Question 12 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 22 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 32 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 42 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 52 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 62 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 72 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 82 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 92 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 102 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 112 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 122 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 132 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 142 Marks
If $\text{A}=\begin{bmatrix}1&2\\3&4 \end{bmatrix}$, find A+AT.
AnswerGiven: $\text{A}=\begin{bmatrix}1&2\\3&4 \end{bmatrix}$
$\text{A}^{\text{T}}=\begin{bmatrix}1&3\\2&4 \end{bmatrix}$
$\text{A+A}^{\text{T}}=\begin{bmatrix}1&2\\3&4 \end{bmatrix}+\begin{bmatrix}1&3\\2&4 \end{bmatrix}$
$\Rightarrow\text{A+A}^{\text{T}}=\begin{bmatrix}1+1&2+3\\3+2&4+4 \end{bmatrix}$
$\Rightarrow\text{A+A}^{\text{T}}=\begin{bmatrix}2&5\\5&8 \end{bmatrix}$
View full question & answer→Question 152 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 162 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 172 Marks
If $\begin{bmatrix}\text{x}&\text{x}-\text{y}\\2\text{x+y}&7 \end{bmatrix}=\begin{bmatrix}3&1\\8&7 \end{bmatrix},$ then find the value of y.
AnswerWe have, $\begin{bmatrix}\text{x}&\text{x}-\text{y}\\2\text{x+y}&7 \end{bmatrix}=\begin{bmatrix}3&1\\8&7 \end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ x = 3
x - y = 1 ...(1)
Putting the value of x in eq. (1)
3 - y = 1
⇒ 3 - 1 = y
$\therefore$ y = 2
View full question & answer→Question 182 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 192 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 202 Marks
Find the value of x from the following: $\begin{bmatrix}2\text{x}-\text{y}&5\\3&\text{y} \end{bmatrix}=\begin{bmatrix}6&5\\3&-2 \end{bmatrix}$
AnswerThe corresponding elements of two equal matrices are equai.
Given: $\begin{bmatrix}2\text{x}-\text{y}&5\\3&\text{y} \end{bmatrix}=\begin{bmatrix}6&5\\3&-2 \end{bmatrix}$
2x - y = 6 $\dots(1)$
y = - 2
Putting the value of y in eq.(1)
2x - (-2) = 6
⇒ 2x + 2 = 6
⇒ 2x = 6 - 2
⇒ 2x = 4
$\Rightarrow\text{x}=\frac{4}{2}=2$
$\therefore$ x = 2
View full question & answer→Question 212 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 222 Marks
Construct a 2 × 2 matrix A = [aij] whose elements aij are give by: $\frac{(\text{i}+\text{j})^2}{2}$
AnswerHere,
$\text{a}_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2,$ $\text{a}_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
$\text{a}_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
So, the required matrix is $\begin{bmatrix}2\frac{9}{2}\\\frac{9}{2}8\end{bmatrix}.$
View full question & answer→Question 232 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 242 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 252 Marks
For what value of x, is the matrix $\text{A}=\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}$ a skew-symmetric matrix?
AnswerSince, A is a skew symmetric matrix.
$\therefore$ aT = -A
$\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}^{\text{T}}=-\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}0&-1&\text{x}\\1&0&-3\\-2&3&0\\ \end{bmatrix}=\begin{bmatrix}0&-1&2\\1&0&-3\\-\text{x}&3&0 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
⇒ x = 2
Hence, the value of x is 2.
View full question & answer→Question 262 Marks
If $\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6 \end{bmatrix}=\begin{bmatrix}2\text{a}+2&\text{b}+2\\8&\text{a}-8\text{b} \end{bmatrix},$ write the value of a - 2b.
Answer$\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6 \end{bmatrix}=\begin{bmatrix}2\text{a}+2&\text{b}+2\\8&\text{a}-8\text{b} \end{bmatrix}$
Form the above equation,
$\therefore$ a + 4 = 2a + 2
⇒ a = 2
$\therefore$ 3b = b + 2
⇒ 2b = 2
⇒ b = 1
a - 2b
= 2 - 2 × 1
= 0
View full question & answer→Question 272 Marks
Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+2+4&\text{z}-3+5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+6&\text{z}+2\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\therefore\ \text{x}+\text{y}=4\ \dots(1)$
Also,
$\text{y}+6=9$
$\Rightarrow\text{y}=3$
$\text{z}+2=12$
$\Rightarrow\text{z}=10$
Putting the value of y in eq. (1), we get
$\text{x}+3=4$
$\Rightarrow\text{x}=4-3$
$\Rightarrow\text{x}=1$
$\therefore\ \text{x}=1,\text{ y}=3$ and $\text{z}=10$
View full question & answer→Question 282 Marks
If $\text{A}=\begin{bmatrix}2&1&4\\4&1&5 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}3&-1\\2&2\\1&3\end{bmatrix}.$ Write the order of AB and BA.
