Question 1012 Marks
Find x, y, z and t, if.
$3\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{t}\end{bmatrix}=\begin{bmatrix}\text{x}&6\\-1&2\text{t}\end{bmatrix}+\begin{bmatrix}4&\text{x}+\text{y}\\\text{z}+\text{t}&3\end{bmatrix}$
Answer$3\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{t}\end{bmatrix}=\begin{bmatrix}\text{x}&6\\-1&2\text{t}\end{bmatrix}+\begin{bmatrix}4&\text{x}+\text{y}\\\text{z}+\text{t}&3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3\text{x}&3\text{y}\\3\text{z}&3\text{t}\end{bmatrix}=\begin{bmatrix}\text{x}+4&6+\text{x}+\text{y}\\-1+\text{z}\text+{t}&2\text{t}+3\end{bmatrix}$
$\therefore\ 3\text{x}=\text{x}+4$
$\Rightarrow3\text{x}-\text{x}=4$
$\Rightarrow2\text{x}=4$
$\Rightarrow\text{x}=2$
Also,
$3\text{y}=6+\text{x}+\text{y}$
$\Rightarrow3\text{y}-\text{y}=6+\text{x}$
$\Rightarrow2\text{y}=6+\text{x}\ \dots(1)$
Putting the value of x in eq. (1), we get
$2\text{y}=6+2$
$\Rightarrow2\text{y}=8$
$\Rightarrow\text{y}=4$
Now,
$3\text{t}=2\text{t}+3$
$\Rightarrow3\text{t}-2\text{t}=3$
$\Rightarrow\text{t}=3$
$3\text{z}=-1+\text{z}+\text{t}$
$\Rightarrow3\text{z}-\text{z}=-1+\text{t}$
$\Rightarrow2\text{z}=-1+\text{t}\ \dots(2)$
Putting the value of t in eq. (2), we get
$2\text{z}=-1+3$
$\Rightarrow2\text{z}=2$
$\Rightarrow\text{z}=1$
$\therefore\ \text{x}=2,\text{ y}=4,\text{ z}=1$ and $\text{t}=3 $
View full question & answer→Question 1022 Marks
Find the value of y, if $\begin{bmatrix}\text{x}-\text{y}&2\\\text{x}&5 \end{bmatrix}=\begin{bmatrix}2&2\\3&5 \end{bmatrix}$
AnswerGiven,
$\begin{bmatrix}\text{x}-\text{y}&2\\\text{x}&5 \end{bmatrix}=\begin{bmatrix}2&2\\3&5 \end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
x = 3
and x - y = 2
⇒ 3 - y = 2
⇒ -y = 2 - 3
⇒ -y = -1
⇒ y = 1
View full question & answer→Question 1032 Marks
Compute the indicated products:
$\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\-3&2&-1\end{bmatrix}$
Answer$\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\-3&2&-1\end{bmatrix}$
$=\begin{bmatrix}(1)(1)+(-2)(-3)&(1)(2)+(-2)(2)&(1)(3)+(-2)(1)\$2)(1)+(3)(-3)&(2)(2)+(3)(2)&(2)(3)+(3)(-1)\end{bmatrix}$
$=\begin{bmatrix}1+6&2-4&3+2\\2-9&4+6&6-3\end{bmatrix}$
$=\begin{bmatrix}7&-2&5\\-7&10&3\end{bmatrix}$
View full question & answer→Question 1042 Marks
If A is a symmetric matrix and n ∈ N, write whether An is symmetric or skew-symmetric or neither of these two.
AnswerIf A is a symmetric matrix, then AT = A.
Now,
(An)T = (AT)n [for all n ∈ N]
⇒ (An)T = (A)n [$\because$ AT = A]
Hence, An is a symmetric matrix.
View full question & answer→Question 1052 Marks
Find a matrix X such that 2A + B + X = 0, where.
$\text{A}=\begin{bmatrix}-1&2\\3&4\end{bmatrix},\text{B}=\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
AnswerGiven: $\text{A}=\begin{bmatrix}-1&2\\3&4\end{bmatrix},\text{B}=\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
and
$2\text{A}+\text{B}+\text{x}=0$
$\Rightarrow\text{x}=-2\text{A}-\text{B}$
$\Rightarrow\text{x}=-2\begin{bmatrix}-1&2\\3&4\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
$\Rightarrow\text{x}=\begin{bmatrix}2&-4\\-6&-8\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
$\Rightarrow\text{x}=\begin{bmatrix}2-3&-4+2\\-6-1&-8-5\end{bmatrix}$
$\Rightarrow\text{x}=\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}$
Hence,
$\text{x}=\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}$
View full question & answer→Question 1062 Marks
Find the values of x and y, if $2\begin{bmatrix}1&3\\0&\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
AnswerGiven,
$2\begin{bmatrix}1&3\\0&\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&6\\0&2\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+\text{y}&6+0\\0+1&2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+\text{y}&6\\1&2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
$2+\text{y}=5$
$\Rightarrow\text{y}=5-2$
$\Rightarrow\text{y}=3$
And $2\text{x}+2=8$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}=3$
Hence,
$\text{x}=3,\text{y}=3$
View full question & answer→Question 1072 Marks
Give example of matrices:
A, B and C such that AB = AC but B ≠ C, A ≠ 0
AnswerLet $\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&0\\-1&0\end{bmatrix},\ \text{C}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
Here,
$\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\-1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$
$\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$\begin{bmatrix}0&0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
LHS = RHS
So,
for A ≠ 0, BC ≠ 0 but AB = AC
We have,
$\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&0\\-1&0\end{bmatrix},\ \text{C}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
View full question & answer→Question 1082 Marks
In a parliament election, a political party hired a public relations firm to promote its candidates in three ways - telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as
$\text{A}=\begin{bmatrix}140&\text{Telephone}\\200&\text{House calls}\\150&\text{Letters}\end{bmatrix}$
The number of contacts of each type made in two cities X and Yis given in the matrix B as
$\begin{matrix}\text{Telephone}&\text{House calls}&\text{Letters}\end{matrix}\\\text{B}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{matrix}\text{City X}\\\text{City Y}\end{matrix}$
Find the total amount spent by the party in the two cities.
