Matrices Test

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Matrices Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$$ such that $$A^{2}-6A+7I=0$$, then $$k=$$

A
$$1$$

B
$$3$$

C
$$2$$

D
$$4$$

Solution

$$A = \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$$
$${A}^{2} = \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} \times \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix}$$
$$\Rightarrow {A}^{2} = \begin{bmatrix} 16+ 1 & -4 - k \\ -4 - k & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix}$$
$$6A = 6 \times \begin{bmatrix} 4 & -1 \\ -1 & k \end{bmatrix} = \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix}$$
$$7I = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
Therefore,
$${A}^{2} - 6A + 7I = 0$$
$$\Rightarrow {A}^{2} = 6A - 7I$$
$$\begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 24 & -6 \\ -6 & 6k \end{bmatrix} - \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$$
$$\begin{bmatrix} 17 & - \left( 4 + k \right) \\ - \left( 4 + k \right) & 1 + {k}^{2} \end{bmatrix} = \begin{bmatrix} 17 & -6 \\ -6 & 6k - 7 \end{bmatrix}$$
Comparing both sides, we get
$$- \left( 4 + k \right) = -6$$
$$\Rightarrow k = 6 - 4 = 2$$
Question 2
1 / -0

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

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

B
$$\begin{bmatrix} 0 & -1\\ -2 & -4\end{bmatrix}$$

C
$$\begin{bmatrix} 0 & 1\\ 2 & -4\end{bmatrix}$$

D
None of these

Solution
Question 3
1 / -0

A is an involuntary matrix given by $$A=\begin{bmatrix} 0 & 1 & -1\\ 4 & -3 & 4\\ 3 & -3 & 4\end{bmatrix}$$ then the inverse of $$\dfrac{A}{2}$$ will be?

A
$$2A$$

B
$$\dfrac{A^{-1}}{2}$$

C
$$\dfrac{A}{2}$$

D
$$A^{-2}$$

Solution
Question 4
1 / -0

A is an involutory matrix given by $$A=\begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix}$$ then the inverse of $$\dfrac{A}{2}$$ will be

A
$$2A$$

B
$$\dfrac{A^{-1}}{2}$$

C
$$\dfrac{A}{2}$$

D
$$A^{2}$$

Solution

Given $$A$$ to be involutory matrix, then according to the definition of involutory matrix we have, $$A^2 =I$$. [ $$I$$ being identity matrix of order $$3$$].
So,
$$A^2=I$$
or, $$A=A^{-1}$$ [ Since involutory matrix is always invertible]
or, $$\dfrac{A}{2}=\dfrac{A^{-1}}{2}$$.
Question 5
1 / -0

Let A=$$\left[ \begin{matrix} 1 \\ 2 \end{matrix}\begin{matrix} 2 \\ 1 \end{matrix} \right] and\quad B=\left[ \begin{matrix} 4 \\ 5 \\ 0 \end{matrix}\begin{matrix} -3 \\ 6 \\ 1 \end{matrix} \right] $$ then

A
$$AB$$ exists

B
$$AB$$ and $$BA$$ both exists

C
Neither $$AB$$ nor $$BA$$ exists

D
$$BA$$ exists, but $$AB$ does not exists

Solution

Given,

Order of $$A=2\times 2$$ matrix

Order of $$B=3\times 2$$ matrix

$$AB$$ does not exist, as the number of columns of matrix $$A$$ is not equal to number of rows of $$B$$
But, $$BA$$ exists, as the number of columns of matrix $$B$$ is equal to number of rows of $$A$$
Question 6
1 / -0

If $$\left[ \begin{matrix} 1 & x & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 1 \\ x \end{matrix} \right] =0,$$ then $$x=$$

A
$$\cfrac{-9\pm\sqrt{35}}{2}$$

B
$$\cfrac{-7\pm\sqrt{53}}{2}$$

C
$$\cfrac{-9\pm\sqrt{53}}{2}$$

D
$$\cfrac{-7\pm\sqrt{35}}{2}$$

Solution

Given,

$$\begin{pmatrix}1&x&1\end{pmatrix}\begin{pmatrix}1&3&2\\ 0&5&1\\ 0&3&2\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$$

Upon multiplying the first 2 matrices, we get,

$$\begin{pmatrix}1\cdot \:1+x\cdot \:0+1\cdot \:0&1\cdot \:3+x\cdot \:5+1\cdot \:3&1\cdot \:2+x\cdot \:1+1\cdot \:2\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$$

Upon further simplification, we get,

$$\begin{pmatrix}1&6+5x&4+x\end{pmatrix}\begin{pmatrix}1\\ 1\\ x\end{pmatrix}=0$$

Again multiplying the above matrices, we get,

$$\begin{pmatrix}1\cdot \:1+\left(6+5x\right)\cdot \:1+\left(4+x\right)x\end{pmatrix}=0$$

Upon further simplification, we get,

$$\begin{pmatrix}x^2+9x+7\end{pmatrix}=0$$

Solving the above quadratic equation, we get,

$$x=\dfrac{-9\pm \sqrt{53}}{2}$$
Question 7
1 / -0

if A=$$\left[ \begin{matrix} 2 & 3 \\ 5 & -7 \end{matrix} \right] then\quad \left( { A }^{ '} \right) ^{ 2 }=$$