AnswerOrder of A = 2×3
Order of B = 3×2
So,
A2×3 × B3×2 has order = 2×2
B3×2 × A2×3 has order = 3×3
Hence,
Order of AB = 2×2
Order of BA = 3×3
View full question & answer→Question 292 Marks
If A and B are square matrices of the same order, explain, why in general:
(A + B)2 ≠ A2 + 2AB + B2
AnswerLHS = (A + B)2
= (A + B)(A + B)
= A(A + B) + B(A + B)
= A2 + AB + BA +B2
We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
(A + B)2 ≠ A2 + 2AB + B2
View full question & answer→Question 302 Marks
If A is a skew-symmetric matrix and n is an even natural number, write whether An is symmetric or skew-symmetric or neither of these two.
AnswerIf A is a skew-symmetric matrix, then AT = -A.
(An)T = (AT)n [For all n ∈ N]
⇒ (An)T = (-A)n [$\because$ AT = -A]
⇒ (An)T = (-1)n An
⇒ (An)T = An, if n is even or -An, if n is odd.
Hence, An is a symmetric when n is an even natural number.
View full question & answer→Question 312 Marks
Find x, y, a and b if $\begin{bmatrix}3\text{x}+4\text{y}&2&\text{x}-2\text{y}\\\text{a}+\text{b}&2\text{a}-\text{b}&^-1\end{bmatrix}=\begin{bmatrix}2&2&4\\5&-5&-1\end{bmatrix}$
AnswerSince the corresponding elements of two equal matrices are equal,
$\begin{bmatrix}3\text{x}+4\text{y}&2&\text{x}-2\text{y}\\\text{a}+\text{b}&2\text{a}-\text{b}&^-1\end{bmatrix}=\begin{bmatrix}2&2&4\\5&-5&-1\end{bmatrix}$
⇒ 3x + 4y = 2 ...(1)
⇒ x - 2y = 4
⇒ x = 4 + 2y ...(2)
Putting the value of x in eq. (1), we get
3(4 + 2y) + 4y = 2
⇒ 12 + 6y + 4y = 2
⇒ 12 + 10y = 2
⇒ 10y = 2 - 12
⇒ 10y = -10
$\Rightarrow\text{y}=\frac{-10}{10}=-1$
Putting the value of y in eq. (2), we get
x = 4 + 2(-1)
⇒ x = 4 - 2 = 2
a + b = 5
⇒ a = 5 - b ...(3)
⇒ 2a - b = -5 ...(4)
Putting the value of a in eq. (4), we get
2(5 - b) - b = -5
⇒ 10 - 2b - b = -5
⇒ 10 - 3b = -5
⇒ -3b = -15
$\Rightarrow\text{b}=\frac{-15}{-3}$
⇒ b = 5
Putting the value of b in eq. (3), we get
a = 5 - 5
⇒ a = 0
$\therefore$ x = 2, y = -1, a = 0 and b = 5
View full question & answer→Question 322 Marks
Construct a 2 × 2 matrix A = [aij] whose elements aij are given by:
$\text{a}_\text{ij}=\frac{|2\text{i}-3\text{j}|}{2}$
AnswerHere,
$\text{a}_{11}=\frac{|2(1)-3(1)|}{2}=\frac{1}{2},$ $\text{a}_{12}=\frac{|2(1)-3(2)|}{2}=2$
$\text{a}_{21}=\frac{|2(2)-3(1)|}{2}=\frac{1}{2},$ $\text{a}_{22}=\frac{|2(2)-3(2)|}{2}=1$
Using equation (i)
$\text{A}=\begin{bmatrix}\frac{1}{2}&2\\\frac{1}{2}&1\end{bmatrix}$
View full question & answer→Question 332 Marks
If $\text{A}=\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}$ is identity matrix, then write the value of a.
AnswerHere,
$\text{A}=\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\text{I}$
$\Rightarrow\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\begin{bmatrix}1&0\\0&1 \end{bmatrix}$
The corresponding elements of equal matrices are equal.
$\therefore \cos\text{a}=1$
$\Rightarrow\text{a}=0^{\circ}$
View full question & answer→Question 342 Marks
If $\text{X}-\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}$ and $\text{X}+\text{Y}=\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix},$ find X and Y.