What should one consider before casting his/ her vote - party's promotional activity of their social activities?
AnswerAccording to the question, Let A be the matrix showing the cost per contact (in paisa). $\text{A}=\begin{bmatrix}140&\text{Telephone}\\200&\text{House calls}\\150&\text{Letters}\end{bmatrix}$ And, B be a matrix showing the number of contacts of each type made in two cities X and Y.
$\begin{matrix}\text{Telephone}&\text{House calls}&\text{Letters}\end{matrix}\\\text{B}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{matrix}\text{City X}\\\text{City Y}\end{matrix}$
Now, The total amount spent by the party in the two cities will shown by BA.
$\text{BA}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{bmatrix}140\\200\\150\end{bmatrix}$
$=\begin{bmatrix}140000+100000+750000\\420000+200000+1500000\end{bmatrix}$
$=\begin{bmatrix}990000\\2120000\end{bmatrix}$
Hence, the total amount spent by the party in the two cities is
X: Rs. 9900
Y: Rs. 21200
One should consider social activities of a party before casting his/ her vote.
View full question & answer→Question 1092 Marks
Three shopkeepers A, B and C go to a store to buy stationary. A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. Cpurchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs 40 paise, a pen costs Rs. 1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual's bill.
Answer | Shopkeepers | Notebooks In dozen | Pens In dozen | Pencils In dozen |
| A | 12 | 5 | 6 |
| B | 10 | 6 | 7 |
| C | 11 | 3 | 8 |
Here,
Cost of notebooks per dozen = (12 × 40) paise = Rs. 4.80
Cost of pens per dozen = Rs. (12 × 1.25) = Rs. 15
Cost of Pencils per dozen = (12 × 35) paise = Rs. 4.20
$\therefore\ \begin{bmatrix}12&5&6\\10&6&7\\11&13&8\end{bmatrix}\begin{bmatrix}4.80\\15\\4.20\end{bmatrix}=\begin{bmatrix}12\times4.80+5\times15+6\times4.20\\10\times4.80+6\times15+7\times4.20\\11\times4.80+13\times15+8\times4.20\end{bmatrix}$
$=\begin{bmatrix}57.60+75+25.20\\48+90+29.40\\52.80+195+33.60\end{bmatrix}$
$=\begin{bmatrix}157.80\\167.40\\281.40\end{bmatrix}$
Thus, the bills of A, B and C are Rs. 157.80, Rs. 167.40 and 281.40, respectively.
View full question & answer→Question 1102 Marks
Let $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix}$ and $\text{C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}.$ Find each of the following:
$3\text{A}-2\text{B}+3\text{C}$
Answer Given, $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{ B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{ C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}$ $3\text{A}-2\text{B}+3\text{C}$
$=3\begin{bmatrix}2&4\\3&2\end{bmatrix}-2\begin{bmatrix}1&3\\-2&5 \end{bmatrix}+3\begin{bmatrix}-2&5\\3&4 \end{bmatrix}$
$=\begin{bmatrix}6&12\\9&6\end{bmatrix}-\begin{bmatrix}2&6\\-4&10\end{bmatrix}+\begin{bmatrix}-6&15\\9&12\end{bmatrix}$
$=\begin{bmatrix}6-2-6&12-6+15\\9+4+9&6-10+12\end{bmatrix}$
$=\begin{bmatrix}-2&21\\22&8\end{bmatrix}$
Hence,
$3\text{A}-2\text{B}+3\text{C}=\begin{bmatrix}-2&21\\22&8\end{bmatrix}$
View full question & answer→Question 1112 Marks
In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways: telephone, house calls and letters. The cost per contact (in paise) is given matrix A as. $\ \ \ \ \ \ \ \ \ \ \ \ \text{Cost per contact}\\\text{A}=\begin{bmatrix}40&\text{Telephone}\\100&\text{House call}\\50&\text{Letter}\end{bmatrix}$ The number of contacts of each type made in two cities X and Y is given in matrix B as
$\text{BA}=\begin{bmatrix}\text{Telephone}&\text{House call}&\text{Letter}\\1000&500&5000\\3000&1000&10000\end{bmatrix} \begin{matrix}\rightarrow\text{X}\\\rightarrow\text{Y}\end{matrix}$
Find the total amount spent by the group in the two cities X and Y.
