Question 1012 Marks
If $A$ and $B$ are square matrices of the same order, explain, why in general:
$(A − B)^2 \neq A^2 − 2AB + B^2$
Answer$(A - B)^2 - (A - B)(A - B)$
$= A(A - B) - B(A - B) \{$using distributive property$\}$
$= A \times A - AB - BA + B \times B$
$= A^2 - AB - BA + B^2 \neq A^2 - 2AB + B^2$
Since, in general matrix multiplication is not commutative $(AB \neq BA),$
So, $(A - B)^2 \neq A^2 - 2AB + B^2$
View full question & answer→Question 1022 Marks
Compute the products AB and BA whichever exists the following cases:
$\text{A}=\begin{bmatrix}1&-2\\2&3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}$
Answer$\text{AB}=\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}1-4&2-6&3-2\\2+6&4+9&6+3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-3&-4&1\\8&13&9\end{bmatrix}$
Since the number of columns in B is greater then the number of rows in A, BA does not exists.
View full question & answer→Question 1032 Marks
If $\begin{bmatrix}1&-1\\-1&1\end{bmatrix},$ satisfies the matrix equation $A^2 = kA,$ write the value of $k.$
AnswerGiven,
$\text{A}=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$
and
$\text{A}^2=\text{kA}$
$\Rightarrow\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}=\text{k}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+1&-1-1\\-1-1&1+1\end{bmatrix}=\begin{bmatrix}\text{k}&-\text{k}\\-1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&-2\\-2&2\end{bmatrix}=\begin{bmatrix}\text{k}&-\text{k}\\-\text{k}&\text{k}\end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so $k = 2$
View full question & answer→Question 1042 Marks
If A is an m × n matrix and B is n × p matrix does AB exist? If yes, write its order.
AnswerGiven: Order of A = m × n
Order of B = n × p
Since the number of columns in A are equal to the number of rows in B, i.e. n, AB exists.
Order of AB = Number of rows in A × Number of columns in B = m × p
View full question & answer→Question 1052 Marks
If $A$ and $B$ are symmetric matrices of the same order, write whether $AB − BA$ is symmetric or skew-symmetric or neither of the two.
AnswerSince $A$ and $B$ are symmetric matrices, $A^T = A$ and $B^T = B.$
Here,
$(AB - BA)^T = (AB)^T - (BA)^T$
$\Rightarrow (AB - BA)^T = B^TA^{T }- A^TB^T [\because (AB)^T = B^TA^T]$
$\Rightarrow (AB - BA)^T = BA - AB [\because B^T = B$ and $A^{T }= A]$
$\Rightarrow (AB - BA)^T = -(AB - BA)$
Therefore, $AB - BA$ is skew - symmetric.
View full question & answer→Question 1062 Marks
If $A$ is a square matrix such that $A^2 = A,$ then write the value of $7A − (I + A)^3,$ where $I$ is the identity matrix.
Answer$A^{2 }= A$
$A^{3 }= A^{2 }= A$
$7A - (I + A)^3$
$= 7A - (I^{3 }+ A^{3 }+ 3A^2I + 3AI^2)$
$= 7A - (I + A + 3A + 3A)$
$= 7A - (I + 7A)$
$= -I$
View full question & answer→Question 1072 Marks
Write a 2×2 matrix which is both symmetric and skew-symmetric.
AnswerA matrix which is both symmetric and skew-symmetric is a null matrix.
Hence, the required matrix is $\begin{bmatrix}0&0\\0&0 \end{bmatrix}$.
View full question & answer→Question 1082 Marks
If $\begin{bmatrix}1&2\\3&4 \end{bmatrix}\begin{bmatrix}3&1\\2&5 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix},$ then write the value of k.
AnswerGiven: $\begin{bmatrix}1&2\\3&4 \end{bmatrix}\begin{bmatrix}3&1\\2&5 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}3+4&1+10\\9+8&3+20 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}7&11\\17&23 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ k = 17
View full question & answer→Question 1092 Marks
Write a square matrix which is both symmetric as well as skew$-$symmetric.
