1 Marks Question

Question 11 Mark

If $\text{A}[\text{a}_{\text{ij}}]=\begin{bmatrix}2&3&-5\\1&4&9\\0&7&-2\end{bmatrix}$ and $\text{B}=[\text{b}_\text{ij}]=\begin{bmatrix}2&-1\\-3&4\\1&-2\end{bmatrix}$
Then find $a_{22} + b_{21}$

Answer

$a_{22} + b_{22} = 4 + (-3) = 1$
Hence, $a_{22} + b_{21} = 1$

View full question & answer→

Question 21 Mark

Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$a_{ij} = i - j$

Answer

Here, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$a_{11} = 1 - 1 = 0, a_{12} = 1 - 2 = -1, a_{13} = 1 - 3 = -2, a_{14} = 1 - 4 = -3$
$a_{21} = 2 - 1 = 1, a_{22} = 2 - 2 = 0, a_{23} = 2 - 3 = -1, a_{24} = 2 - 4 = -2$
$a_{31} = 3 - 1 = 2, a_{32} = 3 - 2 = 1, a_{33} = 3 - 3 = 0$ and $a_{34} = 3 - 4 = -1$
So, the required matrix is $\begin{bmatrix}0&-1&-2&-3\\1&0&-1&-2\\2&1&0&-1\end{bmatrix}.$

View full question & answer→

Question 31 Mark

In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind in all the colleges.

Answer

Number of different types of post in any college is given by,
$\text{X}=\begin{bmatrix}15\\6\\1\\1\end{bmatrix}$
Total number of posts of each kind in all the colleges = 30X
$=\begin{bmatrix}15\\6\\1\\1\end{bmatrix}$
$=\begin{bmatrix}450\\180\\30\\30\end{bmatrix}$

View full question & answer→

Question 41 Mark

Construct a $4 \times 3$ matrix whose element are:
$a_{ij} = i$

Answer

Here,
$a_{11} = 1, a_{12} = 1, a_{13}= 2,$
$a_{21} = 2, a22 = 2, a_{23} = 2$
$a_{31} = 3, a_{32} = 3, a_{33} = 3$
$a_{41} = 4, a_{42} = 4, a_{43} = 4$
Using Equation (i),
$\text{A}=\begin{bmatrix}1&1&1\\2&2&2\\3&3&3\\4&4&4\end{bmatrix}$

View full question & answer→

Question 51 Mark

Construct a 4 × 3 matrix whose element are:
$\text{a}_\text{ij}=\frac{\text{i}-\text{j}}{\text{i}+\text{j}}$

Answer

Here,
$\text{a}_{11}=\frac{1-1}{1+1}=\frac{0}{2},\text{ a}_{12}=\frac{1-2}{1+2}=\frac{-1}{3},$ $\text{a}_{13}=\frac{1-3}{1+3}=\frac{-2}{4}=\frac{-1}{2}$
$\text{a}_{21}=\frac{2-1}{2+1}=\frac{1}{3},\text{ a}_{22}=\frac{2-2}{2-2}=\frac{0}{0}=0,$ $\text{a}_{23}=\frac{2-3}{2+3}=\frac{-1}{5}$
$\text{a}_{31}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2},\text{ a}_{32}=\frac{3-2}{3+2}=\frac{1}{5},$ $\text{a}_{33}=\frac{3-3}{3+3}=\frac{0}{6}=0$
$\text{a}_{41}=\frac{4-1}{4+1}=\frac{3}{5},\text{a}_{42}=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}$ and $\text{a}_{43}=\frac{4-3}{4+3}=\frac{1}{7}$
So, the required matrix is $\begin{bmatrix}0&\frac{-1}{3}&\frac{-1}{2}\\\frac {1}{3}&0&\frac{-1}{5}\\\frac{1}{2}&\frac{1}{5}&0\\ \frac{3}{5}&\frac{1}{3}&\frac{1}{7}\end{bmatrix}.$