AnswerThe cost per contact (in paise) is given by,
$\text{A}=\begin{bmatrix}40&\text{Telephone}\\100&\text{House call}\\50&\text{Letter}\end{bmatrix}$
The number of contacts of each type made in the two cities X and Y is given by,
$\text{BA}=\begin{bmatrix}\text{Telephone}&\text{House call}&\text{Letter}\\1000&500&5000\\3000&1000&10000\end{bmatrix} \begin{matrix}\rightarrow\text{X}\\\rightarrow\text{Y}\end{matrix}$
Total amount spent by the group in the two cities X and Y is given by,
$\text{BA}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{bmatrix}40\\100\\50\end{bmatrix}$
$=\begin{bmatrix}40000+50000+250000\\120000+100000+500000\end{bmatrix}$
$=\begin{bmatrix}340000\\720000\end{bmatrix}\begin{matrix}\text{X}\\\text{Y}\end{matrix}$
Thus, Amount spent on X = Rs. 3400
Amount spent on Y = Rs. 7200
View full question & answer→Question 1122 Marks
To promote making of toilets for women, an organisation tried to generate awarness through (i) house calls, (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below:
- ₹ 50
- ₹ 20
- ₹ 40
The number of attempts made in three villages X, Y and Z are given below:
| | (i) | (ii) | (iii) |
| X | 400 | 300 | 100 |
| Y | 300 | 250 | 75 |
| Z | 500 | 400 | 150 |
Find the total cost incurred by the organisation for three villages separately, using matrices.
AnswerThe cost for each mode per attempt is represented by 3 × 1 matrix:
$\text{A}=\begin{bmatrix}50\\20\\40\end{bmatrix}$
The number of attempts made in the three villages X, Y, and Z are represented by a 3 × 3 matrix:
$\text{B}=\begin{bmatrix}400&300&100\\300&250&75\\500&400&150\end{bmatrix}$
The total cost incurred by the prganization for the three villages seperately is given by matrix multiplication,
$\text{BA}=\begin{bmatrix}400&300&100\\300&250&75\\500&400&150\end{bmatrix}\begin{bmatrix}50\\20\\40\end{bmatrix}$
$\text{BA}=\begin{bmatrix}400\times50+300\times20+100\times40\\300\times50+250\times20+75\times40\\500\times50+400\times20+150\times40\end{bmatrix}$
$=\begin{bmatrix}30,000\\23,000\\39,000\end{bmatrix}$
Note: The answer given in the book is incorrect.
View full question & answer→Question 1132 Marks
Write the number of all possible matrices of order 2×2 with each entry 1, 2 or 3.
AnswerAs matrices is of order 2×2, so there are 4 entries possible.
Each entry has 3 choices that are 1, 2 or 3
So, number of ways to make up such matrices are 3×3×3×3 i.e, 34 times or 81 times
View full question & answer→Question 1142 Marks
Find the values of a, b, c and d from the equation:
$\begin{bmatrix}a-b&2a+c\\2a-b&3c+d\end{bmatrix}=\begin{bmatrix}-1&5\\0&13\end{bmatrix}.$
Answer Equating corresponding entries,
a - b = -1 ...(i)
2a - b = 0 ...(ii)
2a + c = 5 ...(iii)
3c + d = 13 ...(iv)
Eq. (i) - Eq. (ii) = -a = -1 ⇒ a = 1
Putting a = 1 in eq. (i), 1 - b = -1 ⇒ -b = -2 ⇒ b = 2
Putting a = 1 in eq. (iii), 2 + c = 5 ⇒ c = 5 - 2 ⇒ c = 3
Putting c = 3 in eq. (iv), 9 + d = 13 ⇒ d = 13 - 9 ⇒ d = 4
$\therefore$ a = 1, b = 2, c = 3, d = 4
View full question & answer→Question 1152 Marks
If A and B are symmetric matrices, then write the condition for which AB is also symmetric.
AnswerGiven that,
A and B are symmetric matrices, so
⇒ AT = A and BT = B
Now,
$\big(\text{AB}\big)^\text{T}=\text{B}^\text{T}\times\text{A}^\text{T}$ $\big\{\text{since, (AB)}^\text{T}=\text{B}^\text{T}\text{A}^\text{T}\big\}$
$\big(\text{AB}\big)^\text{T}=\text{BA}\ \dots(\text{i})$ $\big\{\text{since, B}^\text{T}=\text{B},\text{A}^\text{T}=\text{A}\big\}$
For AB to be symmetric matrix
(AB)T = AB
From equation (i) and (ii),
AB = BA
So,
For AB to be symmetric matrix we must have AB = BA.
View full question & answer→Question 1162 Marks
If $\begin{bmatrix}\text{x}+3&4\\\text{y}-4&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}5&4\\3&9 \end{bmatrix},$ find x abd y
AnswerThe corresponding elements of two equal matrices are equai. Given:
$\begin{bmatrix}\text{x}+3&4\\\text{y}-4&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}5&4\\3&9 \end{bmatrix}$ x + 3 = 5 and y - 4 = 3
⇒ x = 5 - 3 and y = 3 + 4
⇒ x = 2 and y = 7
$\therefore$ x = 2 and y = 7
View full question & answer→Question 1172 Marks
Construct a 2 × 2 matrix, A = $[\text a_{\text {ij}}]$, whose elements are given by: $\text a_{\text{ij}}=\frac{(\text{i}+\text{j})^2} {2} $
AnswerA = $[\text a_{\text{ ij}}]$ is 2 × 2 matrix where, $\text a_{\text{ij}}=\frac{(\text{i}+\text{j})^2} {2} $ $\therefore\ \ \text{a}_{11}=\frac{(1+1)^2}2=\frac{4}{2}=2$, $\text a_{12}=\frac{(1+2)^2}{2}=\frac{9}{2} $
$\text a_{21}=\frac{(2+1)^{2}}{2}=\frac{9}{2} $, $\text a_{23}=\frac{(2+2)^{2}}{2}=\frac{16}{2}=8 $
$\therefore\ \text A= \begin{bmatrix}2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix} $
View full question & answer→Question 1182 Marks
If $\text{A}=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix},$ then verify that $(\text{A}-\text{B})'=\text{A}'-\text{B}'.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}$ and
$\text{B}=\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix}$ $(\text{A}-\text{B})=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}-\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix}$
$=\begin{bmatrix}0&0\\-2&-3\\-2&3\end{bmatrix}$
and $(\text{A}-\text{B})'=\begin{bmatrix}0&-2&-2\\0&-3&3\end{bmatrix}$
Also, $\text{A}'-\text{B}'=\begin{bmatrix}1&4&5\\2&1&6\end{bmatrix}-\begin{bmatrix}1&6&7\\2&4&3\end{bmatrix}$
$=\begin{bmatrix}0&-2&-2\\0&-3&3\end{bmatrix}$
$=(\text{A}-\text{B})'$
Hence proved.