AnswerLet $\text{A}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
$\text{A}^{\text{T}{}}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
Since $A^T = A, A$ is a symmetric matrix.
Now,
$-\text{A}=-\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
$\Rightarrow-\text{A}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
Since $A^T = -A, A$ is a skew$-$symmetric matrix.
Thus, $\text{A}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$is an exampal of a matrix that is both symmetric and skew$-$symmetric.
View full question & answer→Question 1102 Marks
If $\text{A}=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix},$ find $A^3.$
AnswerGiven,
$\text{A}=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$=\begin{bmatrix}1+0+0&0+0+0&0+0+0\\0+0+0&0+1+0&0+0+0\\0+0+0&0+0+0&0+1+0\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A}^3=\text{A}^2\times\text{A}$
$=\begin{bmatrix}1&0&0\\1&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$=\begin{bmatrix}-1+0+0&0+0+0&0+0+0\\0+0+0&0-1+0&0+0+0\\0+0+0&0+0+0&0+0-1\end{bmatrix}$
$=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$\text{A}^3=\text{A}$
Hence, $\text{A}^3=\text{A}$
View full question & answer→Question 1112 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 1122 Marks
Compute the products AB and BA whichever exists the following cases:
$[\text{a},\text{b}]\begin{bmatrix}\text{c}\\\text{d} \end{bmatrix}+\big[\text{a},\text{b},\text{c},\text{d}\big]\begin{bmatrix}\text{a}\\\text{b}\\\text{c}\\\text{d}\end{bmatrix}$
Answer$[\text{a},\text{b}]\begin{bmatrix}\text{c}\\\text{d} \end{bmatrix}+\big[\text{a},\text{b},\text{c},\text{d}\big]\begin{bmatrix}\text{a}\\\text{b}\\\text{c}\\\text{d}\end{bmatrix}$
$\Rightarrow\big[\text{ac}+\text{bd}\big]+\big[\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2\big]$
$\big[\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2+\text{ac}+\text{bd}\big]$
View full question & answer→Question 1132 Marks
There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommend daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrix. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families. What awareness can you create among people about the planned diet from this question?
AnswerAccording to the question, Let X be the matrix showing number of family members in family A and B. $\text{X}=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}$And, Y be a matrix showing the recommend daily amount of calories.
$\text{Y}=\begin{bmatrix}2400\\1900\\1800\end{bmatrix}$And, Z be a matrix showing the recommend daily amount of proteins.
$\text{Z}=\begin{bmatrix}45\\55\\33\end{bmatrix}$Now, the total requirement of calories of the two families will be shown by XY.
$\text{XY}=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}\begin{bmatrix}2400\\1900\\1800\end{bmatrix}$ $=\begin{bmatrix}9600+11400+3600\\4800+3800+7200\end{bmatrix}$ $=\begin{bmatrix}24600\\15800\end{bmatrix}$ Also, the total requirement of proteins of the two families will be shown by XZ. $\text{XZ}=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}\begin{bmatrix}45\\55\\33\end{bmatrix}$ $=\begin{bmatrix}180+330+66\\90+110+132\end{bmatrix}$ $=\begin{bmatrix}576\\332\end{bmatrix}$Hence, the total requirement of calories and proteins for each of the two families is shown as:
| |
Calories |
Proteins |
| Family A: |
24600 |
576 |
| Family B: |
15800 |
332 |
View full question & answer→Question 1142 Marks
If $\begin{bmatrix}\text{a }-\text{b}&2\text{a}+\text{c}\\2\text{a}-\text{b}&3\text{c}+\text{d} \end{bmatrix}=\begin{bmatrix}-1&5\\0&13 \end{bmatrix},$ fine the value of b.
Answer$\begin{bmatrix}\text{a }-\text{b}&2\text{a}+\text{c}\\2\text{a}-\text{b}&3\text{c}+\text{d} \end{bmatrix}=\begin{bmatrix}-1&5\\0&13 \end{bmatrix}$
From the above matrices,
a - b = -1 ...(1)
2a - b = 0 ...(2)
Solving (1) and (2),
a = 1, b = 2
$\therefore$ b = 2
View full question & answer→Question 1152 Marks
If $\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+2\begin{bmatrix}\text{x}&0\\1&-2\end{bmatrix}=\text{I},$ where I is 2×2 unit matrix. Find x and y.
