Matrices Test

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Matrices Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

If $$\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$$ and $$ A^{2}=I$$, then $$x=$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

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

Also, $$A^2=I$$
$$\Rightarrow \begin{bmatrix} x &1 \\1 & 0 \end{bmatrix} \begin{bmatrix} x & 1 \\ 1& 0\end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
$$\Rightarrow \begin{bmatrix} x^2+1 & x \\ x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$$

For the above two matrices to be equal each of the first matrix must be equal to the corresponding element of the second matrix.

Therefore, $$x^2+1=1 ...(i)$$ and $$x=0...(ii)$$.

From equation $$(i)$$ and equation $$(ii)$$ we get,
$$x=0$$

Hence, the correct option is $$(A)$$
Question 2
1 / -0

If $$[1\ 2\ 3] B = [3\ 4],$$ then order of the matrix $$B$$ is

A
$$3 \times 1$$

B
$$1 \times 3$$

C
$$2 \times 3$$

D
$$3 \times 2$$

Solution

Given, $$[1 \ 2\ 3] $$ $$B = [3 \ 4]$$
or $${[1 \ 2\ 3]}$$ $$_{1\times3}$$ $$B={[3\ 4]}$$ $$_{1\times2}$$
So, order of $$B $$ should be $$3\times2.$$
Hence, option 'D' is correct.
Question 3
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The inverse of the matrix $$\begin{bmatrix}2 & 1\\ 1 & 3\end{bmatrix}$$ is

A
$$\displaystyle \frac{1}{5} \begin{bmatrix}2 & 1\\ 1 & 3\end{bmatrix}$$

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

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

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

Solution

Given, $$A=\begin{bmatrix}2 & 1\\ 1 & 3\end{bmatrix}$$
$$|A|=5$$

Now, $$adj A=C^{T}=\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}^{ T }$$

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

$$A^{-1}=\displaystyle \frac{1}{5}\begin{bmatrix} 3 & -1 \\ -1 & 2 \end{bmatrix}$$
Question 4
1 / -0

If $$A = \begin{bmatrix}1 & -2 & 3\\ -4 & 2 & 5\end{bmatrix}$$ and $$B = \begin{bmatrix}2 & 3\\ 4 & 5\\ 2 & 1\end{bmatrix}$$, then the product of AB and BA is

A
$$\begin{bmatrix}-10 & 2 & 21\\ -16 & 2 & 37\\ -2 & -2 & 11\end{bmatrix}$$

B
$$\begin{bmatrix}0 & -4\\ 10 & 3\end{bmatrix}$$

C
$$\begin{bmatrix}-64 & -8\\ -148 & 26\end{bmatrix}$$

D
$$Cannot\space be\space computed$$

Solution

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

5\end{bmatrix} \begin{bmatrix}2 &3 \\ 4 & 5\\ 2 &

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

+ 10 & -12 + 10 + 5\end{bmatrix}= \begin{bmatrix}0 & -4\\ 10

& 3\end{bmatrix}$$
$$BA = \begin{bmatrix}2 & 3\\ 4 & 5\\ 2

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

& 5\end{bmatrix}$$
$$\begin{bmatrix}2 - 12 & -4 + 6 & 6 +

15\\ 4 - 20 & - 8 + 10 & 12 + 25\\ 2 - 4 & -4 + 2 & 6 +

5\end{bmatrix}= \begin{bmatrix}-10 & 2 & 21\\ -16 & 2 &

37\\ -2 & -2 & 11\end{bmatrix}$$

Thus, order of AB is $$2\times2$$ and order of $$BA$$ is $$3\times3.$$
So, the product AB and BA can not be computed.
Question 5
1 / -0

Let A be a square matrix. Which of the following is/are not symmetric matrix/matrices?

A
$$A+A^{T}$$

B
$$AA^{T}$$

C
$$A-A^{T}$$

D
$$A^{T}A$$

Solution

We know a square matrix $$A$$ is called symmetric if $$A^T = A$$
. Let $$P = A-A^T \Rightarrow P^T = (A-A^T)^T=A^T-A=-P\Rightarrow $$ This is not a symmetric matrix, it is skew symmetric matrix.
All other matrices in the other option are symmetric.
Question 6
1 / -0

If $$A = \begin{bmatrix}1 & 2 & 2\\ 2 & 1 & -2\\ a & 2 & b\end{bmatrix}$$ is a matrix satisfying $$AA^T = 9 I_3$$, then the values of $$a$$ and $$b$$ are

