0 & -1 & 0 \\[0.3em]
-1 & 0 & 0
\end{bmatrix}.$$ Then the only correct statement $$A$$ is

A
$$A = 0$$

B
$$A = (-1) I$$

C
$$A^{-1} $$ does not exist

D
$$A^2 = I$$

Solution

$$A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \, \, \therefore A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \, \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$$
$$A^2 = \begin{bmatrix} 0 + 0 + 1 & 0 & 0 \\ 0 & 0 + 1 + 0 & 0 \\ 0 & 0 & 0 + 0 + 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\therefore A^2 = I $$
Hence it is a identity matrix
Question 7
1 / -0

Using elementary row transformation find the inverse of the matrix A = $$\left[\begin {array}{ll} 3 & -1 & -2\\ 2 & 0 & -1\\ 3 & -5 & 0 \end {array}\right]$$

A
$$\left[\begin {array} {ll} \dfrac{-5}{8} & \dfrac{5}{4} & \dfrac{1}{8} \\
\dfrac{-3}{8} & \dfrac{3}{4} & \dfrac{-1}{8} \\
\dfrac{-5}{4} & \dfrac{3}{2} &\dfrac{1}{4} \end {array}\right]$$

B
$$\dfrac {1}{8}\left [\begin {array} {ll} 5 & -5 & 1 \\ 3 & -3 & 1 \\ 0 & 3 & 1\end {array}\right]$$

C
$$\dfrac{1}{8} \left[\begin {array} {ll} 5 & 5 & 1 \\
3 & 6 & -1 \\
10 & -12 & 2 \end {array}\right]$$

D
None of these

Solution
Question 8
1 / -0

Given $$A= \begin{bmatrix} 3&6 \\ -2&-8 \end{bmatrix}$$ and $$B = \begin{bmatrix} 2 & 16\end{bmatrix}$$, find the matrix $$X$$ such that $$XA=B$$.

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

B
$$\begin{bmatrix} \dfrac{4}{3} & 3 \end{bmatrix}$$

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

D
$$\begin{bmatrix} -\dfrac{4}{3} & 3 \end{bmatrix}$$

Solution

$$XA=B$$
$${ X }_{ 1\times 2 }{ \begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix} }_{ 2\times 2 }={ \begin{bmatrix} 2 & 16 \end{bmatrix} }_{ 1\times 2 }$$
$$X$$ should be of order $$1 \times 2$$
Let $$X$$ be $$\begin{bmatrix} a & b \end{bmatrix}$$
$$\therefore \begin{bmatrix} a & b \end{bmatrix}\begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix}=\begin{bmatrix} 2 & 16 \end{bmatrix}\\ \Rightarrow \begin{bmatrix} 3a-2b & 6a-8b \end{bmatrix}=\begin{bmatrix} 2 & 16 \end{bmatrix}\\ \therefore 3a-2b=2\longrightarrow (1)\times 2\\ \quad 6a-8b=16\longrightarrow (2)\times 1\\ \therefore 6a-4b=4\\ \quad 6a-8b=16\\ \quad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \ 4b=-12\\ \quad \ \Rightarrow b=-3\\ \therefore 3a+6=2\\ \quad a=\dfrac{-4}{3}\\ \therefore X=\begin{bmatrix} \dfrac{-4}{3} & -3 \end{bmatrix}$$
Question 9
1 / -0

Obtain the inverse of the following matrix using elementary operation:
$$A = \begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix}$$

A
$$\begin{bmatrix} -3 & -4 & -3 \\ 2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}$$

B
$$\begin{bmatrix} 3 & 4 & 3 \\ 2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}$$

C
$$\begin{bmatrix} -3 & -4 & 3 \\ 2 & 3 & -2 \\ -8 & -12 & 9 \end{bmatrix}$$

D
$$\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$$

Solution

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

Therefore, $$A^{-1}=\dfrac{1}{|A|}adj(A)$$

$$\implies A^{-1}=\dfrac{1}{3(3-0)-1(8-0)}\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$$

$$\implies A^{-1}=\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$$
Question 10
1 / -0

If $$I =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & \quad \quad 0 & \quad 1 \end{pmatrix},$$ $$P=\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & \quad \quad 0 & \quad -2 \end{pmatrix},$$ then the matrix $$P^3+2P^2$$ is equal to

