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MATRICES · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 71 questions.

Questions · Page 2 of 2

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Question 511 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 521 Mark
Construct a 3 × 4 matrix A = [aij] whose element aij are given by:
aij = 2i
Answer
Here, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
a11 = 2(1) = 2, a12 = 2(1) = 2, a13 = 2(1) = 2, a14 = 2(1) = 2
a21 = 2(2) = 4, a22 = 2(2) = 4, a23 = 2(2) = 4, a24 = 2(2) = 4
a31 = 2(3) = 6, a32 = 2(3) = 6, a33 = 2(3) = 6 and a34 = 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 531 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 541 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 551 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 a11 + b11 + a22b22
Answer
a11 b11 + a22b22 = (2)(2) + (4)(4) = 4 + 16 = 20
Hence, a11b11 + a22b22 = 20
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Question 561 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 571 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 581 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 591 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 601 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 611 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 621 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 631 Mark
Let A be a matrix of order 3 × 4. If R1 denotes the first row of A and C2 denotes its second column, then determine the orders of matrices R1 and C2.
Answer
The order of R1 is 1 × 4 and the order of C2 is 3 × 1.
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Question 641 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 651 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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Question 661 Mark
In the matrix $\text{A}=\begin{bmatrix}2&5 &19 &-7\\ 35 & -2 & \frac{5}{2} &12 \\ \sqrt{3} & 1 &-5 &17\\\end{bmatrix} $, write:
  1. The order of the matrix.
  2. The number of elements.
  3. write the elements a13, a21, a24, a23.
Answer
  1. There are 3 horizontal lines (rows) and 4 vertical lines (columns)in the given matrix A.

Therefore, Order of the matrix is 3 × 4.

  1. The number of elements in the matrix A is 3 × 4 = 12.
  2. $\text a_{13} \rightarrow$Element in first row and third column = 19   

$\text a_{21}\rightarrow$Element in second row and first coiumn = 35

$\text{a}_{33}\rightarrow$Element in third row and third column = -5

$\text{a}_{24}\rightarrow $Element in second row and fourth column = 12

$\text{ a}_{23}\rightarrow$Element in second row and third column = $\frac{5}{2}$

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Question 671 Mark
Compute the indicated products:

$\begin{bmatrix}\text{a} & \text{b} \\-\text{b} & \text{a} \end{bmatrix}\begin{bmatrix}\text{a} & -\text{b} \\\text{b} & \text{a} \end{bmatrix}$

Answer
$\begin{bmatrix}\text{a} & \text{b} \\-\text{b} & \text{a} \end{bmatrix}\begin{bmatrix}\text{a} & -\text{b} \\\text{b} & \text{a} \end{bmatrix}$

$=\begin{bmatrix}\text{a(a)}+\text{b(b)}&\text{a(-b)}+\text{b(a)}\\ -\text{b(a)}+\text{a(b)}&\text{(-b)(-b)}+\text{a(a)}\end{bmatrix}$

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

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Question 681 Mark
Find the transpose of each of the following matrices:

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

Answer
$\text{Let}\ \text{A}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}, \text{then}\ \text{A}^ \text{T}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}$
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Question 691 Mark
Find the transpose of each of the following matrices:

$\begin{bmatrix}5\\ \frac{1}{2}\\-1\end{bmatrix}$

Answer
$\text{Let}\text{A}=\begin{bmatrix}5\\ \frac{1}{2}\\-1\end{bmatrix},\text{then}\ \text{A}^\text{T}=\begin{bmatrix}5&\frac{1}{2}&-1\end{bmatrix} $
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Question 701 Mark
Construct a 2 × 2 matrix, A = $[\text{a}_{\text {ij}}]$, whose elements are given by:

$\text a_{\text {ij}}=\frac{\text i}{\text j} $

Answer
A = $[\text a_{\text {ij}}]$ is 2 × 2 matrix where  $\text a_{\text {ij}}$ = $\frac{\text {i}}{\text{j}}$

$\therefore\ \text a_{\text {11}}=\frac{\text 1}{\text 1}=1 $, $\text a_{\text {12}}=\frac{\text 1}{\text 2} $

 $\text a_{\text {21}}=\frac{\text 2}{\text 1}=2 $, $\text a_{22}=\frac{2}{2}=1 $

$\therefore\ \text A=\begin{bmatrix} 1& \frac{1}{2} \\2& 1 \end{bmatrix}$

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Question 711 Mark
Find the transpose of each of the following matrices:

$\begin{bmatrix}-1&5&6\\ \sqrt{3}&5&6\\2&3&-1\end{bmatrix}$

Answer
$\text{Let}\ \text{A}=\begin{bmatrix}-1&5&6\\ \sqrt{3}&5&6\\2&3&-1\end{bmatrix},\text{then}\ \text{A}^\text{T}=\begin{bmatrix}-1&\sqrt{3}&2\\5&5&3\\6&6&-1\end{bmatrix} $
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1 Marks - Page 2 - Maths STD 12 Science Questions - Vidyadip