Maths M.C.Q (1 Marks)

1. I

Solution:

A² = A

(I + A)³ +A

⇒ I³ - A³ - 3I²A + 3IA² + A

⇒ I - A³ - 3A+ 3A + A [$\therefore$ A² = A)

⇒ I - A.A² + A

⇒ I - A.A + A

⇒ I - A + A

= I

3. $\displaystyle \begin{vmatrix} 1 \\ 10 \end{vmatrix}$

Solution:

$2\text{A+B}=|26|$

1. a symmetric matrix

Solution:

$(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}=\text{BA}=\text{AB}\Rightarrow(\text{AB})^\text{T}=\text{AB}$

So that means the product of the matrices $\text{AB}$ is a symmetric matrix.

3. scalar

Solution:

Given A is a square matrix as the number of rows and columns are same as n

The elements $\text{a}_\text{ij}$​ where $\text{i} = \text{j} $ lie along the diagonal.

and the elements $\text{a}_\text{ij}$ ​ where  $\text{i}\neq\text{j}$ do not lie along the diagonal.

Given, diagonal elements = 2 and the rest of the elements = 0

Such a diagonal matrix where all diagonal elements are equal, is called a scalar matrix.

3. column matrix

Solution:

Transpose of row matrix Let $ \text{A}=\begin{bmatrix}\text{x} &\text{amp; y} &\text{amp; z} \end{bmatrix}$ be a row

matrix $\text{A}^\text{T}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$Clearly $\text{A}^\text{T}$ is a column matrix $\therefore$ Transpose of row.

3.  $\displaystyle \begin{vmatrix} 1 &-4 \\ 4 &1\end{vmatrix}$

Solution:

$2\text{ A-B}=\displaystyle \begin{vmatrix} 4 &\text{amp;}-6 \\ 6 &\text{amp; 4} \end{vmatrix}-\displaystyle \begin{vmatrix} 3 &\text{amp;}-2 \\ 2 &\text{amp; 3} \end{vmatrix}=\displaystyle \begin{vmatrix} 1 &\text{amp;}-4\\ 4 &\text{amp; 1} \end{vmatrix}$ 

4. None of these.

Solution:

Given: A is a square matrix.

Let $\text{A}=\begin{bmatrix}1&2\\1&0\end{bmatrix}$

$\Rightarrow\text{AA}=\begin{bmatrix}1&2\\1&0\end{bmatrix}\begin{bmatrix}1&2\\1&0\end{bmatrix}=\begin{bmatrix}3&2\\1&2\end{bmatrix}$

2. $1\times1$

Solution:

Let $\text{ABC}=\begin{bmatrix}\text{x}&\text{amp;}\text{ y}&\text{amp; }\text{z}\end{bmatrix}\begin{bmatrix}\text{x} &\text{amp;}\text{ h}&\text{amp;}\text{ g} \\\text{h} &\text{amp;}\text{ b}&\text{amp; }\text{f}\\\text{g} &\text{amp;}\text{ f}&\text{amp; }\text{c} \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}$

Here, the order of $\text{A}$ is  $1\times3$

Order of $\text{B}$ is $3\times3$

Since, matrix multiplication satisfies associative property

$\text{i}.\text{e}. (\text{AB})\text{C} = \text{A}(\text{BC})$

Hence, the order of $\text{AB}$ is $1\times3$

Hence, the order of $\text{ABC}$ is $1\times1$

3. 2

Solution:

A matrix is singular if and only if it has a determinant of 0.

$\begin{bmatrix}1 &\text{amp;3}& \text{amp}\lambda+2 \\2& \text{amp;4}&\text{amp;8} \\3&\text{amp;5}&\text{amp;}10 \end{bmatrix}$

$(40-40)-2(20-24)+(\lambda+2)(10-12)=0$

$2\lambda=4$

$\Rightarrow\lambda=2$

4. 7

Solution:

$\text{A}=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$

$\Rightarrow\text{A}^2=\text{A}\times\text{A}$

$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$

$\Rightarrow\text{A}^2=\begin{bmatrix}\cos^2\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}&\Big(-2\cos\frac{2\pi}{7}-\sin\frac{2\pi}{7}\Big)\\2\cos\frac{2\pi}{7}\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}\end{bmatrix}$