AnswerHere,
$\text{X}-\text{Y}+\text{X}+\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}+\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}1+3&1+5&1+1\\1-1&1+1&0+4\\1+11&0+8&0+0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$
Now,
$(\text{X}-\text{Y})-(\text{X}+\text{Y})=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}-\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow\text{X}-\text{Y}-\text{X}-\text{Y}=\begin{bmatrix}1-3&1-5&1-1\\1+1&1-1&0-4\\1-11&0-8&0-0\end{bmatrix}$
$\Rightarrow-2\text{Y}=\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=-\frac{1}{2}\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
View full question & answer→Question 352 Marks
If A and B are square matrices of the same order, explain, why in general:
(A + B)(A - B) ≠ A2 - B2.
AnswerLHS = (A + B)(A - B) = A(A - B) + B(A - B) = A
2 - AB + BA - B
2 We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
(A + B)(A - B) ≠ A2 - B2
View full question & answer→Question 362 Marks
Give example of matrices:
A and B such that AB ≠ BA
AnswerLet $ \text{A}=\begin{bmatrix}\text{a}&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}$
$\text{A}\text{B}=\begin{bmatrix}\text{a}&0\\0&0\end{bmatrix}\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}$
$=\begin{bmatrix}0+0&\text{a}\text{b}+0\\0+0&0+0\end{bmatrix}$
$\text{AB}=\begin{bmatrix}0&\text{a}\text{b}\\0&0\end{bmatrix}$
$\text{BA}=\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}\begin{bmatrix}0&\text{a}\\0&0\end{bmatrix}$
$ =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$ \text{BA}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
From equation (i) and (ii)
AB ≠ BA
When $ \text{A}=\begin{bmatrix}\text{a}&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&\text{b}\\0&0\end{bmatrix}$
View full question & answer→Question 372 Marks
Compute the products AB and BA whichever exists the following cases:
$\text{A}=\begin{bmatrix}1&-1&2&3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}$
Answer$\text{AB}=\begin{bmatrix}1&-1&2&3\end{bmatrix}\begin{bmatrix}0\\1\\3\\2\end{bmatrix}$
$\Rightarrow\text{AB}=[0+(1)+6+6]$
$\Rightarrow\text{AB}=[11]$
Also,
$\text{BA}=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}\begin{bmatrix}1&-1&2&3\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&0&0&0\\1&-1&2&3\\3&-3&6&9\\2&-2&4&6\end{bmatrix}$
View full question & answer→Question 382 Marks
If $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix},$ then find the values of x, y, z and w.
Answer$\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix}$
The corresponding entries of the two equal matrices are equal,
⇒ xy = 8 ...(1),
w = 4 ...(2),
z + 6 = 0 ...(3),
And x + y = 6 ...(4)
From equation (2) and equation (3) we get z = -6 and w = 4.
From equation (4) we have,
x + y = 6
⇒ x = 6 - y,
Subsituting value of x in equation (1) we get,
⇒ (6 - y)y = 8
⇒ y2 - 6y + 8 = 0
⇒ (y - 2)(y - 4) = 0,
⇒ y = 2, 4
Subsituting the value of y in equation (1) we get,
⇒ x = 4, 2
Therefore, value of x, y, z, w are 2, 4, -6, 4 or 4, 2, -6, 4.
View full question & answer→Question 392 Marks
If A is a square matrix such that A2 = A, then write the value of 7A − (I + A)3, where I is the identity matrix.
AnswerA2 = A
A3 = A2 = A
7A - (I + A)3
= 7A - (I3 + A3 + 3A2I + 3AI2)
= 7A - (I + A + 3A + 3A)
= 7A - (I + 7A)
= -I
View full question & answer→Question 402 Marks
If B is a symmetric matrix, write whether the matrix AB AT is symmetric or skew-symmetric.
AnswerIf B is a skew-symmetric matrix, then BT - B.
$(\text{ABA}^\text{T})^\text{T}=(\text{A}^\text{T})^\text{T}\text{B}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{ABC})^\text{T}=\text{C}^\text{T}\text{B}^\text{T}\text{A}^\text{T}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=\text{AB}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{A}^\text{T})^\text{T}=\text{A}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=-\text{ABA}^\text{T}$ $\big[\because\ \text{B}^\text{T}=\text{B}\big]$
Hence, ABAT is a skew-symmetric matrix.