View full question & answer→

Question 61 Mark

Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$a_{ij} = i + j$

Answer

Here, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$a_{11} = 1 + 1 = 2, a_{12} = 1 + 2 = 3, a_{13} = 1 + 3 = 4, a_{14} = 1 + 4 = 5$
$a_{21} = 2 + 1 = 3, a_{22} = 2 + 2 = 4, a_{23} = 2 + 3 = 5, a_{24} = 2 + 4 = 6$
$a_{31} = 3 + 1 = 4, a_{32} = 3 + 2 = 5, a_{33} = 3 + 3 = 6, a_{34} = 3 + 4 = 7$
Using equation (i)
$\text{A}=\begin{bmatrix}2&3&4&5\\3&4&5&6\\4&5&6&7\end{bmatrix}$

View full question & answer→

Question 71 Mark

A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2800 as interest. However, if trust had interchanged money in bonds, they would have got ₹ 100 less as interest. Using matrix method, find the amount invested by the trust.

Answer

Let Rs. x be invested in the first bond and Rs. y be invested in the second bond.
Let A be the investment matrix and B be the interest per rupee matrix. Then,
$\text{A}=\begin{bmatrix}\text{x}&\text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}\frac{10}{100}\\\frac{12}{100}\end{bmatrix}$
Total annual interest $=\text{AB}=\begin{bmatrix}\text{x}&\text{y}\end{bmatrix}\begin{bmatrix}\frac{10}{100}\\\frac{12}{100}\end{bmatrix}=\frac{10\text{x}}{100}+\frac{12\text{y}}{100}$
$\therefore\ \frac{10\text{x}}{100}+\frac{12\text{y}}{100}=2800$
$\Rightarrow10\text{x}+12\text{y}=280000\ ...(1)$
If the rates of interest had been interchanged, then the total interest earned is Rs. 100 less than the previous interest.
$\therefore\ \frac{12\text{x}}{100}+\frac{10\text{y}}{100}=2700$
$\Rightarrow12\text{x}+10\text{y}=270000\ ...(2)$
The system of equations (1) and (2) can be expressed as
PX = Q, where $\text{P}=\begin{bmatrix}10&12\\12&10\end{bmatrix},\ \text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix},\ \text{ Q}=\begin{bmatrix}280000\\270000\end{bmatrix}$
$\big|\text{P}\big|=\begin{vmatrix}10&12\\12&10\end{vmatrix}=100-144=-44\neq0$
Thus, P is invertible.
$\therefore\ \text{X}=\text{P}^{-1}\text{Q}$
$\Rightarrow\text{X}=\frac{\text{adj }\text{P}}{\big|\text{P}\big|}\text{Q}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{(-44)}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}^\text{T}\begin{bmatrix}280000\\270000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{(-44)}\begin{bmatrix}10&-12\\-12&10\end{bmatrix}\begin{bmatrix}280000\\270000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}\frac{2800000-3240000}{-44}\\\frac{-3360000+2700000}{-44}\end{bmatrix}=\begin{bmatrix}10000\\15000\end{bmatrix}$
$\Rightarrow\text{x}=10000$ and $\text{y}=15000$
Therefore, Rs. 10,000 be invested in the first bond and Rs. 15,000 be invested in the second bond.

View full question & answer→

Question 81 Mark

Given an example of:
A triangular matrix.

Answer

$\begin{bmatrix}1&2&3\\0&5&4\\0&0&6\end{bmatrix}$
Here, all elements below the main diagonal in upper triangular matrix are zero.
$\begin{bmatrix}1&0&0\\2&6&0\\3&4&5\end{bmatrix}$
Here, all elements above the main diagonal in lower triangular matrix are zero.