AnswerGiven: $\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+2\begin{bmatrix}\text{x}&0\\1&-1\end{bmatrix}=\text{I}$
$\Rightarrow\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+\begin{bmatrix}2\text{x}&0\\2&-4\end{bmatrix}=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+2\text{x}&0+0\\\text{y}+2&5-4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+2\text{x}&0\\\text{y}+2&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\therefore1+2\text{x}=1$ and $\text{y}+2=0$
$\Rightarrow2\text{x}=1-1$ and $\text{y}=-2$
$\Rightarrow2\text{x}=0$
$\Rightarrow\text{x}=\frac{0}{2}=0$
View full question & answer→Question 1162 Marks
If A is a skew-symmetric and n ∈ N such that $(\text{A}^\text{n})^\text{T}=\lambda\text{A}^\text{n},$ write the value of $\lambda.$
AnswerGiven,
A is skew symmetric matrix
$\Rightarrow\text{A}^\text{T} = -\text{A}$
And
$(\text{A}^\text{n})^\text{T}=\lambda\text{A}^\text{n}$
$\Rightarrow(\text{A}^\text{T})^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow(-\text{A})^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow(-1)^\text{n}\text{A}^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow\lambda=(-1)^\text{n}$
View full question & answer→Question 1172 Marks
Find x, y, a and b if $\begin{bmatrix}2\text{x}-3\text{y}&\text{a}-\text{b}&3\\1&\text{x}+4\text{y}&3\text{a}+4\text{b}\end{bmatrix}=\begin{bmatrix}1&-2&3\\1&6&29\end{bmatrix}$
AnswerSince the corresponding elements of two equal matrices are equal,
$\begin{bmatrix}2\text{x}-3\text{y}&\text{a}-\text{b}&3\\1&\text{x}+4\text{y}&3\text{a}+4\text{b}\end{bmatrix}=\begin{bmatrix}1&-2&3\\1&6&29\end{bmatrix}$
⇒ 2x - 3y = 1 ...(1)
⇒ x + 4y = 6
⇒ x = 6 - 4y ...(2)
Putting the value of x in eq. (1), we get
2(6 - 4y) - 3y = 1
⇒ 12 - 8y - 3y = 1
⇒ 12 + 11y = 1
⇒ -11y = -11
$\Rightarrow\text{y}=\frac{-11}{-11}=1$
Putting the value of y in eq. (2), we get
x = 6 - 4(1)
⇒ x = 6 - 4
⇒ x = 2
Now,
a - b = -2
⇒ a = -2 + b ...(3)
3a + 4b = 29 ...(4)
Putting the value of a in eq. (4), we get
3(-2 + b) + 4b = 29
⇒ -6 + 3b + 4b = 29
⇒ -6 + 7b = 29
⇒ 7b = 29 + 6
⇒ 7b = 35
$\Rightarrow\text{b}=\frac{35}{7}=5$
Putting the value of b in eq. (3), we get
a = -2 + 5
⇒ a = 3
$\therefore$ a = 3, b = 5, x = 2 and y = 1
View full question & answer→Question 1182 Marks
If $\begin{bmatrix}\text{x}&2\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}=2,$ find x.
AnswerGiven: $\begin{bmatrix}\text{x}&2\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}=2$
$\Rightarrow3\text{x}+8=2$
$\Rightarrow3\text{x}=2-8$
$\Rightarrow3\text{x}=-6$
$\Rightarrow\text{x}=\frac{-6}{3}$
$\Rightarrow\text{x}=-2$
View full question & answer→Question 1192 Marks
If $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{ B}=\text{diag}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-6&3&4\end{pmatrix},$ find.