View full question & answer→Question 412 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:
$2\text{A}-3\text{B}$
AnswerGiven, $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{ B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{ C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}$
$2\text{A}-3\text{B}$
$=2\begin{bmatrix}2&4\\3&2\end{bmatrix}-3\begin{bmatrix}1&3\\-2&5 \end{bmatrix}$
$=\begin{bmatrix}4&8\\6&4\end{bmatrix}-\begin{bmatrix}3&9\\-6&15 \end{bmatrix}$
$=\begin{bmatrix}4-3&8-9\\6+6&4-15\end{bmatrix}$
$=\begin{bmatrix}1&-1\\12&-11\end{bmatrix}$
Hence,
$2\text{A}-3\text{B}=\begin{bmatrix}1&-1\\12&-11\end{bmatrix}$
View full question & answer→Question 422 Marks
If $\begin{bmatrix}\text{x}+\text{y}\\\text{x}-\text{y} \end{bmatrix}=\begin{bmatrix}2&1\\4&3 \end{bmatrix}\begin{bmatrix}1\\-2\end{bmatrix},$ then write the value of (x, y).
AnswerAfter doing the matrix multiplication we get,
$\begin{bmatrix}\text{x}+\text{y}\\\text{x}-\text{y} \end{bmatrix}=\begin{bmatrix}0\\-2\end{bmatrix}$
As corresponding entries of two equal matrices are equal so,
x + y = 0,
x - y = -2
Solving simultanecus linear equation gives the value of x = -1 and y = 1
or (x, y) = (-1, 1).
View full question & answer→Question 432 Marks
For a 2×2 matrix A = [aij] whose elements are given by $\text{a}_{\text{ij}}=\frac{\text{i}}{\text{j}}$
, write the value of a12.
AnswerGiven that a 2×2 matrix A = [aij] whose elements are given by $\text{a}_{\text{ij}}=\frac{\text{i}}{\text{j}}.$
We need to find the value of a12.
Thus, $\text{a}_{12}=\frac{1}{2}.$
View full question & answer→Question 442 Marks
Let $\text{A}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\ \text{B}=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}.$ Verify that AB = AC though B ≠ C, A ≠ O.
AnswerHere,
$\text{A}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\ \text{B}=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}$
Now,
$\text{A}\text{B}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$
$\Rightarrow\text{A}\text{B}=\begin{bmatrix}3+5-2&1+2+4\\9+15-6&3+6+12\end{bmatrix}$
$\Rightarrow\text{A}\text{B}=\begin{bmatrix}6&7\\18&21\end{bmatrix}$
$\text{AC}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}$
$\Rightarrow\text{AC}=\begin{bmatrix}4-3+5&2+5+0\\12-9+15&6+15+0\end{bmatrix}$
$\Rightarrow\text{AC}=\begin{bmatrix}6&7\\18&21\end{bmatrix}$
So, AB = AC though B ≠ C, A ≠ O.
View full question & answer→Question 452 Marks
If $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix},$ find A + AT.
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}2&5\\3&7\end{bmatrix}$
Now,
$\text{A}+\text{A}^{\text{T}}=\begin{bmatrix}2&3\\5 &7\end{bmatrix}+\begin{bmatrix}2&5\\3&7\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}2+2&3+5\\5+3&7+7\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}4&8\\8&14\end{bmatrix}$
View full question & answer→Question 462 Marks
Construct a 3 × 4 matrix A = [aij] whose element aij are given by:
$\text{a}_\text{ij}=\frac{1}{2}|-3\text{i}+\text{j}|$
AnswerHere, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$\text{a}_{11}|-3(1)+1|=\frac{1}{2}|-2|=1,$ $\text{a}_{12}=\frac{1}{2}|-3(1)+2|=\frac{1}{2}|-1|=\frac{1}{2},$
$\text{a}_{13}=\frac{1}{2}|-3(1)+3|=\frac{1}{2}|0|=\frac{0}{2}=0,$ $\text{a}_{14}=\frac{1}{2}|-3(1)+4|=\frac{1}{4}|1|=\frac{1}{2}$
$\text{a}_{21}=\frac{1}{2}|-3(2)+1|=\frac{1}{2}|-5|=\frac{5}{2},$ $\text{a}_{22}=\frac{1}{2}|-3(2)+2|=\frac{1}{2}|-4|=2,$
