Question 512 Marks
If $\text{A}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}-4\\3\end{bmatrix},$ write AB.
Answer$\text{AB}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}\begin{bmatrix}-4\\3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-16+9\\-4+6\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix} -7\\2\end{bmatrix}$
View full question & answer→Question 522 Marks
If $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix},$ write $A^2.$
AnswerGiven: $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2+0&0+0\\0+0&0+\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2&0\\0&\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ $(\because\ \text{i} ^2=-1)$
View full question & answer→Question 532 Marks
Find the values of x, y and z from the following equations:
$\begin{bmatrix}\text{x + y+ z}\\ \text{x + z}\\\ \text{y + z} \end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix} $
AnswerWe are given that
$\begin{bmatrix}\text{x + y + z}\\ \text{x+ z}\\ \text{y + z}\end{bmatrix}=\begin{bmatrix}9\\5\\7 \end{bmatrix}$
By defination of equality of matrices.
x + y + z = 9 ...(1)
x + z = 5 ...(2)
y + z = 7 ...(3)
Subtracting (2) from (1), y = 4
Subtracting (3) from (1), x = 2
$\therefore$ from(2), 2 + z = 5, ⇒ z = 3
$\therefore$ x = 2, y = 4, z = 3
View full question & answer→Question 542 Marks
Give example of matrices:
A and B such that AB = O but A ≠ 0, B ≠ 0.
AnswerLet $\text{A}=\begin{bmatrix}0&2\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\therefore\ \text{AB}=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Thus, AB = 0 while A ≠ 0 and B ≠ 0
View full question & answer→Question 552 Marks
Find x, y, z and t, if.
$2\begin{bmatrix}\text{x}&5\\7&\text{y}-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
Answer$2\begin{bmatrix}\text{x}&5\\7&\text{y}-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}&10\\14&2\text{y}-6\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}+3&14\\15&2\text{y}-4\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
Comparing the corresponding elements from both sides,
$2\text{x}+3=7\Rightarrow2\text{x}=4\Rightarrow\text{x}=2$
$2\text{y}-4=14\Rightarrow2\text{y}=18\Rightarrow\text{y}=9$
Hence, x = 2, y = 9
View full question & answer→Question 562 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by$: a_{ij} = i.j$
AnswerHere,
$a_{ij} = i.j. 1 \leq i \leq 2$ and $1 \leq j \leq 3$
$a_{11} = 1 \times 1 = 1, a_{12} = 1 \times 2 = 2, a_{13} = 1 \times 3 = 3$
$a_{21} = 2 \times 1 = 2, a_{22} = 2 \times 2 = 4$ and $a_{23} = 2 \times 3 = 6$
View full question & answer→Question 572 Marks
Let $A$ and $B$ be square matrices of the order $3 \times 3.$ Is $(AB)^2 = A^2B^2$? Give reasons.
AnswerWe are given that, $A$ and $B$ are square matrices of order $3 \times 3.$
Consider, $(AB)^{2 }= AB.AB$
$= \text{ABAB}$
$= \text{AABB} [\because AB = BA]$
$= A^2B^2$
Thus, $AB^2 = A^2B^2$ is true if and only if $AB = BA.$
View full question & answer→Question 582 Marks
If $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix},$ find $AA^T.$
AnswerGiven: $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos^2\text{x}+\sin^2\text{x}&\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}\\\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}&\sin^2\text{x}+\cos^2\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→Question 592 Marks
Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&2-2&-2+2\\4+1&\text{x}+0&6-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&0&0\\5&\text{x}&5\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\text{x}-\text{y}+3=6$
$\Rightarrow\text{x}-\text{y}=6-3$
$\Rightarrow\text{x}-\text{y}=3\ \dots(1)$
Also,
$\text{x}=2\text{x}+\text{y}$
$\Rightarrow-\text{x}=\text{y}\ \dots(2)$
Putting the value of y in eq. (1), we get
$\text{x}-(-\text{x})=3$
$\Rightarrow2\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{2}$
Putting the value of x in eq. (2), we get
$-\Big(\frac{3}{2}\Big)=\text{y}$
$\Rightarrow\text{y}=-\frac{3}{2}$
View full question & answer→Question 602 Marks
If $\text{A}=\begin{bmatrix}9&1\\7&8\end{bmatrix},\text{ B}=\begin{bmatrix}1&5\\7&12\end{bmatrix},$ find matrix C such that 5A + 3B + 2C is a null matrix.
