Maths M.C.Q (1 Marks)

2. square matrix

Solution:

A matrix having mm rows and nn columns with m = n, means number of rows are equal to number of columns.

$\therefore$  given matrix is square matrix.

2. Identity

Solution:

The elements $\text{a}_\text{ij}$ ​ of a matrix where i = j lie along its diagonal and

the elements $\text{a}_\text{ij}$ of a matrix where $\text{i}\neq\text{j}$ are not along the diagonal.

As the diagonal elements are 11 and the rest of the elements are 0, the matrix A is an identity matrix.

3. $\displaystyle \begin{vmatrix}2&0 \\ 1 &1 \end{vmatrix}$

Solution:

$\text{A+B}=\displaystyle \begin{vmatrix} 2 &\text{amp; } 0 \\ 1 &\text{amp; } 1 \end{vmatrix}$

1. -4, 4

Solution:

$\text{A+B =1},\text{ i.e.,} \displaystyle \begin{vmatrix} 1&\text{amp; }\text{x+y} \\\text{y}-4 &\text{amp;} 1 \end{vmatrix}=\begin{vmatrix} 1&\text{amp; } 0 \\ 0 &\text{amp; } 1 \end{vmatrix}$

or $\text{x}=\text{y}=0,\text{ y}-4=0$

$\therefore\text{ x} = -4, \text{ y}=4$

1. Square matrix.

Solution:

Given: $\text{A}=\begin{bmatrix}0&0&4\\0&4&0\\4&0&0\end{bmatrix}$

Since, number of rows is equal to number of columns.

Therefore, A is a square matrix.

Hence, the correct option is (a).

2. A unit matrix

Solution:

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

$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\\text{a}&\text{b}&-1\end{bmatrix}$

$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+0&0+0+0&0+0-0\\0+0+0&0+1+0&0+0-0\\\text{a}+0-\text{a}&0+\text{b}-\text{b}&0+0+1\end{bmatrix}$

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

2. 3 × 1

Solution:

Order of A : 3 × 2 Order of B : 2 × 1 Multiplication of matrices is possible if and only.

if the number of columns of first matrix is equal to the number of rows of second matrix In AB

No.of columns in A is No.of rows in B is 2 $\therefore$ AB exists.

Order of AB is (number of rows of A x number of columns of B)  

$\therefore$ Order of AB is (3 × 1)

2. 3 × 1

Solution:

Order of matrix with mm rows and nn columns is given as $\text{m} \times\text{n}$ Let  $\text{A}=\begin{bmatrix}-1\\3\\4 \end{bmatrix}$

In the given matrix, there are $3 $ rows and $1$ column.

Hence, the order of A is $3\times 1$.

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

3. m = n

For A = [a_ij]_{m × n} to be square matrix.

number of row s = number of columns

$\therefore$ m = n

$\therefore$ (c) is correct.

1. $1\times13$ or $13\times1$

Solution:

If order of matrix $\text{A}=\text{a}\times\text{b}$

Then number of element in $\text{A}=\text{ab}$

Given $\text{ab}=13$

So, $\text{a}=1,\text{b}=13$

or $\text{b}=1,\text{a}=13$

So, $1\times13$ or $13\times1$ are possible order of $\text{A}$

1. $\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

​Solution:​​​​​​

We have, $\begin{bmatrix}4&2\\3&3\end{bmatrix}=\begin{bmatrix}1&2\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

Using elementary row operation R₁ → R₁ - 3R₂

$\begin{bmatrix}-5&-7\\3&3\end{bmatrix}=\begin{bmatrix}1&-7\\0&3\end{bmatrix}\begin{bmatrix}2&0\\1&1\end{bmatrix}$

Since, on using elementary row operation on X = AB, we apply these operation simultaneously on X and on the first matrix A of the product AB on RHS.