View full question & answer→Question 1192 Marks
If a matrix has 18 element, what are the possible orders it can have? What, if it has 5 elements?
AnswerWe know that a matrix of order m × n has m n elements. Therefore, for finding all possible orders of a matrix with 18 elements, we will find all ordered pairs with products of elements as 18. $\therefore$ all possible ordered pair are all possible ordered pairs are all possible ordered pair are all possible ordered pairs are all possoble ordered pairs are all possible ordered pairs are(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3)
$\therefore $ possible orders are 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3.
if number of elements = 5, then possible orders are 1 × 5, 5 × 1.
View full question & answer→Question 1202 Marks
If a matrix has 5 elements, write all possible orders it can have.
AnswerWe know that if a matrix is of order m×n
,then
it has mn
elements.
If the matrix has 5 elements, then the number of elements will be 1×5 or 5×1, i.e. there will be 2 possible orders of the matrix.
View full question & answer→Question 1212 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{AB})^{\text{T}}=\text{B}^{\text{T}}\text{A}^{\text{T}}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2
$\text{AB}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$
$=\begin{bmatrix}4+2&0+10\\-4+3&0+15\end{bmatrix}=\begin{bmatrix}6&10\\-1&15\end{bmatrix}$
$\therefore\ (\text{AB})^{\text{T}}=\begin{bmatrix}6&-1\\10&15\end{bmatrix}$
Now,
$\text{B}^{\text{T}}\text{A}^{\text{T}}=\begin{bmatrix}4&1\\0&5\end{bmatrix}\begin{bmatrix}1&-1\\2&3\end{bmatrix}$ $=\begin{bmatrix}6&-1\\10&15\end{bmatrix}$ $=(\text{AB})^{\text{T}}$ Hence proved.
View full question & answer→Question 1222 Marks
If A = [aij] is a 2×2 matrix such that aij = i + 2j, write A.
AnswerHere,
aij = i + 2j
$\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}$
$=\begin{bmatrix}1+2(1)&1+2(2)\\2+2(1)&2+2(2)\end{bmatrix}$
$=\begin{bmatrix}3&5\\4&6\end{bmatrix}$
Hence,
$\text{A}=\begin{bmatrix}3&5\\4&6\end{bmatrix}$
View full question & answer→Question 1232 Marks
For any square matrix write whether AAT is symmetric or skew-symmetric.
Answer$\big(\text{AA}^\text{T}\big)^\text{T}=\big(\text{A}^\text{T}\big)^\text{T}\times\text{A}^\text{T}$ $\big\{\text{since, (AB)}^\text{T}=\text{B}^\text{T}\text{A}^\text{T}\big\}$
$\therefore\ \big(\text{AA}^\text{T}\big)^\text{T}=\big(\text{AA}^\text{T}\big)\ \dots(\text{i})$ $\big\{\text{since, }(\text{A}^\text{T})^\text{T}=\text{A}\big\}$
We know that, a square matrix A is symmetric if AT = A
So, from equation (i)
(AAT) is a symmetric matric.