AnswerGiven, $5\text{A}+3\text{B}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow5\begin{bmatrix}9&1\\7&8\end{bmatrix}+3\begin{bmatrix}1&5\\7&12\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45&5\\35&40\end{bmatrix}+\begin{bmatrix}3&15\\21&36\end{bmatrix}+2\text{C}\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45+3&5+15\\35+21&40+36\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}48&20\\56&76\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}48&20\\56&76\end{bmatrix}$
$\Rightarrow\text{C}=\frac{1}{2}\begin{bmatrix}-48&-20\\-56&-76\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}$
View full question & answer→Question 612 Marks
If $A$ is $2\times 3$ matrix and $B$ is a matrix such that $A^T B$ and $BA^T$ both are defined, then what is the order of $B$?
AnswerOrder of $A = 2 \times 3$
Order of $A^{T }= 3 \times 2$
Let Order of $B = m \times n$
Given$: A^TB$ and $BA^T$ are defined
If $A^T{}_{3\times 2 }B_{m\times n }$ exists, then the number of columns in $A^T$ must be equal to number of rows in $B.$
$\Rightarrow m = 2$
If $_{ }B_{m\times n} A^T{}_{3\times 2 }$ exists, then the number of columns in $B$ must be equal to number of rows in $A^T$
$\Rightarrow n = 3$
$\therefore$ Order of $B = 2 \times 3$
View full question & answer→Question 622 Marks
$\text{If}\ \text{A}'=\begin{bmatrix}-2&3\\1&2\end{bmatrix}, \text{and}\ \text{B}=\begin{bmatrix}-1&0\\1&2\end{bmatrix}\ \text{then find}\ (\text{A} + 2\text{B})'$
AnswerWe know that A = (A')'
$\therefore\ \text{A}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}$
$\therefore \text{A}+2\text{B}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+2\begin{bmatrix}-1&0\\1&2\end{bmatrix}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+\begin{bmatrix}-2&0\\2&4\end{bmatrix}=\begin{bmatrix}-4&1\\5&6\end{bmatrix}$
$\therefore(\text{A} + 2\text{B})'=\begin{bmatrix}-4&5\\1&6\end{bmatrix}$
View full question & answer→Question 632 Marks
A trust fund has Rs. 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
- Rs. 1800
- Rs. 2000
AnswerIf Rs. x are invested in the first type of bond and Rs. (30000 - x) are invested in the second type of bond, Then the matrix $\text{A}=\begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}$ represents investment and the matrix $\text{B}=\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}$ represents rate of interest.
- $\begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\frac{5\text{x}}{100}+\frac{7(30000-\text{x})}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}$
$\Rightarrow\frac{5\text{x}+210000-7\text{x}}{100}=1800$
$\Rightarrow210000-2\text{x}=180000$
$\Rightarrow2\text{x}=30000$
$\Rightarrow\text{x}=15000$
Thus,
Amount invested in the first bond = Rs. 15000
Amount invested in the second bond = Rs. (30000 - 15000)
= Rs. 15000
- $ \begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\frac{5\text{x}}{100}+\frac{7(30000-\text{x})}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}$
$\Rightarrow\frac{5\text{x}+210000-7\text{x}}{100}=2000$
$\Rightarrow210000-2\text{x}=200000$
$\Rightarrow2\text{x}=10000$
$\Rightarrow\text{x}=5000$
Thus,
Amount invested in the first bond = Rs. 5000
Amount invested in the second bond = Rs. (30000 - 5000)
= Rs. 25000 View full question & answer→Question 642 Marks
If I is the identity matrix and $A$ is a square matrix such that $A^2 = A,$ then what is the value of $(I + A)^2 = 3A?$
AnswerGiven,
A is a square matrix such that $A^2 = A$
Now,
$(I + A)^{2 }- 3A = (I + A)(I + A) - 3A$
$\Rightarrow (I + A)^{2 }- 3A = I \times I + I \times A + A \times I + A \times A - 3A \{$using distributive property$\}$
$\Rightarrow (I + A)^{2 }- 3A = I + A + A + A^{2 }- 3A\{$using $I \times I = I, IA = AI = A\}$
$\Rightarrow (I + A)^{2 }- 3A = I + 2A + A - 3A \{$since, $A^{2 }= A\}$
$\Rightarrow (I + A)^{2 }- 3A = I$
View full question & answer→Question 652 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.
$\therefore a^{T }= -A$
$\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}^{\text{T}}=-\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}0&-1&\text{x}\\1&0&-3\\-2&3&0\\ \end{bmatrix}=\begin{bmatrix}0&-1&2\\1&0&-3\\-\text{x}&3&0 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
$\Rightarrow x = 2$
Hence, the value of $x$ is $2.$
View full question & answer→Question 662 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 672 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 682 Marks
Verify that $A^2 = I,$ when $\text{A}=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}.$
AnswerWe have, $\text{A}=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}.\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$
$[\because\ \text{A}^2=\text{A}.\text{A}]$$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\text{I}$
Hence proved.