4. m × n

Solution:

Let A = [a_ij]_m×n and B = [b_ij]_p×q

B' = [b_ji]_q×p

Now, AB’ is defined, so n = q

and B’A is also defined, so p = m

$\therefore$ Order of B' = [b_ji]_n×m

And order of B = B = [b_ij]m_×n

1. [1, 2, 3]

Solution:

Since, $ \text{A}=\begin{bmatrix} 1 &\text{amp; } 2 &\text{amp; }3 \end{bmatrix}$ represents a matrix with three,

elements 1, 2, 3 $\therefore$ Elements of A = (1, 2, 3)

2. Symmetric matrix.

Solution:

We have $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$

$\therefore\ \text{A}'=\text{A}$

So, the given matrix is a symmetric matrix.

3. Skew-symmetric matrix.

Solution:

We have $\text{B}=\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$

$\Rightarrow\ \text{B}'=\begin{bmatrix}0&5&8\\-5&0&-12\\8&12&0\end{bmatrix}$

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

$=-\text{B}$

Since, B' = -B,

Thus, B is a skew-symmetric matrix.

4. x = y

Solution:

We know that the reflection matrix through a line $\text{y}=\text{mx}$ making an $\angle \theta$ with x - axis is given as.

$\begin{bmatrix} \cos { 2\theta }&\text{amp;}\sin { 2\theta } \\ \sin { 2\theta } &\text{amp;} -\cos { 2\theta }\end{bmatrix}$

Given transformation matrix is $\begin{bmatrix}0&\text{amp; }1\\1&\text{amp; }0\end{bmatrix}$

$\Rightarrow\cos2\theta=0\sin2\theta=1$

$\Rightarrow 2\theta ={90}^{0}$

$\Rightarrow \theta={45}^{0}$

$\Rightarrow \tan \theta=1$

Hence, the line of reflection is $\text{y}=\text{x}$

1. A skew-symmetric matrix.

Solution:

Here,

$\text{A}=\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$

$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}0&-5&7\\5&0&-11\\-7&11&0\end{bmatrix}$

$\Rightarrow\text{A}^\text{T}=-\begin{bmatrix}0&5&-7\\-5&0&11\\7&-11&0\end{bmatrix}$

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

Thus, A is a skew-symmetric matrix.

1. None of these.

Solution:

Let us consider a matrix $\begin{bmatrix}\text{a}&\text{b}&\text{c}\\\text{d}&\text{e}&\text{f}\\\text{g}&\text{h}&\text{i}\end{bmatrix}$

The element a can have two values 0 or 2 in two ways.

Similarly all other elements can also have two values 0 or 2 in two ways each.

So, the total number of combinations is 2⁹.

So, total no of matrices will be 2⁹.

2. a = 1, b = 4

Solution:

Here,

$(\text{A+B})^2=\text{A}^2+\text{B}^2$

$\Rightarrow\text{A}^2+\text{AB}+\text{BA}+\text{B}^2=\text{A}^2+\text{B}^2$

$\Rightarrow\text{AB}+\text{BA}=0$

$\Rightarrow\text{AB}=-\text{BA}$

$\Rightarrow\begin{bmatrix}1&-1\\2&-1\end{bmatrix}\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}=-\begin{bmatrix}\text{a}&1\\\text{b}&-1\end{bmatrix}\begin{bmatrix}1&-1\\2&-1\end{bmatrix}$

$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=-\begin{bmatrix}\text{a}+2&-\text{a}-1\\\text{b}-2&-\text{b}+1\end{bmatrix}$

$\Rightarrow\begin{bmatrix}\text{a}-\text{b}&2\\2\text{a}-\text{b}&3\end{bmatrix}=\begin{bmatrix}-\text{a}-2&\text{a}+1\\\text{b}+2&\text{b}-1\end{bmatrix}$

The corresponding elements of two equal matrices are equal.