View full question & answer→Question 1242 Marks
Solve the equation for x, y, z and t if:$2\begin{bmatrix}\text{x} & \text{z} \\\text{y} & \text{t} \end{bmatrix} + 3\begin{bmatrix}1 & -1\\0 & 2 \end{bmatrix} = 3\begin{bmatrix}3 & 5 \\4 & 6 \end{bmatrix}.$
AnswerGiven: $2\begin{bmatrix}\text{x} & \text{z} \\\text{y} & \text{t} \end{bmatrix} + 3\begin{bmatrix}1 & -1\\0 & 2 \end{bmatrix} = 3\begin{bmatrix}3 & 5 \\4 & 6 \end{bmatrix}.$
$\Rightarrow \begin{bmatrix}2\text{x}&2\text{z}\\2\text{y}&2\text{t}\end{bmatrix}+\begin{bmatrix}3&-3\\0&6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2\text{x} + 3&2\text{z}-3\\2\text{y} + 0&2\text{t} + 6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix} $
Equation corresponding entries, we have
2x + 3 = 9 ⇒ 2x = 9 - 3 ⇒ 2x = 6 ⇒ x = 3
And 2z - 3 = 15
⇒ 2z = 15 + 3
⇒ 2z = 18 ⇒ z = 9
And 2y = 12 ⇒ y = 6
And 2t + 6 = 18 ⇒ 2t = 18 - 6
⇒ 2t = 12 ⇒ t = 6
$\therefore$ x = 3, y = 6, z = 9, t = 6
View full question & answer→Question 1252 Marks
Find the values of x, y and z from the following equations: $\begin{bmatrix}4 & 3 \\ \text x & 5 \end{bmatrix}=\begin{bmatrix}\text y& \text z\\1 & 5 \end{bmatrix} $
AnswerWe are given that $\begin{bmatrix}4 & 3 \\ \text x & 5 \end{bmatrix}=\begin{bmatrix}\text y& \text z \\1 & 5 \end{bmatrix} $
By defination of equality of matrices $4 = \text y,\ 3 = \text z,\ \text x = 1$
$\therefore\ \text{ x = 1},\ \text{ y = 4},\text{ z = 3 } $ View full question & answer→Question 1262 Marks
Construct a 2 × 3 matrix A = [aij] whose elements aij are give by: $\text{a}_\text{ij}=\frac{(\text{i}+\text{j})^2}{2}$
AnswerHere,
$\text{a}_\text{ij}=\frac{(\text{i}+\text{j})^2}{2}$
$\text{a}_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2,$ $\text{a}_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{13}=\frac{(1+3)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
$\text{a}_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$ and $\text{a}_{23}=\frac{(2+3)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2}$
Required matrix = $\text{A}=\begin{bmatrix}2&\frac{9}{2}&8\\\frac{9}{2}&8&\frac{25}{2}\end{bmatrix}$
View full question & answer→Question 1272 Marks
If $\text A = \begin{bmatrix}\frac{2}{3}&1&\frac{5}{3}\\ \frac{1}{3}&\frac{2}{3}&\frac{4}{3}\\ \frac{7}{3}&2&\frac{2}{3}\end{bmatrix} \text{and}\ \text{B} = \begin{bmatrix}\frac{2}{5}&\frac{3}{5}&1\\ \frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\ \frac{7}{5}&\frac{6}{5}&\frac{2}{3}\end{bmatrix}, $ then compute 3A - 5B.
Answer$3\text{A - 5B}=3\begin{bmatrix}\frac{2}{3}&1&\frac{5}{3}\\ \frac{1}{3}& \frac{2}{3} &\frac{4}{3}\\ \frac{7}{3}&2&\frac{2}{3}\end{bmatrix}-5\begin{bmatrix}\frac{2}{5}&\frac{3}{5}&1\\ \frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\ \frac{7}{5}&\frac{6}{5}&\frac{2}{5}\end{bmatrix}$
$=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}-\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}$
$=\begin{bmatrix}2-2&3-3&5-5\\1-1&2-2&4-4\\7-7&6-6&2-2\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
View full question & answer→Question 1282 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{a}+\text{b})\text{B}=\text{aB}+\text{bB}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2
$(\text{a}+\text{b})\text{B}=\begin{bmatrix}4&-2\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$ $[\because$ a = 4, b = -2$]$
$=\begin{bmatrix}8&0\\2&10\end{bmatrix}$
and
$\text{aB}+\text{bB}=4\text{B}-2\text{B}$ $=\begin{bmatrix}16&0\\4&20\end{bmatrix}-\begin{bmatrix}8&0\\2&10\end{bmatrix}$ $=\begin{bmatrix}8&0\\2&10\end{bmatrix}$
$=(\text{a}+\text{b})\text{B}$
Hence proved.
View full question & answer→Question 1292 Marks
Construct a 2 × 3 matrix A = [a
ij] whose elements a
ij are give by:
aij = i + j
AnswerHere,
aij = i + j
a11 = 1 + 1 = 2, a12 = 1 + 2 = 3, a13 = 1 + 3 = 4
a21 = 2 + 1 = 3, a22 = 2 + 2 = 4 and a23 = 2 + 3 = 5
Required matrix = $\text{A}=\begin{bmatrix}2&3&4\\3&4&5\end{bmatrix}$
View full question & answer→Question 1302 Marks
If F(x) =$\begin{bmatrix}\cos x& -\sin x& 0\\ \sin x& \cos x&0\\0&0&1\end{bmatrix}, $ show that F(x) F(y) = F(x + y).
AnswerGiven: F(x) = $\text{F(x)}=\begin{bmatrix} \cos x &-\sin x&0\\ \sin x&\cos x&0\\0&0&1\end{bmatrix}.....(1)$
Changing x to y in eq. (i), $\text{F(y)}=\begin{bmatrix} \cos y &-\sin y&0\\ \sin y&\cos y&0\\0&0&1\end{bmatrix} $
$\text{L.H.S}=\begin{bmatrix} \cos x &-\sin x&0\\ \sin x&\cos x&0\\0&0&1\end{bmatrix}.\begin{bmatrix} \cos y &-\sin y&0\\ \sin y&\cos y&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}\cos x \cos y- \sin x \sin y+0&- \cos x \sin y- \sin x \cos y+0&0-0+0\ &\\\sin x \cos y+ \cos x \sin y +0&- \sin x \sin y+ \cos x \cos y+0&0+0+0\\0+0+0&0+0+0&0+0+1\end{bmatrix}$
$=\begin{bmatrix} \cos(x+y)&-\sin(x+y)&0\\ \sin(x+y)&\ \cos(x+y)&0\\0&0&1\end{bmatrix}=\text F(x+y)= \text{R.H.S}$ [changing x to to (x + y) in eq. (i)]
View full question & answer→Question 1312 Marks
Show by an example that for $\text{A}\neq0,\ \text{B}\neq0,\ \text{AB}=0.$
AnswerLet $\text{A}=\begin{bmatrix}0&1\\0&2\end{bmatrix}\neq0$ and $\text{B}=\begin{bmatrix}-1&1\\0&0\end{bmatrix}\neq0$
$\therefore\ \text{AB}=\begin{bmatrix}0&1\\0&2\end{bmatrix}\begin{bmatrix}-1&1\\0&0\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 1322 Marks
If $\text{A}=\begin{pmatrix}3&5\\7&9 \end{pmatrix}$ is written as A = P + Q, where as A = P + Q, where P is a symmetric matrix and Qis skew symmetric matrix, then write the matrix P.