A - 2B
AnswerGiven, $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{B}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-\text{b}&3&4\end{pmatrix}$
$\text{A}-2\text{B}$
$=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix}-2\text{diag}\begin{pmatrix}1&1&-4\end{pmatrix}$
$=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix}-\text{diag}\begin{pmatrix}2&2&-8\end{pmatrix}$
$=\text{diag}\begin{pmatrix}2-2&-5-2&9+8\end{pmatrix}$
$=\text{diag}\begin{pmatrix}0&-7&-17\end{pmatrix}$
So, $\text{A}-2\text{B}=\text{diag}\begin{pmatrix}0&-7&-17\end{pmatrix}$
View full question & answer→Question 1202 Marks
If $A$ is a skew-symmetric matrix and $n$ is an odd natural number, write whether $A^n$ is symmetric or skew-symmetric or neither of the two.
AnswerGiven,
$n$ is odd natural number and $A$ is kew symmetric matrix.
$\Rightarrow A^{T }= -A$
Now,
$(A^n)^{T }= (A^T)^{n }$
$\Rightarrow (A^n)^{T }= (-A)^{n }\{$since, $a^{T }= -A\}$
$\Rightarrow (A^n)^T = (-1)^n A^n$
$\Rightarrow (A^n)^T = -A^n \{$since, $n$ is odd natural number$\}$
We know that, a square matrix $A$ is skew symmetric if $A^T = -A$
So,
$A^n$ is a skew symmetric matrix.
View full question & answer→Question 1212 Marks
Find the value of $\lambda,$ non-zero scalar, if $\lambda\begin{bmatrix}1&0&2\\3&4&5\end{bmatrix}+2\begin{bmatrix}1&2&3\\-1&-3&2\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
AnswerGiven: $\lambda\begin{bmatrix}1&0&2\\3&4&5\end{bmatrix}+2\begin{bmatrix}1&2&3\\-1&-3&2\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\lambda&0&2\lambda\\3\lambda&4\lambda&5\lambda\end{bmatrix}+\begin{bmatrix}2&4&6\\-2&-6&4\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\lambda+2&0+4&2\lambda+6\\3\lambda-2&4\lambda-6&5\lambda+4\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
$\Rightarrow\lambda+2=4$
$\Rightarrow\lambda=4-2$
$\therefore\ \lambda=2$
View full question & answer→Question 1222 Marks
construct a 3 × 4 matrix, whose elements are given by:
$\text a_{\text {ij}}=2{\text{i}}-{\text{j}} $
Answer$\text{Let A}=\left[\text a_{\text{ij}}\right]\text {be required}\ 3\times4\ \text{matrix where}\ {\text a_{\text {ij}}} =2{\text{i - j}}$ $\therefore\ \text a_{11}=2-1=1,\ \text a_{12}=2-2=0,$ $ \text a_{13}=2-3=-1,\ \text a_{14}=2-4=-2 $ $\text a_{21}=4-1=3,\ \text{a}_{22}=4-2=2,$ $ \text{a}_{23}=4-3=1,\ \ \text{a}_{24}=4-4=0 $ $\text a_{31}=6-1=5,\ \ \text a_{32}=6-2=4,$ $\text a_{33}=6-3=3,\ \ \text a_{34}=6-4=2 $ $\therefore\ \text A = \begin{bmatrix}1 & 0 &-1 & -2 \\3 & 2&1&0\\\ 5&4&3&2 \end{bmatrix} $
View full question & answer→Question 1232 Marks
Find a matrix X such that 2A + B + X = 0, where.
If $\text{A}=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix},$ then find the matrix X of order 3 × 2 such that 2A + 3X = 5B.