A
$$a = -2, b = - 1$$

B
$$a = -2, b = 1$$

C
$$a = 2, b = - 1$$

D
No values of a,b satisfy given conditions

Solution

We have,
$$A = \begin{bmatrix}1 & 2 & 2\\ 2 & 1 &

-2\\ a & 2 & b\end{bmatrix} \Rightarrow A^T = \begin{bmatrix}1

& 2 & a\\ 2 & 1 & 2\\ 2 & -2 & b\end{bmatrix}$$

$$\therefore AA^T = 9I_3$$

$$\Rightarrow \begin{bmatrix}1

& 2 & 2\\ 2 & 1 & -2\\ a & 2 &

b\end{bmatrix} \begin{bmatrix}1 & 2 & a\\ 2 & 1 & 2\\ 2

& -2 & b\end{bmatrix} = 9 \begin{bmatrix}1 & 0 & 0\\ 0

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

$$\begin{bmatrix}9 & 0 & a+2b + 4\\ 0 & 9 & 2a + 2 -

2b\\ a+2b +4 & 2a + 2 - 2b & a^2 + 4 + b^2\end{bmatrix}

= \begin{bmatrix}9 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 &

9\end{bmatrix}$$

On comparing above matrices, we get
$$a + 2b + 4 = 0, 2a + 2 - 2b = 0$$ and $$a^2 + 4 + b^2 = 9$$
or $$a + 2b + 4 = 0, a - b + 1= 0$$ and $$a^2 + b^2 = 5$$
Solving $$a + 2b + 4 =0$$ and $$a - b + 1 = 0$$, we get
$$a = -2, b = -1$$. Clearly, these values satisfy $$a^2 + b^2 = 5$$.
$$\therefore$$ $$a = - 2$$ and $$b = - 1$$.
Question 7
1 / -0

Using elementary transformation, find the inverse of the matrix $$A =\begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a} \right )\end{bmatrix}$$.

A
$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ -c & a\end{bmatrix} $$

B
$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& - b\\ c & a\end{bmatrix} $$

C
$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ c & a\end{bmatrix} $$

D
None of these.

Solution

$$A = \begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a} \right ) \end{bmatrix}$$
We

write, $$A = I.A$$
$$\Rightarrow \begin{bmatrix}a & b\\ c & \left ( \frac{1 + bc}{a}

\right ) \end{bmatrix} = \begin{bmatrix}1 & 0\\ 0 & 1

\end{bmatrix}A$$
$$\Rightarrow \begin{bmatrix}1 & \frac{b}{a}\\ c

& \left ( \frac{1 + bc}{a} \right ) \end{bmatrix} =

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

$$\left ( R_1 \rightarrow \frac{R_1}{a} \right )$$
or $$

\begin{bmatrix}1 & \frac{b}{a}\\ 0 & \frac{1}{a} \end{bmatrix} =

\begin{bmatrix} \frac{1}{a} & 0\\ \frac{-c}{a} &

1\end{bmatrix} A$$ $$(R_2 \rightarrow R_2 - c R_1)$$
or

$$ \begin{bmatrix}1 & \frac{b}{a}\\ 0 & 1\end{bmatrix} =

\begin{bmatrix}\frac{1}{a} & 0\\ -c & a\end{bmatrix}A $$

$$(R_2 \rightarrow aR_2)$$
or $$\begin{bmatrix}1 & 0\\ 0 &

1\end{bmatrix} = \begin{bmatrix} \frac{1 + bc}{a}& -b\\ -c &

a\end{bmatrix} A$$ $$\left ( R_1 \rightarrow R_1 -

\frac{b}{a} R_2 \right )$$
$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& - b\\ -c & a\end{bmatrix} $$
Question 8
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The value of x is such that matrix product $$\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix} \begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$$ equals an identity matrix. Then the value of 20x is

A
4

B
5

C
1

D
7

Solution

$$\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 &

-2 & 1\end{bmatrix} \begin{bmatrix}-x & 14x & 7x\\ 0 & 1

& 0\\ x & -4x & -2x\end{bmatrix}$$ $$=\begin{bmatrix}5x

& 0 & 0\\ 0 & 1 & 0\\ 0 & 10x-2 &

5x\end{bmatrix}= \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0

& 0 & 1\end{bmatrix}$$ (given)
Comparing elements we get,
$$\Rightarrow \displaystyle 5x = 1, 10 x - 2 = 0, \therefore x = \frac{1}{5}$$
Question 9
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Given $$A, B, C$$ are three matrices such that
$$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}x \\ y \\ z\end{bmatrix}.$$ Evaluate $$ABC$$.