View full question & answer→Question 692 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,
$\Rightarrow 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$
$\Rightarrow x = 6 - y,$
Subsituting value of $x$ in equation $(1)$ we get,
$\Rightarrow (6 - y)y = 8$
$\Rightarrow y^2 - 6y + 8 = 0$
$\Rightarrow (y - 2)(y - 4) = 0,$
$\Rightarrow y = 2, 4$
Subsituting the value of y in equation $(1)$ we get,
$\Rightarrow 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 702 Marks
Let A and B be matrices of orders 3×2 and 2×4 respectively. Write the order of matrix AB.
AnswerSince, the order of matrix A is 3×2 and order of matrix B is 2×4
So, the order of AB will be the "number of rows of A × number of columns of B" = 3×4
View full question & answer→Question 712 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 722 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\times 3$
Order of $B = 3\times 2$
So,
$A_{2\times 3} \times B_{3\times 2}$ has order $= 2\times 2$
$B_{3\times 2} \times A_{2\times 3}$ has order $= 3\times 3$
Hence,
Order of $AB = 2\times 2$
Order of $BA = 3\times 3$
View full question & answer→Question 732 Marks
If $x\begin{bmatrix}2\\3\end{bmatrix} +y\begin{bmatrix}-1\\1\end{bmatrix} = \begin{bmatrix}10\\5\end{bmatrix}, $find the values of x and y.
AnswerGiven: $x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2x\\3x\end{bmatrix}+\begin{bmatrix}-y\\ y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
Equating corresponding entries, we have
2x - y = 10 ...(i)
3x + y = 5 ...(ii)
Adding eq. (i) and (ii), we have 5x = 15 ⇒ x = 3
Putting x = 3 in eq. (ii), 9 + y = 5 ⇒ y = -4
View full question & answer→Question 742 Marks
If $A$ is a skew-symmetric matrix and $n$ is an even natural number, write whether $A^n$ is symmetric or skew-symmetric or neither of these two.
AnswerIf $A$ is a skew-symmetric matrix, then $A^T = -A.$
$(A^n)^T = (A^T)^n [$For all $n \in N]$
$\Rightarrow (A^n)^T = (-A)^n [\because A^T = -A]$
$\Rightarrow (A^n)^T = (-1)^n A^n$
$\Rightarrow (A^n)^T = A^n,$ if $n$ is even or $-A^n,$ if $n$ is odd.
Hence, $A^n$ is a symmetric when $n$ is an even natural number.
View full question & answer→Question 752 Marks
If $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix},$ show that $A - A^T$ is a skew symmetric matrix.
AnswerGiven:$\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$
$\text{A}-\text{A}^{\text{T}}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}-\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}^{\text{T}}$
$=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}-\begin{bmatrix}3 &1 \\-4 & -1 \end{bmatrix}$
$=\begin{bmatrix}3-3 & -4-1 \\1+4 & -1+1 \end{bmatrix}$
$\text{A}-\text{A}^{\text{T}}=\begin{bmatrix}0 & -5 \\5 & 0 \end{bmatrix}\ \dots( \text{i})$
$-(\text{A}-\text{A}^{\text{T}})^\text{T}=-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}^{\text{T}}$
$=-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}$
$-(\text{A}-\text{A}^{\text{T}})^{\text{T}}-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}\ \dots(\text{ii})$
From equation $(i)$ and $(ii)$
$(\text{A}-\text{A}^{\text{T}})^{\text{T}}=-(\text{A}-\text{A}^{\text{T}})^{\text{T}}$
We know that, $x$ is skewsym metric matrix if $x = -x^T$
So, $(A - A^T)$ is skewsym metric matrix.
View full question & answer→Question 762 Marks
If $\begin{bmatrix}\text{x}+3&\text{z}+4&2\text{y}-7\\4\text{x}+6&\text{a}-1&0\\\text{b}-3&3\text{b}&\text{z}+2\text{c}\end{bmatrix}=\begin{bmatrix}0&6&3\text{y}-2\\2\text{x}&-3&2\text{c}-2\\2\text{b}+4&-21&0\end{bmatrix}$ Obtain the values of a, b, c, x, y and z.