Answer$2\text{A}+3\text{X}=5\text{B}$
$\Rightarrow2\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}+3\text{X}=5\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}16&0\\8&-4\\6&12\end{bmatrix}+3\text{X}=\begin{bmatrix}10&-10\\20&10\\-25&5\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}10&-10\\20&10\\-25&5\end{bmatrix}-\begin{bmatrix}16&0\\8&-4\\6&12\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}10-16&-10-0\\20-8&10+4\\-25-6&5-12\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}-6&-10\\12&14\\-31&-7\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{3}\begin{bmatrix}-6&-10\\12&14\\-31&-7\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}-2&\frac{-10}{3}\\4&\frac{14}{3}\\\frac{-31}{3}&\frac{-7}{3}\end{bmatrix}$
View full question & answer→Question 1242 Marks
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are given by$: \text{a}_\text{ij}=\frac{(2\text{i}-\text{j})^2}{2}$
AnswerHere,
$\text{a}_{11}=\frac{[2(1)+1]^2}{2}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{12}=\frac{[2(1)+2]^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
$\text{a}_{21}=\frac{[2(2)+1]^2}{2}=\frac{(4+1)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2},$ $\text{a}_{22}=\frac{[2(2)+2]^2}{2}=\frac{(4+2)^2}{2}=\frac{(6)^2}{2}=\frac{36}{2}=18$
So, the required matrix is $\begin{bmatrix}\frac{9}{2}&8\\\frac{25}{2}&18\end{bmatrix}.$
View full question & answer→Question 1252 Marks
Given an example of two non-zero 2×2 matrices A and B such that AB = 0.
AnswerLet,
$\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\neq0$
$\text{B}=\begin{bmatrix}0&0\\0&1\end{bmatrix}\neq0$
$\text{AB}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
So,
$\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\text{B}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
View full question & answer→Question 1262 Marks
A matrix $X$ has $a + b$ rows and $a + 2$ columns while the matrix $Y$ has $b + 1$ rows and $a + 3$ columns. Both matrices $XY$ and $YX$ exist. Find $a$ and $b.$ Can you say $XY$ and $YX$ are of the same type? Are they equal.
AnswerHere, $[X]_{(a+b) \times (a+2)} [Y]_{(b+1) \times (a+3)} $ Since $XY$ exists, the number of columns in $X$ is equal to the number of rows in $Y.$
$\Rightarrow a + 2 = b + 1 ...(1)$
Similarly, since $YX$ exists, the number of columns in $Y$ is equal to the number of rows in $X.$
$\Rightarrow a + b = a + 3 $
$\Rightarrow b = 3$
Putting the value of $b$ in $(1),$ we get $a + 2 ≈ 3 + 1 $
$\Rightarrow a = 2$
Since the order of the matrices $XY$ and $YX$ is not same,
$XY$ and $YX$ are not of the same type and they are unequal.
View full question & answer→Question 1272 Marks
Find x, y and z so that A = B, where.
$\text{A}=\begin{bmatrix}\text{x}-2&3&2\text{z}\\18\text{z}&\text{y}+2&6\text{z}\end{bmatrix},\text{B}=\begin{bmatrix}\text{y}&\text{z}&6\\6\text{y}&\text{x}&2\text{y}\end{bmatrix}$
AnswerSince all the corresponding elements of a matrix are equal,
$\text{A}=\begin{bmatrix}\text{x}-2&3&2\text{z}\\18\text{z}&\text{y}+2&6\text{z}\end{bmatrix},\text{B}=\begin{bmatrix}\text{y}&\text{z}&6\\6\text{y}&\text{x}&2\text{y}\end{bmatrix}$
Here,
x - 2 = y ...(1)
z = 3 ...(2)
18z = 6y ...(3)
Putting the value of z in eq. (3), we get
18(3) = 6y
⇒ 54 = 6y
$\Rightarrow\text{y}=\frac{54}{6}=9$
Putting the value of y in eq. (1), we get
x - 2 = 9
⇒ x = 9 + 2
⇒ x = 11
$\therefore$ x = 11, y = 9 and z = 3
View full question & answer→Question 1282 Marks
Find the values of $x, y$ and $z$ from the following equations:
$\begin{bmatrix}\text{x+y} & 2 \\5+\text{z} & \text {xy} \end{bmatrix}=\begin{bmatrix}6 & 2 \\5& 8 \end{bmatrix}$
AnswerWe are given that
$\begin{bmatrix}\text{x+y} & 2 \\5+\text{z} & \text {xy} \end{bmatrix}=\begin{bmatrix}6 & 2 \\5& 8 \end{bmatrix}$
By defination of equality of matrices,
$x + y = 6 ...(1)$
$5 + z = 5 ...(2)$
$xy = 8 ...(3)$
Form $(2), z = 0$
Form $y = 6 - x ...(4)$
Putting $y = 6 - x$ in $(3),$ we get.