A
$$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$$

B
$$ABC = \begin{bmatrix}ax^2 - by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$$

C
$$ABC = \begin{bmatrix}ax^2 - by^2 + cz^2 - 2hxy + 2gzx + 2fyz\end{bmatrix}$$

D
$$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 - 2hxy + 2gzx - 2fyz\end{bmatrix}$$

Solution

$$A = \begin{bmatrix}x & y & z\end{bmatrix}$$, $$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c\end{bmatrix}$$, $$C = \begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
$$AB=\begin{bmatrix}ax+hy+gz & hx+by+z & gx+fy+cz\end{bmatrix}$$
$$ABC=\begin{bmatrix}ax+hy+gz & hx+by+z & gx+fy+cz\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
$$=\begin{bmatrix}ax^2 + by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$$
Hence, option A.
Question 10
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If $$A = \begin{bmatrix} 1 & -2 & 3 \\ -4 & 2 & 5 \end{bmatrix}$$ and $$B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{bmatrix}$$. Find $$AB$$ and show that $$AB \ne BA$$

A
$$AB = \begin{bmatrix} 0 & -4 \\ 10 & -3 \end{bmatrix}$$

B
$$AB = \begin{bmatrix} 0 & 4 \\ 10 & 3 \end{bmatrix}$$

C
$$AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$$

D
None of these.

Solution

Given,
$$A=\begin{bmatrix} 1\quad & -2\quad & 3 \\ -4 & 2 & 5 \end{bmatrix},B=\begin{bmatrix} 2\quad & 3 \\ 4\quad & 5 \\ 2\quad & 1 \end{bmatrix}\\ AB=\begin{bmatrix} 1\quad & -2\quad & 3 \\ -4 & 2 & 5 \end{bmatrix}\begin{bmatrix} 2\quad & 3 \\ 4\quad & 5 \\ 2\quad & 1 \end{bmatrix}=\begin{bmatrix} 2-8+6\quad & 3-10+3 \\ -8+8+10\quad & -12+10+5 \end{bmatrix}=\begin{bmatrix} 0\quad & -4 \\ 10\quad & 3 \end{bmatrix}\\ BA=\begin{bmatrix} 2\quad & 3 \\ 4\quad & 5 \\ 2\quad & 1 \end{bmatrix}\begin{bmatrix} 1\quad & -2\quad & 3 \\ -4 & 2 & 5 \end{bmatrix}=\begin{vmatrix} 2-12\quad & -4+6\quad & 6+15 \\ 4-20 & -8+10\quad & 12+25 \\ 2-4 & -4+2 & -6+5 \end{vmatrix}\begin{bmatrix} -10\quad & 2\quad & 21 \\ -16 & 2 & 37 \\ -2 & -2 & -1 \end{bmatrix}$$
Hence $$ AB\neq BA$$

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Matrices Test - 29

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Matrices Test - 29

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Question 1

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 2

$$3 \times 1$$

$$1 \times 3$$

$$2 \times 3$$

$$3 \times 2$$

Solution

Question 3

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

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

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

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

Solution

Question 4

$$\begin{bmatrix}-10 & 2 & 21\\ -16 & 2 & 37\\ -2 & -2 & 11\end{bmatrix}$$

$$\begin{bmatrix}0 & -4\\ 10 & 3\end{bmatrix}$$

$$\begin{bmatrix}-64 & -8\\ -148 & 26\end{bmatrix}$$

$$Cannot\space be\space computed$$

Solution

Question 5

$$A+A^{T}$$

$$AA^{T}$$

$$A-A^{T}$$

$$A^{T}A$$

Solution

Question 6

$$a = -2, b = - 1$$

$$a = -2, b = 1$$

$$a = 2, b = - 1$$

No values of a,b satisfy given conditions

Solution

Question 7

$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ -c & a\end{bmatrix} $$

$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& - b\\ c & a\end{bmatrix} $$

$$\Rightarrow A^{-1} = \begin{bmatrix} \frac{1 + bc}{a}& b\\ c & a\end{bmatrix} $$

None of these.

Solution

Question 8

4

5

1

7

Solution

Question 9

$$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$$

$$ABC = \begin{bmatrix}ax^2 - by^2 + cz^2 + 2hxy + 2gzx + 2fyz\end{bmatrix}$$

$$ABC = \begin{bmatrix}ax^2 - by^2 + cz^2 - 2hxy + 2gzx + 2fyz\end{bmatrix}$$

$$ABC = \begin{bmatrix}ax^2 + by^2 + cz^2 - 2hxy + 2gzx - 2fyz\end{bmatrix}$$

Solution

Question 10

$$AB = \begin{bmatrix} 0 & -4 \\ 10 & -3 \end{bmatrix}$$

$$AB = \begin{bmatrix} 0 & 4 \\ 10 & 3 \end{bmatrix}$$

$$AB = \begin{bmatrix} 0 & -4 \\ 10 & 3 \end{bmatrix}$$

None of these.

Solution

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