$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos^2\theta-\sin^2\theta=\cos2\theta\\2\sin\theta\cos\theta=\sin2\theta\end{bmatrix}$

$\Rightarrow\text{A}^3=\text{A}^2\times\text{A}$

$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$

$\Rightarrow\text{A}^3=\begin{bmatrix}\Big(\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}-\sin\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}-\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\\\Big(\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\end{bmatrix}$

$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos\text{(A+B)}=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}\\\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\end{bmatrix}$

Now we check if the pattern is same for k = 6.

Here,

$\text{A}^6=\text{A}^3.\text{A}^3$

$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$

$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{12\pi}{7}&-\sin\frac{12\pi}{7}\\\sin\frac{12\pi}{7}&\cos\frac{12\pi}{7}\end{bmatrix}$

Now, we check if the pattern is same for k = 7.

Here,

$\text{A}^7=\text{A}^6\times\text{A}$

$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$

$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{14\pi}{7}&-\sin\frac{14\pi}{7}\\\sin\frac{14\pi}{7}&\cos\frac{14\pi}{7}\end{bmatrix}$

$\Rightarrow\text{A}^7=\begin{bmatrix}\cos2\pi&-\sin2\pi\\\sin2\pi&\cos2\pi\end{bmatrix}$$\begin{bmatrix}\because\ \frac{14\pi}{7}=2\pi\end{bmatrix}$

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

So, the least positive integral value of k is 7.

4. None of these.

Solution:

Let $\text{A}=\begin{bmatrix}0&-3\\3&0\end{bmatrix}$$\big[\because\ \text{a}_\text{ij} = \text{i}^2 -\text{j}^2\big]$ 

$|\text{A}|=0-(-9)=9\neq0$

2. Not possible to find

We are given that

$\begin{bmatrix}3\text{x}+7&5\\ \text{y}+1&2-3\text{x}\end {bmatrix},\ \begin{bmatrix}0& \text {y}-2\\8&4 \end{bmatrix}$

By defination of equality of matrices.

3x + 7 = 0, 5 = y - 2, y + 1 = 8, 2 - 3x = 4

$\therefore \text x = \frac{7}{3},\ \text y = 7,\ \text x = -\frac {2}{3}$

$\therefore$ (b) is correct answer.

1. $\frac{1}{2}\text{I}$

Solution:

Given $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&\cot^{-1}(\pi\text{x})\end{bmatrix}$ $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\pi\text{x})&\tan^{-1}(\frac{\text{x}}{\pi})\\\sin^{-1}(\frac{\text{x}}{\pi})&-\tan^{-1}(\pi\text{x})\end{bmatrix}$

$\text{A}-\text{B}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)+\cos^{-1}(\pi\text{x})&0\\0&\cot^{-1}(\pi\text{x})+\tan^{-1}(\pi\text{x})\end{bmatrix}$

$=\frac{1}{\pi}\begin{bmatrix}\frac{\pi}{2}&0\\0&\frac{\pi}{2}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac{1}{2}\text{I}$

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

Solution:

A diagonal matrix with all diagonal elements are equal is a scalar matrix.

1. It is not necessary that either A = 0 or, B = 0

Solution:

Let $\text{A}=\begin{bmatrix}0&2\\0&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$

$\therefore\ \text{AB}=\begin{bmatrix}0&2\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

3. $\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$

Solution:

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

$\text{A}^\text{n}=\text{a}^\text{n}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\text{a}^\text{n}&0&0\\0&\text{a}^\text{n}&0\\0&0&\text{a}^\text{n}\end{bmatrix}$

3. $1-\alpha^2-\beta\gamma=0$

Solution:

Given $\text{A}=\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}$ and A² = I, then

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

$\Rightarrow\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}\begin{bmatrix}\alpha&\beta\\\gamma&-\alpha\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$\Rightarrow\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

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

$\Rightarrow1-\alpha^2-\beta\gamma=0$

1. Skew-symmetric matrix.

Solution:

Given $\text{A}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$

$\text{A}^\text{T}=\begin{bmatrix}0&5&-8\\-5&0&-12\\8&12&0\end{bmatrix}$

$\Rightarrow\text{A}=-\text{A}^\text{T}$

So, A is skew-symmetric matrix.