$x(6 - x) =$ or $6 x - x^{2 } - 8 = 0$
$\therefore x^2 - 6x + 8 = 0 ? (x - 2) (x - 4) = 0;$
$? x = 2.4$
$\therefore$ from $(4), y = 6 - 2, 6 - 4 = 4, 2$
$\therefore$ we have
$x =2, y = 4, z = 0; x = 4, y = 2, z = 0$
View full question & answer→Question 1292 Marks
If $2\begin{bmatrix}3&4\\5&\text{x} \end{bmatrix}+\begin{bmatrix}1&\text{y}\\0&1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix},$ find x - y.
Answer$2\begin{bmatrix}3&4\\5&\text{x} \end{bmatrix}+\begin{bmatrix}1&\text{y}\\0&1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}6+1&8+\text{y}\\10+0&2\text{x}+1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}7&8+\text{y}\\10&2\text{x}+1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix}$
⇒ 8 + y = 0 and 2x + 1 = 5
⇒ y = - 8 and 2x = 4
⇒ y = - 8 and x = 2
Hence, x - y = 2 - (- 8) = 10.
View full question & answer→Question 1302 Marks
Let $A$ and $B$ be square matrices of the same order. Does $(A + B)^2 = A^2 + 2AB + B^2$ hold? If not, why?
Answer$\text{LHS} = (A + B)^2$
$= (A + B)(A + B)$
$= A(A + B) + B(A + B)$
$= A^2 + AB + BA + B^2$
We know that a matrix does not have commutative property.
So, $AB \neq BA$
Thus, $(A + B)^2 \neq A^2 + 2AB + B^2$
View full question & answer→Question 1312 Marks
If $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6 \end{bmatrix},$ write the value of (x + y + z).
Answer$\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
$\therefore$ z + 6 = 0 and x + y = 6
⇒ z = -6 and x + y = 6
Therefore, x + y + z = 6 - 6 = 0.
View full question & answer→Question 1322 Marks
Give example of matrices:
A and B such that AB = 0 but BA ≠ 0
AnswerLet $\text{A}=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$ \therefore\ \text{AB}=0$
and $\text{BA}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0+0&1+0\\0+0&0+0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&1\\0&0\end{bmatrix}$
Thus, AB = 0 but BA ≠ 0
View full question & answer→Question 1332 Marks
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs. 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
AnswerLet the monthly incomes of Aryan and Babban be 3x and 4x, respectively.
Suppose their monthly expenditures are 5y and 7y, respectively
Since each saves Rs. 15,000 per month,
Monthly saving of Aryan: 3x - 5y = 15,000
Monthly saving of Babban: 4x - 7y = 15,000
The above system of equations can be written in the matrix form as follows:
$\begin{bmatrix}3&-5\\4&-7\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y} \end{bmatrix}=\begin{bmatrix}15000\\15000 \end{bmatrix}$
Or,
AX = B, where $\text{A}=\begin{bmatrix}3&-5\\4&-7\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y} \end{bmatrix}$ and $\text{B}=\begin{bmatrix}15000\\15000 \end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}3&-5\\4&-7\end{vmatrix}=-21-(-20)=-1$
$\text{Adj A}=\begin{bmatrix}-7&-4\\5&3\end{bmatrix}^\text{T}=\begin{bmatrix}-7&5\\-4&3\end{bmatrix}$
So, $\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=-1=\begin{bmatrix}-7&5\\-4&3\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}\begin{bmatrix}15000\\15000 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}105000-75000\\60000-45000 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y} \end{bmatrix}=\begin{bmatrix}30000\\15000\end{bmatrix}$
$\Rightarrow\text{x}=30,000$ and $\text{y}=15,000$
Therefore,
Monthly income of Aryan = 3 × Rs. 30,000 = Rs. 90,000
Monthly income of Babban = 4 × Rs. 30,000 = Rs. 1,20,000
From this problem, we are encouraged to understand the power of savings.
