Matrices Test

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Matrices Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$A + B$$ exists

B
$$AB$$ exists

C
$$BA$$ exists

D
none of these

Solution

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

Order of $$A$$ is $$3\times2$$
Order of $$B$$ is $$3\times3$$

Order of $$A$$ and $$B$$ are not same. So , $$A+B$$ does not exists.

Again , $$AB$$ also does not exists as the number of columns of $$A$$ are not equal to the number of rows of $$ B.$$

Here, $$BA$$ exists as the number of columns of $$B$$ is equal to the number of rows of $$A$$.
Question 2
1 / -0

Use the method of elementary row transformation to compute the inverse of
$$\quad \begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{bmatrix}$$

A
$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

B
$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{1}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{11}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{2}{21}\end{bmatrix}$$

C
$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{16}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{4}{21}\end{bmatrix}$$

D
$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{2}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{4}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

Solution

Let $$\quad A = \begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{bmatrix}$$

$$\Rightarrow \quad Write \space A A^{-1}= I$$

$$\quad \begin{bmatrix} 1 & 2 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{bmatrix} A^{-1}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

$$\quad \begin{matrix}R_{21}(-2)\\ \mbox{~}\\ R_{31}(1)\end{matrix}\begin{bmatrix}1 & 0 & 5 \\ 2 & 3 & 1 \\ -1 & 1 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \\ -2 & 1 & 0 \\ 1 & 0 & 1\end{bmatrix}$$

$$\quad \begin{matrix}R_2(-1) \\ \mbox{~} \\ R_3(1/3)\end{matrix}\begin{bmatrix}1 & 2 & 5 \\ 0 & 1 & 9 \\ 0 & 1 & 2\end{bmatrix}A^{-1} = \begin{bmatrix}1& 0 & 0 \\ 2 & -1 & 0 \\ \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3}\end{bmatrix}$$

$$\quad \begin{matrix}R_{12}(-2) \\ \mbox{~} \\ R_{32}(-1)\end{matrix}\begin{bmatrix}1 & 0 & -13 \\ 0 & 1 & 9 \\ 0 & 0 & -7\end{bmatrix} A^{-1}= \begin{bmatrix}-3 & 2 & 0 \\ 2 & -1 & 0 \\ -\displaystyle\frac{5}{3} & 1 & \displaystyle\frac{1}{3}\end{bmatrix}$$

$$\quad \begin{matrix}R_3(-1/7)\\ \mbox{~}\end{matrix}\begin{bmatrix}1 & 0 & -13 \\ 0 & 1 & 9 \\ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}-3 & 2 & 0 \\ 2 & -1 & 0 \\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

$$\quad \begin{matrix}R_{13}(13) \\ \mbox{~} \\ R_{23}(-9)\end{matrix}\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

Hence, $$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$
Question 3
1 / -0

If $$A = \begin{bmatrix}1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}, \space B = \begin{bmatrix}2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \end{bmatrix}$$, then $$AB$$ =

A
$$\begin{bmatrix}5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}$$

B
$$\begin{bmatrix}5 & 9 & 13 \\ -1 & -2 & 4 \\ 2 & -2 & -4\end{bmatrix}$$

C
$$\begin{bmatrix}5 & -9 & 13 \\ 1 & -2 & 4 \\ 2 & -2 & 4\end{bmatrix}$$

D
$$\begin{bmatrix}5 & -9 & 13 \\ 1 & -2 & 4 \\ -2 & 2 & 4\end{bmatrix}$$

Solution

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

$$AB=\begin{bmatrix}1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}\begin{bmatrix}2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \end{bmatrix}=\begin{bmatrix}5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}$$

Hence, option A is correct.
Question 4
1 / -0

Given $$A = \begin{bmatrix}1 & -1 \\ 2 & -1\end{bmatrix}$$, which of the following result is true?

