Maths M.C.Q (1 Marks)

3. 6

Solution:

Let the order of the matrix is $ (\text{a}\times\text{b})$ There are 18 elements in the matrix.

So, $\text{a}\times\text{b} = 18$

Possible orders can be $ (1\times18),( 18\times1),( 2\times9),( 9\times2),( 3\times6),( 6\times3)$

There are 6 possible orders.

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

Solution:

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

On using C₂ → C₂ - 2C₁; $\begin{bmatrix}1&-5\\2&0\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3&-5\\2&0\end{bmatrix}$

Since, on using elementary column operation on X = AB, we apply these operations simultaneously on X and on the second matrix B of the product AB on RHS.

2. $1\times1$

Solution:

If two matrix of order $\text{x}\times\text{m}$ and $\text{m}\times\text{n}$ are multiplied then the order of resultant matrix is $\text{x}\times\text{n}$

Now in the given equation going from left, matrices of order $1\times3$ and $3\times3$ are multiplied.So, the order of resultant matrix is  $1\times3 $ And now this is multiplied by matrix of order $3\times1.$

This will give resultant matrix of order $1\times1$

3. -2

Solution:

A scalar matrix has all the elements of the diagonals same.For example:$ \begin{bmatrix} 3 &\text{amp; } 0\\ 0&\text{amp; } 3 \end{bmatrix}$

In our case A is given to be a scalar matrix hence all the diagonal elements must be same.

So, $\text{x} = \text{m} = -1$

And $\text{x}+\text{m} = -1 -1 = -2$

1. $\text{I}\cos\theta+\text{J}\sin\theta$

Solution:

Here,

$\text{I}\cos\theta+\text{J}\sin\theta$

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

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

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

3. x = 2, y = 4

Solution:

$\begin{bmatrix}2\text{x}+\text{y}&4\text{x}\\5\text{x}-7&4\text{x}\end{bmatrix}=\begin{bmatrix}7&7\text{y}-13\\\text{y}&\text{x}+6\end{bmatrix},$

Equating the terms, we get

4x = x + 6

⇒ x = 2

And

2x + y = 7

⇒ y = 3

2. 6

Solution:

Given a matrix $2\times3\Rightarrow \begin{bmatrix} { \text{a} }_{11} &\text{amp; } {\text{a} }_{12} &\text{amp; } { \text{a} }_{13} \\ { \text{a} }_{21} &\text{amp; } {\text{a} }_{22} &\text{amp; } {\text{a} }_{23} \end{bmatrix}$ Clearly there are 6 elements.

1. 0

Solution:

Given: AB = I₃

$\Rightarrow\begin{bmatrix}1&2&\text{x}\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&-2&\text{y}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

$\Rightarrow\begin{bmatrix}1&0&\text{y+x}\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

The corresponding elements of two equal matrices are equal.

$\therefore\ \text{y}+\text{x}=0$

2. n^B

Solution:

Here,

$\text{A}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$ 

$\therefore\ \text{AB}=\begin{bmatrix}\text{n}&0&0\\0&\text{n}&0\\0&0&\text{n}\end{bmatrix}\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$

$\Rightarrow\text{AB}=\begin{bmatrix}\text{na}_1&\text{na}_2&\text{na}_3\\\text{nb}_1&\text{nb}_2&\text{nb}_3\\\text{nc}_1&\text{nc}_2&\text{nc}_3\end{bmatrix}$

$\Rightarrow\text{AB}=\text{n}\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3\end{bmatrix}$

$\Rightarrow\text{AB}=\text{n}\text{B}$

3. AB and BA both are defined.

Soluton:

Given: $\text{A} = \begin{bmatrix}2& -1 &3\\-4 & 5 & 1 \end{bmatrix} \ \text{B} = \begin{bmatrix}2& 3\\4 & -2 \\1 & 5 \end{bmatrix}$

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

$ \begin{bmatrix}3 & 23\\13 & -17 \end{bmatrix}$

So, AB is defined as of columns in A is equal to number of rows in B.

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

$ = \begin{bmatrix}-8& 13 &9\\16 & -14 & 10\\-18 & 24 & 8 \end{bmatrix}$

So, BA is also defined of columns in B is equal to number of rows in A.

4. 512

Solution:

The given matrix of the order $3\times3$ has 9 elements and each of these elements can be either 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Therefore, by the multiplication principle, the required number of possible matrices is $2^9=512$

3. kA

Solution:

Here,

$\text{S}=\big[\text{S}_{\text{ij}}\big]$

$\Rightarrow\text{S}=\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}$$\big[\because\ \text{S}_\text{ij} = \text{k}\big]$

Let $\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}$$\big[\because\ \text{A is square matrix}\big]$

Now,

$\text{AS}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}=\begin{bmatrix}\text{ka}_{11}&\text{ka}_{12}\\\text{ka}_{21}&\text{ka}_{22}\end{bmatrix}$$=\text{k}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\text{kA}$

$\text{SA}=\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\begin{bmatrix}\text{ka}_{11}&\text{ka}_{12}\\\text{ka}_{21}&\text{ka}_{22}\end{bmatrix}$$=\text{k}\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\text{kA}$

$\therefore\ \text{AS}=\text{SA}=\text{kA}$

1. I

Solution:

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

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

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

1. A

Solution:

We have, A² = I

Now, (A - I)³ + (A + I)³ - 7A = [(A - I) + (A + I)][(A - I)² + (A + I)² - (A - I)(A + I)] - 7A

$[\because$ a³ + b³ = (a + b)(a² + b² - ab)$]$

= [(2A){A² + I² - 2AI + A² + I² + 2AI - (A² - I²)}] - 7A

= [(2A){AI + I² - 2AI + AI + I² + 2AI - AI +I²}] - 7A $[\because$ A² = AI$]$

= 2A[I + I² + I + I² - I + I²] - 7A

= 2A[5I - I] - 7A

= 8AI - 7AI $[\because$ A = AI$]$

= AI

= A

3. $\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$

Solution:

A diagonalmatrixwith all its main diagonal entries equal is ascalar matrix, that is, ascalarmultiple of the identitymatrix

$\therefore \begin{bmatrix} 4 &\text{amp; 0} \\ 0 &\text{amp; } 4 \end{bmatrix}$ is a scalar matrix.

4. $|\text{A}|=\pm1$

Solution:

Given, $\text{A}^2=\text{I}$

Take determinant both sides,

$|\text{A}^2|=|\text{I}|\Rightarrow|\text{A}^2|=1\Rightarrow|\text{A}|=\pm1$

1. 4

Solution:

Since, given matrix A is of order $2\times2 = 4\therefore$  Number of elements in $\text{A} = 4$

3. Non - singular matrix

Solution:

$\text{A} = \big[1\big] $ is an identity matrix with order $1\times1.|\text{A}|\neq0$

So A is nonsingular.

1. nk

Solution:

$\text{Trace}=\sum\limits^\text{n}_{\text{i}-1}\text{a}_{\text{ij}}=\text{nk}$

3. $|\text{A}|^8$

Solution:

KEY :  3

$|\text{Adj}(\text{Adj}\text{A}^2)|$

$\text{Q}=|\text{A}^2|^{(3-1)^2}=|\text{A}^2|^4=|\text{A}|^8$

1. Square matrix

Solution:

Given, $\text{A}=\begin{bmatrix}0&\text{amp};0&\text{amp};4\\0&\text{amp};4&\text{amp};0\\4&\text{amp};0&\text{amp};0\end{bmatrix}$

The matrix is a square as it has same no. of rows and columns,

But it is not a diagonal matrix as there are elements other than diagonal ones which are not zero.

4. $\text{m}\times\text{n}$

Solution:

A matrix has mm rows and n columns then its order is $\text{m}\times\text{n}$

3. column matrix

Solution:

Matrix $ \displaystyle \begin{bmatrix}-12\\10 \\13 \\4 \end{bmatrix}$ is a column matrix.Hence, the answer is column matrix.

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

Solution:

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

$\text{A}^4=\text{i}^4\begin{bmatrix}1&0\\0&1\end{bmatrix}$

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

$\text{A}^{4\text{n}}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

4. m = n

Solution:

Matrix $\text{A} = [\text{a}_\text{ij}]_{\text{m} \times \text{n}}$ is a square matrix Number of columns = number of rows = m

1. Symmetric matrix.

Solution:

Let $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}3&2\\2&3\end{bmatrix}$

$\text{AB}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}3&2\\2&3\end{bmatrix}=\begin{bmatrix}7&8\\8&7\end{bmatrix}$

$\text{ABA}=\begin{bmatrix}7&8\\8&7\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}=\begin{bmatrix}23&22\\22&23\end{bmatrix}$

1.  $\displaystyle \begin{vmatrix} 2 & 7 \\ 0 & 1\end{vmatrix} $

Solution:

$ \text{A}=\displaystyle \begin{vmatrix} 6 &\text{amp; } 9 \\ 1 &\text{amp; } 4\end{vmatrix}-\displaystyle \begin{vmatrix} 4 &\text{amp; } 2 \\ 1 &\text{amp; } 3 \end{vmatrix}=\displaystyle \begin{vmatrix} 2 &\text{amp; } 7 \\ 0 &\text{amp; } 1 \end{vmatrix}$ 

1. B + A

Solution:

Yes, matrices are commutative.

We can see it as follows, Let element of A matrix be denoted by $\text{a}_\text{ij}$​ and B matrix be denoted by $\text{b}_\text{ij},$​

Then corresponding elements of$\text{ A + B}$ matrix will be $(\text{a}_\text{ij}​ +\text{b}_\text{ij}​) $and corresponding

elements of $\text{B + A}$ matrix will be $(\text{b}_\text{ij}​ +\text{ a}_\text{ij}​) $ But since addition is commutative, corresponding elements

of$\text{ A + B}$and $\text{B + A}$ matrices are the same, So they are equal.

2. 17

Solution:

x = 1.y = 8

$\therefore$ x + 2y = 17

4. 4(6!)

Solution:

No.of numbers in Matrix is 6

The possible orientations of Matrix is.

1 × 6, 2 × 3, 3 × 2, 6 × 1

The numbers in each orientation can be arranged in 6! ways.

$\implies$The total possibilities are 4(6!).

4. q × p

Solution:

$\begin{bmatrix}\text{a}_{11} &\text{amp;}\text{ a}_{12} \\\text{a}_{21}& \text{amp;}\text{ a}_{22}\\\text{a}_{31}&\text{amp; }\text{a}_{32} \end{bmatrix}$

Hence order of $\text{A}$ is $3\times2$

Row contains pp elements

So number of columns $=\text{P}$

Each column contains $\text{q}:$ element

So number of rows $=\text{q}$

Therefore, order $=\text{q}\times\text{p}$