We should save certain part of our monthly income for the future.
View full question & answer→Question 1342 Marks
If $B$ is a skew$-$symmetric matrix, write whether the matrix $AB \ A^T$ is symmetric or skew$-$symmetric.
AnswerIf $B$ is a skew$-$symmetric matrix, then $B^T = -B.$
$(\text{ABA}^\text{T})^\text{T}=(\text{A}^\text{T})^\text{T}\text{B}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{ABC})^\text{T}=\text{C}^\text{T}\text{B}^\text{T}\text{A}^\text{T}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=\text{AB}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{A}^\text{T})^\text{T}=\text{A}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=\text{A}(-\text{B})\text{A}^\text{T}$ $\big[\because\ \text{B}^\text{T}=-\text{B}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=-\text{ABA}^\text{T}$
$\therefore (\text{ABA}^\text{T})$ is a skew$-$symmetric matrix.
View full question & answer→Question 1352 Marks
If $\begin{bmatrix}2\text{x}+1&5\text{x}\\0&\text{y}^2+1\end{bmatrix}=\begin{bmatrix}\text{x}+3&10\\0&26\end{bmatrix},$ find the value of (x + y).
AnswerAs the given matrices are equal, therefore, their corresponding elements must be equal.
Comparing the corresponding elements, we get
$\begin{bmatrix}2\text{x} + 1 = \text{x} + 3&5\text{x}=10\\0=0&\text{y}^2+1=26\end{bmatrix}$
On simplifying, we get
x = 2 and $\text{y}=\pm5$
Therefore, x + y = 2 + 5 = 7
or x + y = 2 - 5 = -3
Hence, the value of (x + y) is 7, -3.
View full question & answer→Question 1362 Marks
If $\text{A}=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix},$ find $A^2.$
AnswerHere,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+0&0+0+0&0+0+0\\0+0+0&0+1+0&0+0+0\\0+0+0&0+0+0&0+0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
View full question & answer→Question 1372 Marks
Find the value of a, b, c and d from the following equations:
$\begin{bmatrix}2\text{a}+\text{b}&\text{a}-2\text{b}\\5\text{c}-\text{d}&4\text{c}+3\text{d}\end{bmatrix}=\begin{bmatrix}4&-3\\11&24\end{bmatrix}$
AnswerAs the given m atrices are equal, therefore their corresponding elements must be equal.
Comparing the corresponding elements, we get
2a + b = 4 ...(i)
a - 2b = -3 ...(ii)
5c - d = 11 ...(iii)
4c + 3d = 24 ...(iv)
Multiplying (i) by 2 and adding to (ii)
5a = 5 ⇒ a = 1
(i) ⇒ b = 4 - 2 × 1 = 2
Multiplying (iii) by 3 and adding to (iv)
19c = 57 ⇒ c = 3
(iii) ⇒ d = 5 × 3 - 11 = 4
Hence, a = 1, b = 2, c = 3, d = 4
View full question & answer→Question 1382 Marks
If $\text{A}=\begin{bmatrix}1\\2\\3\end{bmatrix},$ write $AA^T.$
AnswerIf $\text{A}=\begin{bmatrix}1\\2\\3\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2&3\end{bmatrix}$
Now,
$\text{AA}^\text{T}=\begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}1&2&3\end{bmatrix}$
$\Rightarrow\ \text{AA}^\text{T}=\begin{bmatrix}1&2&3\\2&4&6\ \\3&6&9\end{bmatrix}$
View full question & answer→Question 1392 Marks
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
AnswerLet the monthly incomes of Aryan and Babban be 3x and 4x, respectively.
Suppose their monthly expenditures are 5y and 7y, respectively.