A
$$A^2 = I$$

B
$$A^2 = -I$$

C
$$A^2 = 2I$$

D
None of these

Solution

Given, $$A = \begin{bmatrix}1 & -1 \\ 2 & -1\end{bmatrix}$$

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

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

$$\Rightarrow A^{2}=-I$$
Question 5
1 / -0

$$A = \begin{bmatrix}4 & x+2 \\ 2x-3 & x+1\end{bmatrix}$$ is symmetric, then $$x$$ =

A
$$3$$

B
$$5$$

C
$$2$$

D
$$4$$

Solution

$$A = \begin{bmatrix}4 & x+2 \\ 2x-3 & x+1\end{bmatrix}$$
$$A^{T}=\begin{bmatrix}4 & 2x-3 \\ x+2 & x+1\end{bmatrix}$$
$$A$$ is symmetric.
$$\therefore\quad A=A^{T}$$
$$\Rightarrow 2x-3=x+2$$
$$\therefore x=5$$
Hence, option B.
Question 6
1 / -0

Let $$\displaystyle A = \begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}$$. The only correct statement about the matrix $$A$$ is

A
$$\displaystyle A$$ is a zero matrix

B
$$\displaystyle A = \left ( -1 \right )I_{3}$$

C
$$A^2=-I$$

D
$$\displaystyle A^{2} = I$$

Solution

$$\displaystyle A = \begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}$$
$$\displaystyle A^2=\begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
$$\therefore A^2=I$$
Hence, option D.
Question 7
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If $$\displaystyle A=\begin{bmatrix}0 &c &-b \\-c &0 &a \\b &-a &0 \end{bmatrix}$$ and $$\displaystyle B=\begin{bmatrix}a^{2} &ab &ac \\ab &b^{2} &bc \\ac &bc &c^{2} \end{bmatrix},$$ then $$AB=$$

A
$$\displaystyle A^{3}$$

B
$$\displaystyle B^{2}$$

C
$$I$$

D
$$O$$

Solution

Given : $$\displaystyle A=\begin{bmatrix}0 &c &-b \\-c &0 &a \\b &-a &0 \end{bmatrix}$$ $$\displaystyle B=\begin{bmatrix}a^{2} &ab &ac \\ab &b^{2} &bc \\ac &bc &c^{2} \end{bmatrix}$$

Then, $$\displaystyle AB= \begin{bmatrix} 0& c & -b\\ -c & 0 & a\\b &-a &0 \end{bmatrix}\begin{bmatrix} a^{2} & ab & ac\\ ab & b^{2} & bc \\ ac & bc & c^{2}\end{bmatrix}$$

$$=\begin{bmatrix} abc-abc & b^{2}c-b^{2}c & bc^{2}-bc^{2} \\ -a^{2}c+a^{2}c & -abc+abc & -ac^{2}+ac^{2} \\ a^2b-a^2b & ab^2-ab^2 & abc-abc \end{bmatrix}$$

$$=O$$
Question 8
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Let $$\displaystyle A=\begin{bmatrix}-1\\2\\3 \end{bmatrix}$$ and $$\displaystyle B=\begin{bmatrix} -2 & -1 & -4 \end{bmatrix},$$ then matrix $$(AB) A$$ equals

A
$$12A$$

B
$$-12A$$

C
$$4A$$

D
$$3A$$

Solution

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

$$AB=\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}$$

$$(AB)A=\begin{bmatrix} 2 & 1 & 4 \\ -4 & -2 & -8 \\ -6 & -3 & -12 \end{bmatrix}\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$$

$$\Rightarrow (AB)A=\begin{bmatrix} 12 \\ -24 \\ -36 \end{bmatrix}=-12\begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}=-12A$$
Question 9
1 / -0

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

A
$$\displaystyle \alpha=a^{2}+b^{2},\beta =2ab$$

B
$$\displaystyle \alpha=a^{2}+b^{2},\beta =a^{2}-b^{2}$$

C
$$\displaystyle \alpha=2ab ,\beta =a^{2}+b^{2}$$

D
$$\displaystyle \alpha=a^{2}+b^{2},\beta =ab $$

Solution

$$\displaystyle A^{2}=AA=\begin{pmatrix}a &b \\b &a \end{pmatrix}\begin{pmatrix}a &b \\b &a \end{pmatrix}$$ $$\displaystyle =\begin{pmatrix}a^{2}+b^{2} &2ab \\2ab &a^{2}+b^{2} \end{pmatrix}$$ $$\displaystyle =
\begin{pmatrix} \alpha &\beta \\\beta &\alpha \end{pmatrix}$$
Question 10
1 / -0