A
$$P$$

B
$$1-P$$

C
$$2I+P$$

D
$$2I-P$$

Solution

Given the matrix $$P=\begin{pmatrix}1&0&0\\0&-1&0\\ 0&0&-2\end {pmatrix}$$.
Now,
$$P^2=\begin{pmatrix}1^2&0&0\\0&(-1)^2&0\\ 0&0&(-2)^2\end {pmatrix}$$ [ Since $$P$$ is a diagonal matrix]
$$\Rightarrow P^2=\begin{pmatrix}1&0&0\\0&1&0\\ 0&0&4\end {pmatrix}$$......(1).
Again,
$$P^3=\begin{pmatrix}1^3&0&0\\0&(-1)^3&0\\ 0&0&(-2)^3\end {pmatrix}$$
$$\Rightarrow P^3=\begin{pmatrix}1&0&0\\0&-1&0\\ 0&0&-8\end {pmatrix}$$.....(2).
So,
$$P^3+2P^2$$ $$=\begin{pmatrix}1+2&0&0\\0&-1+2&0\\ 0&0&-8+8\end {pmatrix}$$ $$=\begin{pmatrix}3&0&0\\0&1&0\\ 0&0&0\end {pmatrix}$$ $$=2I+P$$.

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Re-Attempt Weekly Quiz Competition

Matrices Test - 38

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Matrices Test - 38

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SHARING IS CARING

Question 1

$$4$$

$$5$$

$$-5$$

$$-4$$

Solution

Question 2

Not defined

a $$1\times1$$ matrix

a $$3\times3$$ matrix

a $$3\times2$$ matrix

Solution

Question 3

$$243$$A

$$81$$A

$$27$$A

$$729$$A

Solution

Question 4

$$A^{2} = -2A$$

$$A^{2} = -4A$$

$$A^{2} = -3A$$

$$A^{2} = 4A$$

Solution

Question 5

$$1$$

$$-1$$

$$4$$

no real value of $$\alpha$$

Solution

Question 6

$$A = 0$$

$$A = (-1) I$$

$$A^{-1} $$ does not exist

$$A^2 = I$$

Solution

Question 7

$$\left[\begin {array} {ll} \dfrac{-5}{8} & \dfrac{5}{4} & \dfrac{1}{8} \\\dfrac{-3}{8} & \dfrac{3}{4} & \dfrac{-1}{8} \\\dfrac{-5}{4} & \dfrac{3}{2} &\dfrac{1}{4} \end {array}\right]$$

$$\dfrac {1}{8}\left [\begin {array} {ll} 5 & -5 & 1 \\ 3 & -3 & 1 \\ 0 & 3 & 1\end {array}\right]$$

$$\dfrac{1}{8} \left[\begin {array} {ll} 5 & 5 & 1 \\3 & 6 & -1 \\10 & -12 & 2 \end {array}\right]$$

None of these

Solution

Question 8

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

$$\begin{bmatrix} \dfrac{4}{3} & 3 \end{bmatrix}$$

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

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

Solution

Question 9

$$\begin{bmatrix} -3 & -4 & -3 \\ 2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}$$

$$\begin{bmatrix} 3 & 4 & 3 \\ 2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}$$

$$\begin{bmatrix} -3 & -4 & 3 \\ 2 & 3 & -2 \\ -8 & -12 & 9 \end{bmatrix}$$

$$\begin{bmatrix}3& -2 &8 \\ -4& 3&-12\\3 &-2&9\end{bmatrix}$$

Solution

Question 10

$$P$$

$$1-P$$

$$2I+P$$

$$2I-P$$

Solution

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$$\left[\begin {array} {ll} \dfrac{-5}{8} & \dfrac{5}{4} & \dfrac{1}{8} \\
\dfrac{-3}{8} & \dfrac{3}{4} & \dfrac{-1}{8} \\
\dfrac{-5}{4} & \dfrac{3}{2} &\dfrac{1}{4} \end {array}\right]$$

$$\dfrac{1}{8} \left[\begin {array} {ll} 5 & 5 & 1 \\
3 & 6 & -1 \\
10 & -12 & 2 \end {array}\right]$$