$\text{a}_{23}=\frac{1}{2}|-3(2)+3|=\frac{1}{2}|-3|=\frac{3}{2},$ $\text{a}_{24}=\frac{1}{2}|-3(2)+4|=\frac{1}{2}|-2|=1$
$\text{a}_{31}=\frac{1}{2}|-3(3)+1|=\frac{1}{2}|-8|=4,$ $\text{a}_{32}=\frac{1}{2}|-3(3)+2|=\frac{1}{2}|-7|=\frac{7}{2},$
$\text{a}_{33}=\frac{1}{2}|-3(3)+3|=\frac{1}{2}|-6|=3$ and $\text{a}_{34}=\frac{1}{2}|-3(3)+4|=\frac{1}{2}|-5|=\frac{5}{2}$
So, the required matrix is $\begin{bmatrix}1&\frac{1}{2}&0&\frac{1}{2}\\\frac{5}{2}&2&\frac{3}{2}&1\\4&\frac{7}{2}&3&\frac{5}{2}\end{bmatrix}.$
View full question & answer→Question 472 Marks
Construct a 2 × 2 matrix A = [aij] whose elements aij are given by:
$\text{a}_\text{ij}=\text{e}^{2\text{ix}}\sin(\text{xj})$
AnswerHere,
$\text{a}_{11}=\text{e}^{2\times1\times\text{x}}\sin(\text{x}\times1)=\text{e}^{2\text{x}}\sin(\text{x}),$ $\text{a}_{12}=\text{e}^{2\times1\times\text{x}}\sin(\text{x}\times2)=\text{e}^{2\text{x}}\sin(2\text{x})$
$\text{a}_{21}=\text{e}^{2\times2\times\text{x}}\sin(\text{x}\times1)=\text{e}^{4\text{x}}\sin(\text{x}),$ $\text{a}_{22}=\text{e}^{2\times2\times\text{x}}\sin(\text{x}\times2)=\text{e}^{4\text{x}}\sin(2\text{x})$
So, the required matrix is $\begin{bmatrix}\text{e}^{2\text{x}}\sin(\text{x})&\text{e}^{2\text{x}}\sin(2\text{x})\\\text{e}^{4\text{x}}\sin(\text{x})&\text{e}^{4\text{x}}\sin(2\text{x})\end{bmatrix}.$
View full question & answer→Question 482 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:
$3\text{A}-\text{C}$
AnswerGiven, $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{ B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{ C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}$
$3\text{A}-\text{C}$
$\Rightarrow3\text{A}-\text{C}=3\begin{bmatrix}2&4\\3&2\end{bmatrix}-\begin{bmatrix}-2&5\\3&4 \end{bmatrix}$
$\Rightarrow3\text{A}-\text{C}=\begin{bmatrix}6&12\\9&6\end{bmatrix}-\begin{bmatrix}-2&5\\3&4 \end{bmatrix}$
$\Rightarrow3\text{A}-\text{C}=\begin{bmatrix}6+2&12-5\\9-3&6-4\end{bmatrix}$
$\Rightarrow3\text{A}-\text{C}=\begin{bmatrix}8&7\\6&2\end{bmatrix}$
View full question & answer→Question 492 Marks
For what valuse of x and y are the following matrices equal?
$\text{A}=\begin{bmatrix}2\text{x}+1&2\text{y}\\0&\text{y}^2-5\text{y}\end{bmatrix}\text{B}=\begin{bmatrix}\text{x}+3&\text{y}^2+2\\0&-6\end{bmatrix}$
AnswerGiven,
$\text{A}=\begin{bmatrix}2\text{x}+1&2\text{y}\\0&\text{y}^2-5\text{y}\end{bmatrix}\text{B}=\begin{bmatrix}\text{x}+3&\text{y}^2+2\\0&-6\end{bmatrix}$
Since equal matrics has all corresponding elements equal,
So,
2x + 1 = x + 3 ...(i)
2y = y2 + 2 ...(ii)
y2 - 5y = -6 ...(iii)
Solving equation (i),
2x + 1 = x + 3
2x - x = 3 - 1
x = 2
Solving equation (ii),
2y = y2 + 2
y2 - 2y + 2 = 0
D = b2 - 4ac
= (-2)2 - 4
= 4 - 8
= -2
So, There is no real value of y from equation (ii),
Solving equation (iii),
y2 - 5y = -6
y2 - 5y + 6 = 0
y2 - 3y - 2y + 6 = 0
y(y - 3) - 2(y - 3) = 0
(y - 3)(y - 2) = 0
y = 3 or y = 2
From solution of equation (i), (ii) and (iii), we can say that A and B can not be equal for any value of y.
View full question & answer→Question 502 Marks
If A and B are square matrices of the same order such that AB = BA, then show that (A + B)2 = A2 + 2AB + B2.
AnswerGiven,
A and B two square matrices of same order such that AB = BA
To prove: (A + B)2 = A2 + 2AB + B2
Now, solving LHS gives,
(A + B)2 = (A + B)(A + B)
= A(A + B) + B(A + B) [by dist, of matrix multiplication over addition]
= A2 + AB + BA + B2 [by dist, of matrix multiplication over addition]
= A2 + 2AB + B2 [As, AB = BA]
= RHS
Hence proved.
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