View full question & answer→

Question 91 Mark

Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$a_{ij} = j$

Answer

Here, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$a_{11} = 1, a_{12} = 2, a_{13} = 3, a_{14} = 4$
$a_{21} = 1, a_{22} = 2, a_{23} = 3, a_{24} = 4$
$a_{31} = 1, a_{32} = 2, a_{33} = 3, a_{34} = 4$
Using Equation (i),
$\text{A}=\begin{bmatrix}1&2&3&4\\1&2&3&4\\1&2&3&4\end{bmatrix}$

View full question & answer→

Question 101 Mark

For what values of a and b if A = B, where$\text{A}=\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6\end{bmatrix},\text{ B}=\begin{bmatrix}2\text{a}+2&\text{b}^2+2\\8&\text{b}^2-5\text{b}\end{bmatrix}$

Answer

As the given matrices A and B are equal, therefore, their corresponding elements must be equal.
Comparing the corresponding elements, we get,
$\begin{bmatrix}\text{a}+4=2\text{a}+2&3\text{b}=\text{b}^2+2\\8=8&-6=\text{b}^2-5\text{b} \end{bmatrix}$
On simplifying, we get a = 2 and the common value of b = 2.
Hence, the values of a and b are 2, 2.

View full question & answer→

Question 111 Mark

Given an example of:
A diagonal matrix which is not scalar.

Answer

$\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}$
For a diagonal matrix which is not scalar, all elements except those in the leading diagonal should be zero and the elements in the diagonal should not be equal.

View full question & answer→

Question 121 Mark

Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$a_{ij} = 2i$

Answer

Here, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$a_{11} = 2(1) = 2, a_{12} = 2(1) = 2, a_{13} = 2(1) = 2, a_{14} = 2(1) = 2$
$a_{21} = 2(2) = 4, a_{22} = 2(2) = 4, a_{23} = 2(2) = 4, a_{24} = 2(2) = 4$
$a_{31} = 2(3) = 6, a_{32} = 2(3) = 6, a_{33} = 2(3) = 6$ and $a_{34} = 2(3) = 6$
So, the required matrix is $\begin{bmatrix}2&2&2&2\\4&4&4&4\\6&6&6&6\end{bmatrix}.$

View full question & answer→

Question 131 Mark

If $\text{A}[\text{a}_{\text{ij}}]=\begin{bmatrix}2&3&-5\\1&4&9\\0&7&-2\end{bmatrix}$ and $\text{B}=[\text{b}_\text{ij}]=\begin{bmatrix}2&-1\\-3&4\\1&-2\end{bmatrix}$
Then find $a_{11} + b_{11} + a_{22}b_{22}$

Answer

$a_{11} b_{11} + a_{22}b_{22} = (2)(2) + (4)(4) = 4 + 16 = 20$
Hence, $a_{11}b_{11} + a_{22}b_{22} = 20$

View full question & answer→

Question 141 Mark

Compute the following sums:
$\begin{bmatrix}3&-2\\1&4\end{bmatrix}+\begin{bmatrix}-2&4\\1&3\end{bmatrix}$

Answer

$\begin{bmatrix}3&-2\\1&4\end{bmatrix}+\begin{bmatrix}-2&4\\1&3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3-2&-2+4\\1+1&4+3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&2\\2&7\end{bmatrix}$

View full question & answer→

Question 151 Mark

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.
$2\text{X}+3\text{Y}=\begin{bmatrix}2&3\\4&0\end{bmatrix},\ 3\text{X}+2\text{Y}\begin{bmatrix}-2&2\\1&-5\end{bmatrix}$