Since each saves Rs. 15,000 per month,
Monthly saving of Aryan: 3x - 5y = 15,000
Monthly saving of Babban: 4x - 7y = 15,000
The above system of equations can be written in the matrix form as follows:
$\begin{bmatrix}3&-5\\4&-7\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}15000\\15000\end{bmatrix}$
or,
AX = B, where $\text{A}=\begin{bmatrix}3&-5\\4&-7\end{bmatrix},\ \text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}15000\\15000\end{bmatrix}$
Now,
$\big|\text{A}\big|=\begin{vmatrix}3&-5\\4&-7\end{vmatrix}=-21-(-20)=-1$
$\text{Adj}\ \text{A}=\begin{bmatrix}-7&-4\\5&3\end{bmatrix}^\text{T}=\begin{bmatrix}-7&5\\-4&3\end{bmatrix}$
So, $\text{A}^{-1}=\frac{1}{\big|\text{A}\big|}\text{ adj }\text{A}=-1\begin{bmatrix}-7&5\\-4&3\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$ \Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}\begin{bmatrix}15000\\15000\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}105000-75000\\60000-45000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}30000\\15000\end{bmatrix}$
$\Rightarrow\text{x}=30,000$ and $\text{y}=15,000$
Therefore,
Monthly income of Aryan = 3 × Rs. 30,000 = Rs. 90,000
Monthly income of Babban = 4 × Rs. 30,000 = Rs. 1,20,000
From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.
View full question & answer→Question 1402 Marks
If $\begin{bmatrix}\text{x}-\text{y}&\text{z}\\2\text{x}-\text{y}&\text{w}\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix},$ find x, y, z, w.
AnswerGiven, $\begin{bmatrix}\text{x}-\text{y}&\text{z}\\2\text{x}-\text{y}&\text{w}\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix}$ Since corresponding entries of equal matrices are equal, So x - y = -1 ...(i) z = 4 ...(ii) 2x - y = 0 ...(iii) w = 5 ...(iv) Solving equation (i) and (iii)
x = 1 Put x = 1 in equation (i), x - y = -1 1 - y = -1 -y = -1 - 1 -y = -2 y = 2 Equation (ii) and (iv) given the values of z and w respectively, So z = 4, w = 5 Hence, x = 1, y = 2, z = 4, w = 5 View full question & answer→Question 1412 Marks
If $\begin{bmatrix}2\text{x+y}&3\text{y}\\0&4 \end{bmatrix}=\begin{bmatrix}6&0\\6&4\end{bmatrix}$, then find x.
AnswerGiven,
$\begin{bmatrix}2\text{x+y}&3\text{y}\\0&4 \end{bmatrix}=\begin{bmatrix}6&0\\6&4\end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
3y = 0
⇒ y = 0
And 2x + y = 6
⇒ 2x + 0 = 6
⇒ 2x = 6
⇒ x = 3
So,
x = 3, y = 0
View full question & answer→Question 1422 Marks
Show that:
$\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}\neq\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}$
Answer$\text{L.H.S}=\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}$$=\begin{bmatrix}5(2) + (-1)3&5(1) + (-1)4\\6(2) + 7(3)&6(1) + 7(4)\end{bmatrix}$$ = \begin{bmatrix}7&1\\33&34\end{bmatrix} $
$\text{R.H.S} = \begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix} $$= \begin{bmatrix}2(5) + 1(6)&2(-1) + 1(7)\\3(5) + 4(6)&3(-1) + 4(7)\end{bmatrix} $$= \begin{bmatrix}16&5\\39&25\end{bmatrix} $
$\therefore \text{L.H.S.} \neq \text{R.H.S.}$
View full question & answer→Question 1432 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 1442 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 1452 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 1462 Marks
If $A$ is a symmetric matrix and $n \in N,$ write whether $A^n$ is symmetric or skew-symmetric or neither of these two.
AnswerIf $A$ is a symmetric matrix, then $A^T = A.$
Now,
$(A^n)^T = (A^T)^n [$for all $n \in N]$
$\Rightarrow (A^n)^T = (A)^n [\because A^T = A]$
Hence, $A^n$ is a symmetric matrix.
View full question & answer→Question 1472 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 1482 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 1492 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 1502 Marks
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ 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→