If $$A =\begin{bmatrix}\alpha &\beta \\\gamma &-\alpha \end{bmatrix}$$ is such that $$A^2 = I$$, then

A
$$1 +\alpha^2+\beta\gamma=0$$

B
$$1 -\alpha^2-\beta\gamma=0$$

C
$$1 -\alpha^2+\beta\gamma=0$$

D
$$1 +\alpha^2-\beta\gamma=0$$

Solution

$$A =\begin{bmatrix}\alpha &\beta \\\gamma &-\alpha \end{bmatrix}$$
$$A^2=\begin{bmatrix}\alpha^2+\beta \gamma &\alpha \beta-\alpha \beta \\\alpha \gamma - \alpha \gamma&\beta \gamma +\alpha^2 \end{bmatrix}$$
but, $$A^2=I$$
$$\Rightarrow \alpha ^2+\beta \gamma = 1$$
$$\Rightarrow 1 -\alpha ^2-\beta \gamma=0$$
Hence, option B is correct.

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Re-Attempt Weekly Quiz Competition

Matrices Test - 31

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Matrices Test - 31

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Question 1

$$A + B$$ exists

$$AB$$ exists

$$BA$$ exists

none of these

Solution

Question 2

$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{2}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{1}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{1}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{11}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{2}{21}\end{bmatrix}$$

$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{1}{7} & -\displaystyle\frac{16}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{5}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{4}{21}\end{bmatrix}$$

$$\quad A^{-1} = \begin{bmatrix}\displaystyle\frac{4}{21} & \displaystyle\frac{2}{7} & -\displaystyle\frac{13}{21} \\ -\displaystyle\frac{1}{7} & \displaystyle\frac{2}{7} & \displaystyle\frac{3}{7}\\ \displaystyle\frac{4}{21} & -\displaystyle\frac{2}{7} & -\displaystyle\frac{1}{21}\end{bmatrix}$$

Solution

Question 3

$$\begin{bmatrix}5 & 9 & 13 \\ -1 & 2 & 4 \\ -2 & 2 & 4\end{bmatrix}$$

$$\begin{bmatrix}5 & 9 & 13 \\ -1 & -2 & 4 \\ 2 & -2 & -4\end{bmatrix}$$

$$\begin{bmatrix}5 & -9 & 13 \\ 1 & -2 & 4 \\ 2 & -2 & 4\end{bmatrix}$$

$$\begin{bmatrix}5 & -9 & 13 \\ 1 & -2 & 4 \\ -2 & 2 & 4\end{bmatrix}$$

Solution

Question 4

$$A^2 = I$$

$$A^2 = -I$$

$$A^2 = 2I$$

None of these

Solution

Question 5

$$3$$

$$5$$

$$2$$

$$4$$

Solution

Question 6

$$\displaystyle A$$ is a zero matrix

$$\displaystyle A = \left ( -1 \right )I_{3}$$

$$A^2=-I$$

$$\displaystyle A^{2} = I$$

Solution

Question 7

$$\displaystyle A^{3}$$

$$\displaystyle B^{2}$$

$$I$$

$$O$$

Solution

Question 8

$$12A$$

$$-12A$$

$$4A$$

$$3A$$

Solution

Question 9

$$\displaystyle \alpha=a^{2}+b^{2},\beta =2ab$$

$$\displaystyle \alpha=a^{2}+b^{2},\beta =a^{2}-b^{2}$$

$$\displaystyle \alpha=2ab ,\beta =a^{2}+b^{2}$$

$$\displaystyle \alpha=a^{2}+b^{2},\beta =ab $$

Solution

Question 10

$$1 +\alpha^2+\beta\gamma=0$$

$$1 -\alpha^2-\beta\gamma=0$$

$$1 -\alpha^2+\beta\gamma=0$$

$$1 +\alpha^2-\beta\gamma=0$$

Solution

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