Answer$\text{A}=\begin{bmatrix}3&5\\7&9 \end{bmatrix}$
P is symmetric matrix. so,
$\text{P}=\frac{1}{2}(\text{A}+\text{A}^{\text{T}})$
Q is skew symmetric matrix. so, $\text{Q}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})$
$\text{A}^{\text{T}}=\begin{bmatrix}3&7\\5&9 \end{bmatrix}$
$\text{P}=\frac{1}{2}\begin{bmatrix}6&12\\12&18 \end{bmatrix}=\begin{bmatrix}3&6\\6&9 \end{bmatrix}$
View full question & answer→Question 1332 Marks
If a matrix has 24 elements, what are the possible orders it can have? What, if has 13 elements?
AnswerWe know that a matrix of order m × n has mn elements. Therefore, for finding all possible orders of a matrix with 24 elements, we will find all ordered pairs with products of elements as 24. $\therefore$ all possible ordered pairs are
(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)
$\therefore$ possible orders are
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4
if number of elements = 13, then possible orders are 1 × 13, 13 × 1.
View full question & answer→Question 1342 Marks
Write matrix satisfying $\text{A}+\begin{bmatrix}2&3\\-1&4\end{bmatrix}=\begin{bmatrix}3&-6\\-3&8\end{bmatrix}.$
AnswerGiven: $\text{A}+\begin{bmatrix}2&3\\-1&4\end{bmatrix}=\begin{bmatrix}3&-6\\-3&8\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}3&-6\\-3&8\end{bmatrix}-\begin{bmatrix}2&3\\-1&4\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}3-2&-6-3\\-3+1&8-4\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}1&-9\\-2&4\end{bmatrix}$
View full question & answer→Question 1352 Marks
If $\begin{bmatrix}\text{a+b}&2\\5&\text{b} \end{bmatrix}=\begin{bmatrix}6&5\\2&2 \end{bmatrix}$, then find a.
AnswerThe corresponding elements of two equal matrices are equal.
$\Rightarrow\begin{bmatrix}\text{a+b}&2\\5&\text{b} \end{bmatrix}=\begin{bmatrix}6&5\\2&2 \end{bmatrix}$
⇒ a + b = 6 ...(1)
$\therefore$ b = 2
Putting the value of b in eq.(1)
a + b = 6
⇒ a = 6 - 2
$\therefore$ a = 4
View full question & answer→Question 1362 Marks
Construct a 2 × 3 matrix A = [a
ij] whose elements a
ij are give by:
aij = 2i - j
AnswerHere,
a11 = 2(1) -1 = 1, a12 = 2(1) -2 = 0, a13 = 2(1) -3 = -1
a21 = 2(2) -1 = 3, a22 = 2(2) -2 = 2, a23 = 2(2) -3 = 1
Using equation (i)
$\text{A}=\begin{bmatrix}1 &0&-1\\3&2&1\end{bmatrix}$
View full question & answer→Question 1372 Marks
Matrix $\text{A}=\begin{bmatrix}0&2\text{b}&-2\\3&1&3\\3\text{a}&3&-1 \end{bmatrix}$ is given to be symmetric, find values of a and b.
AnswerWe have
$\text{A}=\begin{bmatrix}0&2\text{b}&-2\\3&1&3\\3\text{a}&3&-1 \end{bmatrix}$
$\text{A}'=\begin{bmatrix}0&3&3\text{a}\\2\text{b}&1&3\\-2&3&-1\end{bmatrix}$
We know thet a matrix is symmetric if A = A'.