$\Rightarrow\text{a}+1=2\text{ and b}-1=3$

$\therefore\ \text{a}=1\text{ and b}=4$

4. 512

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

Solution:

Here,

$\text{A}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}$

$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$

Now,

$\text{A}+\text{A}^\text{T}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}+\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$

$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}2+2&0+4&-3-5\\4+0&3+3&1+7\\-5-3&7+1&2+2\end{bmatrix}$

$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$

$\text{A}-\text{A}^\text{T}=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}-\begin{bmatrix}2&4&-5\\0&3&7\\-3&1&2\end{bmatrix}$

$\Rightarrow\text{A}-\text{A}^\text{T}=\begin{bmatrix}2-2&0-4&-3+5\\4-0&3-3&1-7\\-5+3&7-1&2-2\end{bmatrix}$

$\Rightarrow\text{A}-\text{A}^\text{T}=\begin{bmatrix}0&-4&2\\4&0&-6\\-2&6&0\end{bmatrix}$

Let $\text{P}=\frac{1}{2}(\text{A}+\text{A}^\text{T})=\frac{1}{2}\begin{bmatrix}4&4&-8\\4&6&8\\-8&8&4\end{bmatrix}$ $=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}$

$\text{Q}=\frac{1}{2}(\text{A}-\text{A}^\text{T})=\frac{1}{2}\begin{bmatrix}0&-4&2\\4&0&-6\\-2&6&0\end{bmatrix}$$=\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}$

Now,

$\text{P}^\text{T}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}^\text{T}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}=\text{P}$

$\text{Q}^\text{T}=\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}=\begin{bmatrix}0&2&-1\\-2&0&3\\1&-3&0\end{bmatrix}$$=-\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}=-\text{Q}$

$\text{P+Q}=\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}+\begin{bmatrix}0&-2&1\\2&0&-3\\-1&3&0\end{bmatrix}$

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

$=\begin{bmatrix}2&0&-3\\4&3&1\\-5&7&2\end{bmatrix}=\text{A}$

Thus, we have expressed A is the sum of a symmetric and a skew-symmetric matrix.

Hence,the symmetric matrix is $\begin{bmatrix}2&2&-4\\2&3&4\\-4&4&2\end{bmatrix}.$

1. A

Solution:

(A - I)³ + (A + I)³ - 7A

= A³ - I³ - 3A²I + 3AI² + A³ + I³ + 3A²I + 3AI² - 7A

= 2A³ + 6AI² - 7A

= 2A.A² + 6A - 7A

= 2A.I - A ($\because$ A² = I)

= 2A - A

= A

Hence, the correct option is (a).

2. $\text{A}$

Solution:

$\text{A}^2=\begin{bmatrix}\text{a}^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}\begin{bmatrix}\text{a}^2 &\text{amp; ab}&\text{amp; ac} \\\text{ab}&\text{amp; }\text{b}^2&\text{amp;}\text{ bc}\\\text{ac}&\text{amp;}\text{bc}&\text{amp;}\text{c}^2 \end{bmatrix}=\text{A}$

2. 0

Solution:

$\begin{bmatrix}\text{x}\\\text{x}\\\text{y}\end{bmatrix}+\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}+\begin{bmatrix}\text{z}\\0\\0\end{bmatrix}=\begin{bmatrix}10\\5\\5\end{bmatrix}\therefore\text{x}+\text{y}+\text{z}=10,\text{ x}+\text{y}=5$

$\text{y}+\text{z}=5$ Replacing $\text{x}+\text{y}=5$ in $\text{x}+\text{y}=\text{z}=10$

We have, $\text{z}=5$

Also, $\text{y}+\text{z}=5$

$\therefore\text{y}=5-\text{z}=0$​​​​

4. None of the above.

Solution:

For any two matrices A and S, we may have AB = BA = I, $\text{AB}\neq\text{BA}$ and AB = O but it is not always true.

1. square matrix

Solution:

Since, order of given matrix is 3 × 3.

$\therefore$ No of rows = No. of columns

So, given matrix is a square matrix.

1. I

​​​​​​​Solution:

We have, A = [a_ij]_2×2, where a_ij = 1, if $\text{i}\neq\text{j}$ and 0 if i = j

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

And $\text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$