1 Marks Question

Question 511 Mark

Compute the following:
$\begin{bmatrix}-1 & 4&-6 \\8 & 5&16\\ 2&8&5\end{bmatrix}+\begin{bmatrix}12 & 7&6 \\8 & 0&5\\3&2&4 \end{bmatrix}$

Answer

$\begin{bmatrix}-1&4&-6\\8&5&16\\2&8&5\end{bmatrix}+\begin{bmatrix}12&7&6\\8&0&5\\3&2&4\end{bmatrix}$
$=\begin{bmatrix}-1+12&4+7&-6+6\\8+8&5+0&16+5\\2+3&8+2&5+4\end{bmatrix}=\begin{bmatrix}11&11&0&\\16&5&21\\5&10&9\end{bmatrix} $

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Question 521 Mark

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

Answer

Here$,$
$a_{11} = 1, a_{12} = 1, a_{13}= 2,$
$a_{21} = 2, a_{22} = 2, a_{23} = 2$
$a_{31} = 3, a_{32} = 3, a_{33} = 3$
$a_{41} = 4, a_{42} = 4, a_{43} = 4$
Using Equation $(i),$
$\text{A}=\begin{bmatrix}1&1&1\\2&2&2\\3&3&3\\4&4&4\end{bmatrix}$

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Question 531 Mark

Compute the following:
$\begin{bmatrix}\text{a}^2+\text{b}^2 & \text{b}^2+\text{c}^2 \\ \text{a}^2+\text{c}^2 & \text{a}^2+\text{b}^2 \end{bmatrix}+\begin{bmatrix}2\text{ab} & 2\text{bc} \\-2\text{ac} &-2\text{ab}\end{bmatrix}$

Answer

$\begin{bmatrix}\text{a}^2+\text{b}^2 &\text{b}^2+\text{c}^2\\ \text{a}^2+\text{c}^2&\text{a}^{2}+\text{b}^2\end{bmatrix}+\begin{bmatrix}\text{2ab } &\text{2bc}\\ \text{-2ac}&\text{-2ab}\end{bmatrix}$
$=\begin{bmatrix}\text{a}^2+\text{b}^2+2\text{ab} &\text{b}^2+\text{c}^2+2\text{bc}\\ \text{a}^2+\text{c}^2-2\text{ac}&\text{a}^{2}+\text{b}^2-\text{2ab}\end{bmatrix}$
$=\begin{bmatrix}(\text{a + b})^2 &(\text{b + c})^2\\ (\text{a - c})^2&(\text{a - b})^2\end{bmatrix}$

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Question 541 Mark

Let $\text{A} = \begin{bmatrix}2&4\\3&2\end{bmatrix}, \text{B} = \begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{C} = \begin{bmatrix}-2&5\\3&4\end{bmatrix}.$Find each of the following:$\text{3A - C}$

Answer

$\text {3A - C}=3\begin{bmatrix}2&4\\3&2 \end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}=\begin{bmatrix}6& 12\\ 9&6 \end{bmatrix}-\begin{bmatrix}-2&5\\3&4\end{bmatrix}$
$=\begin{bmatrix}6+2&12-5\\9-3&6-4\end{bmatrix}=\begin{bmatrix}8&7\\6&2 \end{bmatrix}$

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Question 551 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.

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Question 561 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 $(i),$
$\text{A}=\begin{bmatrix}1&2&3&4\\1&2&3&4\\1&2&3&4\end{bmatrix}$

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Question 571 Mark

Let $\text{A} = \begin{bmatrix}2&4\\3&2\end{bmatrix}, \text{B} = \begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{C} = \begin{bmatrix}-2&5\\3&4\end{bmatrix}.$Find each of the following:
$\text{BA}$

Answer

$\text {BA}=\begin{bmatrix}1&3\\-2&5 \end{bmatrix}\begin{bmatrix}2&4\\3&2\end{bmatrix}=\begin{bmatrix}1(2)+3(3)& 1(4)+3(2)\\ (-2)2+5(3)&(-2)4+5(2) \end{bmatrix}=\begin{bmatrix}11&10\\11&2\end{bmatrix} $

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Question 581 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 $\begin{bmatrix}2&2&2&2\\4&4&4&4\\6&6&6&6\end{bmatrix}.$

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Question 591 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.