Thus,
$\begin{bmatrix}0&2\text{b}&-2\\3&1&3\\3\text{a}&3&-1 \end{bmatrix}=\begin{bmatrix}0&3&3\text{a}\\2\text{b}&1&3\\-2&3&-1\end{bmatrix}$
Now,
2b = 3
$\Rightarrow\text{b}=\frac{3}{2}$
Also,
3a = -2
$\Rightarrow\text{a}=\frac{-2}{3}$
Therefore,
$\text{a}=\frac{-2}{3}$ and $\text{b}=\frac{3}{2}$
View full question & answer→Question 1382 Marks
Construct a 3 × 4 matrix, whose elements are given by: $\text{a}_{\text{ij}}=\frac{1}{2} \left|-3{\text{i}+\text{j}}\right| $
Answer$\text{Let A}=\left[\text {a}_{\text {ij}}\ \text {be required}\ 3\times4\ {\text {matrix where}}\ {\text a_{\text {ij} }}={\frac{1}{2}}\left|-3{\text{i+j}}\right|\right] $ $\therefore\ \text a_{11}=\frac{1}{2}\left|{-3+1}\right|=\frac{1}{2}(2)=1, $ $\text a_{12}=\frac{1}{2}\left|{-3+2}\right|=\frac{1}{2}(1)=\frac{1}{2} $ $\text a_{13}=\frac{1}{2}\left|{-3+3}\right|=\frac{1}{2}(0)=0, $
$\text a_{14}=\frac{1}{2}\left|{-3+4}\right|=\frac{1}{2}(1)=\frac {1}{2} $ $\text a_{21}=\frac{1}{2}\left|{-6+1}\right|=\frac{1}{2}(5)=\frac {5}{2}, $ $\text a_{22}=\frac{1}{2}\left|-6+2 \right|=\frac {1}{2}(4)=2 $
$\text a_{23}=\frac{1}{2}\left|-6+3\right|=\frac {1}{2}(3)=\frac {3}{2}, $ $\text a_{24}=\frac{1}{2}\left|-6+4\right|=\frac{1}{2}|-2|=\frac{1}{2}\times2=1 $ $\text a_{31}=\frac{1}{2}\left|-9+1\right|=\frac{1}{2}(8)=4, $
$\text a_{32}=\frac{1}{2}\left|-9+2\right|=\frac{1}{2}(7)=\frac{7}{2} $
$\text a_{33}=\frac{1}{2}\left|-9+3\right|=\frac{1}{2}(6)=3,$ $\text a_{34}=\frac{1}{2}\left|-9+4\right|=\frac {1}{2}(5)=\frac{5}{2} $ $\therefore\ \text{A}=\begin{bmatrix}1 & \frac{1}{2}&0&\frac{1}{2}\\ \frac{5}{2}&2 &\frac{3}{2} & 1\\ 4&\frac{7}{2}&3& \frac{5}{2}\end{bmatrix}$ View full question & answer→Question 1392 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{bA})^{\text{T}}=\text{bA}^{\text{T}}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2
$(\text{bA})^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}\ [\because\ \text{b}=-2]$
$=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}$
and
$\text{A}^{\text{T}}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$ $\therefore\ \text{bA}^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}=(\text{bA})^{\text{T}}$ Hence proved.
View full question & answer→Question 1402 Marks
Show that if A and B are square matrices such that AB = BA, then (A + B)2 = A2 + 2AB + B2.
AnswerSince, A and B are square matrices such that AB = BA $\therefore$ (A + B)2 = (A + B).(A + B)
= A2 + AB + BA + B2
= A2 + AB + AB + B2 $[\because$ AB = BA$]$
= A2 + 2AB + B2
Hence proved.
View full question & answer→Question 1412 Marks
Construct a 2 × 2 matrix, A = $[\text{a}_{\text{ ij}}]$, whose elements are given by: $\text {a}_\text {ij}=\frac {(\text{i}+2 \text{j})^{2}}{2} $
Answer$\text A=\left[\text {a}_{\text {ij}} \right]\text{is}\ 2\times2\ {\text {matrix where}}\ \text {a}_{\text {ij}}=\frac{(\text{i}+2 \text{j})^{2}}{2}$ $\therefore\ \text{a}_{11}=\frac{(1+2)^2}{2}=\frac{9}{2}$, $\text{a}_{12}=\frac{(1+4)^2}{2}=\frac{25}{2}$
$\text{a}_{21}=\frac{(2+2)^2}{2}=\frac{16}{2}=8,$ $\text{a}_{22}=\frac{(2+4)^2}{2}=\frac{36}{2}=18 $
$\therefore\ \ \text{A}=\begin{bmatrix}\frac{9}{2}& \frac{25}{2} \\8 & 18 \end{bmatrix} $
View full question & answer→Question 1422 Marks
If A = [aij] is a square matrix such that aij = i2 - j2, then write whether A is symmetric or skew-symmetric.
AnswerHere,
$\text{a}_{\text{ij}}=\text{i}^2-\text{j}^2,1\leq\text{i}\leq2$ and $1\leq\text{j}\leq2$
$\therefore\ \text{a}_{11}=1^2-1^2=1-1=0,$ $\text{a}_{12}=1^2-2^2=1-4=-3$
$\text{a}_{21}=2^2-1^2=4-1=3$ and $\text{a}_{22}=2^2-2^2=4-4=0$
$\therefore\ \text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\begin{bmatrix}0&-3\\3&0\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}0&3\\-3&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\begin{bmatrix}0&-3\\3&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\text{A}$
Since AT = -A, A is skew-symmetric.
View full question & answer→Question 1432 Marks
If possible, find the sum of the matrices A and B, where $\text{A}=\begin{bmatrix}\sqrt{3}&1\\2&3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}$
AnswerWe have, $\text{A}=\begin{bmatrix}\sqrt{3}&1\\2&3\end{bmatrix}_{2\times2}$ and
$\text{B}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}_{2\times3}$ Here, A and B are of different Orders. Also, we know that the addition of two matrices A and B is possible only if order of both the matrices A and B should be same.
Hence, the sum of matrices A and B is not possible.