Answer

We have,
$3(2\text{X}+3\text{Y})-2(3\text{X}+2\text{Y})=3\begin{bmatrix}2&3\\4&0\end{bmatrix}-2\begin{bmatrix}-2&2\\1&-5\end{bmatrix}$
$\Rightarrow6\text{X}+9\text{Y}-6\text{X}-4\text{Y}=\begin{bmatrix}6&9\\12&0\end{bmatrix}+\begin{bmatrix}4&-4\\-2&10\end{bmatrix}$
$\Rightarrow5\text{Y}=\begin{bmatrix}6+4&9-4\\12-2&0+10\end{bmatrix}$
$\Rightarrow\text{Y}=\frac{1}{5}\begin{bmatrix}10&5\\10&10\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}2&1\\2&2\end{bmatrix}\ \dots(1)$
Also,
$2(2\text{X}+3\text{Y})-3(3\text{X}+2\text{Y})=2\begin{bmatrix}2&3\\4&0\end{bmatrix}-3\begin{bmatrix}-2&2\\1&-5\end{bmatrix}$
$\Rightarrow4\text{X}+6\text{Y}-9\text{X}-6\text{Y}=\begin{bmatrix}4&6\\8&0\end{bmatrix}+\begin{bmatrix}6&-6\\-3&15\end{bmatrix}$
$\Rightarrow-5\text{X}=\begin{bmatrix}6+4&6-6\\8-3&0+15\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{-5}\begin{bmatrix}10&0\\5&15\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix}\ \dots(2)$
From (1) and (2), we get
$\text{X}=\begin{bmatrix}-2&0\\-1&-3\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}2&1\\2&2\end{bmatrix}$

View full question & answer→

Question 161 Mark

Construct a 4 × 3 matrix whose element are:
$\text{a}_\text{ij}=2\text{i}+\frac{\text{i}}{\text{j}}$

Answer

Here,
$\text{a}_{11}=2(1)+\frac{1}{1}=\frac{2+1}{1}=\frac{3}{1}=3,$ $\text{a}_{12}=2(1)+\frac{1}{3}=\frac{4+1}{2}=\frac{5}{2},$ $\text{a}_{13}=2(1)+\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}$
$\text{a}_{21}=2(2)+\frac{2}{1}=\frac{4+2}{1}=\frac{6}{1}=6,$ $\text{a}_{22}=2(2)+\frac{2}{2}=\frac{8+2}{2}=\frac{10}{2}=5,$ $\text{a}_{23}=2(2)+\frac{2}{3}=\frac{12+2}{3}=\frac{14}{3}$
$\text{a}_{31}=2(3)+\frac{3}{1}=\frac{6+3}{1}=\frac{9}{1}=9,$ $\text{a}_{32}=2(3)+\frac{3}{2}=\frac{12+3}{2}=\frac{15}{2},$ $\text{a}_{33}=2(3)+\frac{3}{5}=\frac{18+3}{3}=\frac{21}{3}=7$
$\text{a}_{41}=2(4)+\frac{4}{1}=\frac{8+4}{1}=\frac{12}{1}=12,$ $\text{a}_{42}=2(4)+\frac{4}{2}=\frac{16+4}{2}=\frac{20}{2}=10$ and $\text{a}_{43}=2(4)+\frac{4}{3}=\frac{24+4}{3}=\frac{28}{3}$
So, the required matrix is $\begin{bmatrix}3&\frac{5}{2}&\frac{7}{3}\\6&5&\frac{14}{3}\\9&\frac{15}{2}&7\\12&10&\frac{28}{3}\end{bmatrix}.$

View full question & answer→

Question 171 Mark

Given an example of:
A row matrix which is also a column matrix,

Answer

[6]
This is a matrix that contains only one element.

View full question & answer→

Question 181 Mark

Compute the following sums:
$\begin{bmatrix}2&1&3\\0&3&5\\-1&2&5\end{bmatrix}+\begin{bmatrix}1&-2&3\\2&6&1\\0&-3&1\end{bmatrix}$

Answer

$\begin{bmatrix}2&1&3\\0&3&5\\-1&2&5\end{bmatrix}+\begin{bmatrix}1&-2&3\\2&6&1\\0&-3&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+1&1-2&3+3\\0+2&3+6&5+1\\-1+0&2-3&5+1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3&-1&6\\2&9&6\\-1&-1&6\end{bmatrix}$

View full question & answer→

Question 191 Mark

Let $A$ be a matrix of order $3 \times 4$. If $R_1$ denotes the first row of $A$ and $C_2$ denotes its second column, then determine the orders of matrices $R_1$ and $C_2$.

Answer

The order of $R_1$ is $1 \times 4$ and the order of $C_2$ is $3 \times 1$.

View full question & answer→

Maths 1 Marks Question

Take a timed test