A
$$\left[ \begin{matrix} 5 & -7 & 12 \\ 1 & 4 & 22 \end{matrix} \right] $$

B
$$\left[ \begin{matrix} 1 & 17 \\ 1 & -4 \\ 0 & 2 \end{matrix} \right] $$

C
$$\left[ \begin{matrix} -19 & -25 \\ -15 & 64 \end{matrix} \right] $$

D
$$\left[ \begin{matrix} 19 & -25 \\ -15 & 64 \end{matrix} \right] $$

Solution

Given,

$$A=\begin{pmatrix}2&3\\ \:5&-7\end{pmatrix}$$

$$\Rightarrow A'=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}$$

Now,$$(A')^2$$

$$=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}^2$$

$$=\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}\begin{pmatrix}2&5\\ 3&-7\end{pmatrix}$$

$$=\begin{pmatrix}2\cdot \:2+5\cdot \:3&2\cdot \:5+5\left(-7\right)\\ 3\cdot \:2+\left(-7\right)\cdot \:3&3\cdot \:5+\left(-7\right)\left(-7\right)\end{pmatrix}$$

$$=\begin{pmatrix}19&-25\\ -15&64\end{pmatrix}$$
Question 8
1 / -0

If $$A$$ and $$B$$ are two matrices such that $$AB$$ and $$A+B$$ are both defined then $$A$$ and $$B$$ are

A
Square matrices of the same order

B
Square matrices of different order

C
Rectangular matrices of same order

D
Rectangular matrices of different order

Solution

For the addition of two matrices,it is required that
the matrices should be of the same order.
For multiplication of two matrices A of order $$m\times$$ n and
B of order $$p\times q,$$ it requires that
$$ n=p----(1) $$
$$ m = p----(2) $$
$$ n = q----(3) $$
form (1) and (3)
$$ p = q-------(4) $$
from (1)and (2)
$$ n = m-----(5) $$
From (4)and (5),we can conclude that
A and B are square matrices and their orders
are one and same .
$$\therefore $$ The addition and multiplication of two matrices
should be of same order and they should be
square matrices.
$$\therefore $$ so, the answer is A. square matrices of the same
order.
Question 9
1 / -0

If $$A=\left[ \begin{matrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{matrix} \right] $$, then value of $$A^{-1}$$ is equal to

A
$$A$$

B
$$A^{2}$$

C
$$A^{3}$$

D
$$A^{4}$$

Solution
Question 10
1 / -0

If $$U = [ 2, -3 , 4 ]$$ , $$V = \begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$$ , $$X = [0 , 2 , 3]$$ and $$Y = \begin{bmatrix} 2 \\ 2 \\ 4 \end{bmatrix}$$ , then $$UV + XY$$ =

A
$$20$$

B
$$[ -20 ]$$

C
$$-20$$

D
$$[ 20 ]$$

Solution

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Matrices Test - 42

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Matrices Test - 42

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SHARING IS CARING

Question 1

$$1$$

$$3$$

$$2$$

$$4$$

Solution

Question 2

$$\begin{bmatrix} 0 & -1\\ 2 & -4\end{bmatrix}$$

$$\begin{bmatrix} 0 & -1\\ -2 & -4\end{bmatrix}$$

$$\begin{bmatrix} 0 & 1\\ 2 & -4\end{bmatrix}$$

None of these

Solution

Question 3

$$2A$$

$$\dfrac{A^{-1}}{2}$$

$$\dfrac{A}{2}$$

$$A^{-2}$$

Solution

Question 4

$$2A$$

$$\dfrac{A^{-1}}{2}$$

$$\dfrac{A}{2}$$

$$A^{2}$$

Solution

Question 5

$$AB$$ exists

$$AB$$ and $$BA$$ both exists

Neither $$AB$$ nor $$BA$$ exists

$$BA$$ exists, but $$AB$ does not exists

Solution

Question 6

$$\cfrac{-9\pm\sqrt{35}}{2}$$

$$\cfrac{-7\pm\sqrt{53}}{2}$$

$$\cfrac{-9\pm\sqrt{53}}{2}$$

$$\cfrac{-7\pm\sqrt{35}}{2}$$

Solution

Question 7

$$\left[ \begin{matrix} 5 & -7 & 12 \\ 1 & 4 & 22 \end{matrix} \right] $$

$$\left[ \begin{matrix} 1 & 17 \\ 1 & -4 \\ 0 & 2 \end{matrix} \right] $$

$$\left[ \begin{matrix} -19 & -25 \\ -15 & 64 \end{matrix} \right] $$

$$\left[ \begin{matrix} 19 & -25 \\ -15 & 64 \end{matrix} \right] $$

Solution

Question 8

Square matrices of the same order

Square matrices of different order

Rectangular matrices of same order

Rectangular matrices of different order

Solution

Question 9

$$A$$

$$A^{2}$$

$$A^{3}$$

$$A^{4}$$

Solution

Question 10

$$20$$

$$[ -20 ]$$

$$-20$$

$$[ 20 ]$$

Solution

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