View full question & answer→Question 1442 Marks
If $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix},$ then verify that $(\text{A}')'=\text{A}.$
AnswerWe have, $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and
$\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix}$ We have to verify that, $(\text{A}')'=\text{A}$
$\therefore\ \text{A}'=\begin{bmatrix}0&4\\-1&3\\2&-4\end{bmatrix}$
and $(\text{A}')'=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}=\text{A}$ Hence Proved. View full question & answer→Question 1452 Marks
Find the values of x and y if. $\begin{bmatrix}\text{x}+10&\text{y}^2+2\text{y}\\0&-4\end{bmatrix}=\begin{bmatrix}3\text{x}+4&3\\0&\text{y}^2-5\text{y}\end{bmatrix}$
AnswerHere,
x + 10 = 3x + 4 [$\because$ All the corresponding elements of the matrix are equal]
⇒ x - 3x = 4 - 10
⇒ -2x = -6
$\therefore$ x = 3
Also,
y2 + 2y = 3
⇒ y2 + 2y - 3 = 0
⇒ y2 + 3y - y - 3 = 0
⇒ y(y + 3) - 1(y + 3) = 0
⇒ (y + 3)(y - 1) = 0
⇒ y + 3 = 0 or y - 1 = 0
⇒ y = -3 or y = 1
Now
-4 = y2 - 5y
⇒ y2 - 5y + 4 = 0
⇒ y2 - 4y - y + 4 = 0
⇒ y(y - 4) - 1(y - 4) = 0
⇒ (y - 4)(y - 1) = 0
⇒ y - 4 = 0 or y - 1 = 0
⇒ y = 4 or y = 1
Since y2 + 2y = 3 and y2 - 5y = -4 must hold good simultaneously, we take the common solution of these two equations.
Thus,
y = 1, x = 3 and y = 1
View full question & answer→Question 1462 Marks
Construct a 3 × 2 matrix whose elements are given by $\text{a}_{\text{ij}}=\text{e}^{\text{i.x}}=\sin\text{jx}.$
AnswerWe have, $\text{A}=[\text{a}_{\text{ij}}]_{3\times2},$ Such that, $\text{a}_{\text{ij}}=\text{e}^{\text{i.x}}=\sin\text{jx};$ where
$1\leq\text{i}\leq3$ and $1\leq\text{j}\leq2,\ \text{i, j}\in\text{N}$ | $\therefore\ \text{a}_{11}=\text{e}^\text{x}\sin\text{x}$ | $\text{a}_{12}=\text{e}^{\text{x}}\sin2\text{x}$ |
| $\text{a}_{21}=\text{e}^{2\text{x}}\sin\text{x}$ | $\text{a}_{22}=\text{e}^{2\text{x}}\sin2\text{x}$ |
| $\text{a}_{31}=\text{e}^{3\text{x}}\sin\text{x}$ | $\text{a}_{32}=\text{e}^{3\text{x}}\sin2\text{x}$ |
$\therefore\ \text{A}=\begin{bmatrix}\text{e}^{\text{x}}\sin\text{x}&\text{e}^{\text{x}}\sin2\text{x}\\\text{e}^{2\text{x}}\sin\text{x}&\text{e}^{2\text{x}}\sin2\text{x}\\\text{e}^{3\text{x}}\sin\text{x}&\text{e}^{3\text{x}}\sin2\text{x}\end{bmatrix}$
View full question & answer→Question 1472 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{A}^{\text{T}})^{\text{T}}=\text{A}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2
$\text{A}^{\text{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}^{\text{T}}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$
Now, $(\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}^{\text{T}}$
$=\text{A}$
Hence proved.
View full question & answer→Question 1482 Marks
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ then prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{yb}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}.$
Answer$\text{PQ}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$ $=\begin{bmatrix}\text{xa}&0&0\\0&\text{yb}&0\\0&0&\text{zc}\end{bmatrix}\ ....(\text{i})$
and $\text{QP}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$=\begin{bmatrix}\text{ax}&0&0\\0&\text{by}&0\\0&0&\text{cz}\end{bmatrix}\ ....(\text{ii})$
Thus, we see that, PQ = QP [using Eq. (i) and (ii)]
Hence proved.
View full question & answer→Question 1492 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $\text{a}(\text{C}-\text{A})=\text{aC}-\text{aA}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2
$(\text{C}-\text{A})=\begin{bmatrix}2-1&0-2\\1+1&-2-3\end{bmatrix}=\begin{bmatrix}1&-2\\2&-5\end{bmatrix}$
and
$\text{a}(\text{C}-\text{A})=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}\ [\because\ \text{a}=4]$ Also,
$\text{aC}-\text{aA}=\begin{bmatrix}8&0\\4&-8\end{bmatrix}-\begin{bmatrix}4&8\\-4&12\end{bmatrix}$ $=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}$
$=\text{a}(\text{C}-\text{A})$
Hence proved.
View full question & answer→Question 1502 Marks
For the matrix $\text{A}=\begin{bmatrix}1&5\\6&7\end{bmatrix}$, verify that
- (A + A') is a symmetric matrix.
- (A - A') is a skew symmentric matrix.
Answer$\text{A}'=\begin{bmatrix}1&6\\5&7\end{bmatrix}$ - $\text{A}+\text{A}'=\begin{bmatrix}1&5\\6&7\end{bmatrix}+\begin{bmatrix}1&6\\5&7\end{bmatrix}=\begin{bmatrix}2&11\\11&14\end{bmatrix}$
$\therefore\ (\text{A}+\text{A})'=\begin{bmatrix}2&11\\11&14\end{bmatrix} = \text{A}+\text{A}'$
Hence, $(\text{A}+\text{A}')$ is a symmentric matrix.
- $\text{A} - \text{A}'=\begin{bmatrix}1&5\\6&7\end{bmatrix}-\begin{bmatrix}1&6\\5&7\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$
$(\text{A} - \text{A}')'=\begin{bmatrix}0&1\\-1&0\end{bmatrix}=-\begin{bmatrix}0&-1\\1&0\end{bmatrix}=-(\text{A} -\text{A}')$
Hence, $(\text{A} - \text{A}')$ is a skew - symmentric matrix.
View full question & answer→