3. |A| = 0 or |B| = 0

Soluton:

From the properties of the matrices, if A, B are non - zero square matrices of same order such that AB = 0 then the either of the matrices must be singular matrix.

A singular matrix is a matrix whose determinant is zero.

$\therefore$ |A| = 0 or |B| = 0

3. Multiplicative inverse matrix of A

Solution:

$\text{AB}=\begin{bmatrix}\text{I}&\text{amp; }\end{bmatrix}\text{BA}=\text{I}$ is the multiplicative inverse of A.

Hence, the answer is multiplicative inverse matrix of A.

3. -6, -4, -9

Solution:

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

$\text{kA}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$

$\Rightarrow\begin{bmatrix}0&2\text{k}\\3\text{k}&-4\text{k}\end{bmatrix}=\begin{bmatrix}0&3\text{a}\\2\text{b}&24\end{bmatrix}$

$\Rightarrow-4\text{k}=24$

$\Rightarrow\text{k}=-6$

$2\text{k}=3\text {a}$

$\Rightarrow\text{a}=-4$

$3\text{k}=2\text{b}$

$\Rightarrow\text{b}=-9$

4. 1 × 3

Solution:

Since, Order of a matrix is represented by m × n, where mm is the number of rows and nn is the number of columns.

Given, $\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is a matrix in which number of row is 1 and number of columns are 3.

$\therefore\begin{bmatrix}2&\text{amp};5 &\text{amp};7 \end{bmatrix}$ is a matrix of order 1 × 3

4. $\text{Not possible to find}$

Solution:

$\begin{bmatrix}3\text{x}+7&5\\\text{y}+1&2-3\text{x}\end{bmatrix}=\begin{bmatrix}0&\text{y}-2\\8&4\end{bmatrix}$

$\Rightarrow3\text{x}+7=0$

$\Rightarrow\text{x}=\frac{-7}{3}$

$5=\text{y}-2$

$\Rightarrow\text{y}=7$

$\text{y}+1=8$

$\Rightarrow\text{y}=7$

$2-3\text{x}=4$

$\Rightarrow\text{x}=\frac{-2}{3}$

We are getting two values of x.

So, it is not possible to find.

4. 1

Solution:

Given m[-3 + 4] + n[4 - 3] = [10 - 11]

⇒ -3m + 4n = 10 and 4m - 3n = -11

by solving we get m= -2 and n=1

$\therefore$ 3m + 7n = -6 + 7 = 1

3. |A|⁸

^Solution:

$∣​\text{adj}(\text{adj}\text{A}^2)​∣​=\text{Q}=\begin{vmatrix}\text{A}^2\end{vmatrix}^{(3-1)^2} =\begin{vmatrix}\text{ A}^2 \end{vmatrix} ^4 =\begin{vmatrix} \text{A}\end{vmatrix}^8$

3. ​​​​​$\theta=2\text{n}\pi+\frac{\pi}{3},\text{n}\in\text{Z}$

Solution:

Here,

$\text{A}=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$

$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$

Now,

$\text{A}^\text{T}+\text{A}=\text{I}_2$

$\Rightarrow\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}+\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$\Rightarrow\begin{bmatrix}2\cos\theta&0\\0&2\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$\Rightarrow2\cos\theta=1$

$\Rightarrow\cos\theta=\frac{1}{2}$

$\Rightarrow\cos\theta=\cos\frac{\pi}{3}$

$\Rightarrow\theta=2\text{n}\pi\pm\frac{\pi}{3}$ $(\text{n}\in\text{Z})$

1. 3 × 4

Solution:

Given that matrix A is 3 x 4.

Let the B matrix be P x Q.

$\therefore$ A is 4 x 3.

Since AB is defined, so number of columns of A must be equal to number of rows of B, therefore, P = 3.

Also, BA is defined, so the number of columns of B must be equal to number of rows of A, then Q = 4.

Therefore, matrix B is 3 x 4.