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Question 601 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 $\begin{bmatrix}0&\frac{-1}{3}&\frac{-1}{2}\\\frac {1}{3}&0&\frac{-1}{5}\\\frac{1}{2}&\frac{1}{5}&0\\ \frac{3}{5}&\frac{1}{3}&\frac{1}{7}\end{bmatrix}.$

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Question 611 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, $a_{11}b_{11} + a_{22}b_{22} = 20$

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Question 621 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 $\text{Y}=\begin{bmatrix}2&1\\2&2\end{bmatrix}$

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Question 631 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 $\begin{bmatrix}3&\frac{5}{2}&\frac{7}{3}\\6&5&\frac{14}{3}\\9&\frac{15}{2}&7\\12&10&\frac{28}{3}\end{bmatrix}.$

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Question 641 Mark

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer

Let the number of books as a 1 x 3 matrix B = $\begin{bmatrix}10\text { dozen}& 8 \text{ dozen}&10\text{ dozen}\\10 \times 12 = 120&8\times 12 = 96& 10 \times12= 120 \end{bmatrix}$
Let the selling prices of each book as a 3 x 1 matrix S = $\begin{bmatrix}80\\60\\40\end{bmatrix}$
$\therefore$ Total amount received by selling all books = BS = $\begin{bmatrix}120&96&120\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}$
= [120(80) + 96(60) + 120(40)] = [9600 + 5760 + 4800] = [20160]
Therefore, Total amount received by selling all the books = ₹ 20160

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Question 651 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.

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Question 661 Mark

Compute the following:
$\begin{bmatrix}\cos^2\text{x} & \sin^2\text{x} \\ \sin^2 \text{x} & \cos^2 \text{x} \end{bmatrix}+\begin{bmatrix}\sin^2\text{x}&\cos^2\text{x}\\ \cos^2\text{x}&\sin^2\text{x}\end{bmatrix} $

Answer

$\begin{bmatrix}\cos^2\text{x} & \sin^2\text{x} \\ \sin^2 \text{x} & \cos^2 \text{x} \end{bmatrix}+\begin{bmatrix}\sin^2\text{x}&\cos^2\text{x}\\ \cos^2\text{x}&\sin^2\text{x}\end{bmatrix} $
$=\begin{bmatrix}\text{cos}^2\text{x}+\sin^2\text{x}&\sin^2\text{x}+\cos^2\text{x}\\ \sin^2\text{x}+\cos^2\text{x}&\cos^2\text{x}+\sin^2\text{x}\end{bmatrix}=\begin{bmatrix}1&1\\1&1\end{bmatrix} $

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Question 671 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

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

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Question 681 Mark

Show that the matrix $\text{A}=\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}$is a symmentic matrix.

Answer

We have:
$\text{A}'=\begin{bmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{bmatrix}=-\begin{bmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{bmatrix}=-\text{A}$
$\therefore\ \text{ A}'=-\text{A} $
Hence, A is a skew-symmetric matrix.

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Question 691 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 $3 \times 1.$

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Question 701 Mark

Compute the following:
$\begin{bmatrix}\text{a} & \text{b} \\-\text{b} & \text{a} \end{bmatrix}+\begin{bmatrix}\text{a} & \text{b} \\\text{b} & \text{a} \end{bmatrix}$

Answer

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

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Question 711 Mark

Let $\text{A} = \begin{bmatrix}2&4\\3&2\end{bmatrix}, \text{B} = \begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{C} = \begin{bmatrix}-2&5\\3&4\end{bmatrix}.$ Find each of the following:
$\text{AB}$

Answer

$\text {AB}=\begin{bmatrix}2&4\\3&2 \end{bmatrix}\begin{bmatrix}1&3\\-2&5\end{bmatrix}=\begin{bmatrix}2(1)+4(-2)& 2(3)+4(5)\\ 3(1)+2(-2)&3(3)+2(5) \end{bmatrix}=\begin{bmatrix}-6&26\\-1&19\end{bmatrix} $

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