AnswerSince all the corresponding elements of a matrix are equal, x + 3 = 0 ⇒ x = -3 Also, 2y - 7 = 3y - 2 ⇒ 2y - 3y = -2 + 7 ⇒ -y = 5 ⇒ y = -5⇒ z + 4 = 6
⇒ z = 6 - 4 ⇒ z = 2⇒ a - 1 = -3
⇒ a = -3 + 1 ⇒ a = -2 3b = -21 ⇒ b = -7⇒ z + 2c = 0
⇒ 2 = -2c ⇒ c = -1 Thus, x = -3, y = -5, a = -2, b = -7 and c = -1
View full question & answer→Question 772 Marks
If $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix},$ prove that $A - A^T$ is a skew symmetric matrix.
AnswerGiven: $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix}$
$\text{A}^{\text{T}}=\begin{bmatrix}2 & 4 \\3 & 5 \end{bmatrix}$
Now,
$(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix}-\begin{bmatrix}2 & 4 \\3 & 5 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}2-2 & 3-4 \\4-3 & 5-5 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^\text{T})=\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}\ ...(\text{i})$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}^\text{T}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}0 & 1 \\-1 & 0 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=-\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})=(\text{A}-\text{A}^{\text{T}})^{\text{T}} [$Using eq.$(i)]$
Thus, $(A - A^T)$ is a skew$-$symmetric matrix.
View full question & answer→Question 782 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 792 Marks
Find x, y satisfying the matrix equation.
$\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=0$
AnswerGiven: $\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}+3\text{y}-8\\\text{x}+5\text{y}-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow2\text{x}+3\text{y}-8=0$
$\Rightarrow2\text{x}+3\text{y}=8\ \dots(1)$
Also,
$\text{x}+5\text{y}-11=0$
$\Rightarrow\text{x}+5\text{y}=11$
$\Rightarrow\text{x}=11-5\text{y}\ \dots(2)$
Putting the value of x in eq. (1), we get
$2(11-5\text{y})+3\text{y}=8$
$\Rightarrow22-10\text{y}+3\text{y}=8$
$\Rightarrow-7\text{y}=8-22$
$\Rightarrow-7\text{y}=-14$
$\Rightarrow\text{y}=2$
Putting the value of y in eq. (2), we get
$\text{x}=11-5(2)$
$\Rightarrow\text{x}=11-10$
$\Rightarrow\text{x}=1$
$\therefore\ \text{x}=1$ and $\text{y}=2$
View full question & answer→Question 802 Marks
If $A$ and $B$ are square matrices of the same order, explain, why in general:
$(A + B)(A - B) \neq A^2 - B^2.$
Answer$LHS = (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) \neq A^2 - B^2$
View full question & answer→Question 812 Marks
If $B$ is a 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{ABA}^\text{T}$ $\big[\because\ \text{B}^\text{T}=\text{B}\big]$
Hence, $(\text{ABA}^\text{T})$ is a skew$-$symmetric matrix.
View full question & answer→Question 822 Marks
If $\text{A}=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ satisfies $\text{A}^4=\lambda\text{A},$ then write the value of $\lambda.$
Answer$\text{A}^2=\text{A}\cdot\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+1&1+1\\1+1&1+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&2\\2&2\end{bmatrix}$
Now,
$\text{A}^4=\text{A}^2\cdot\text{A}^2$
$\Rightarrow\text{A}^4=\begin{bmatrix}2&2\\2&2\end{bmatrix}\begin{bmatrix}2&2\\2&2\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}4+4&4+4\\4+4&4+4\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}8&8\\8&8\end{bmatrix}$
Also,
$\text{A}^4=\lambda\text{A}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\lambda\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\begin{bmatrix}\lambda&\lambda\\\lambda&\lambda\end{bmatrix}$
$\therefore\ \lambda=8$
View full question & answer→Question 832 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 842 Marks
For a $2\times 2$ matrix $A = [a_{ij}] $ whose elements are given by $\text{a}_{\text{ij}}=\frac{\text{i}}{\text{j}},$ write the value of $a_{12}.$
AnswerGiven that a $2\times 2$ matrix $A = [a_{ij}]$ whose elements are given by $\text{a}_{\text{ij}}=\frac{\text{i}}{\text{j}}.$
We need to find the value of $a_{12}.$
Thus, $\text{a}_{12}=\frac{1}{2}.$
View full question & answer→Question 852 Marks
The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs. 8.30, Rs. 3.45 and Rs. 4.50 each respectively. Find the total amount the store will receive from selling all the items.
AnswerMatrix representation of stock of various types of book in the store is given by, $\text{X}=\begin{bmatrix}\text{Physics}&\text{Chemistry}&\text{Mathematics}\\120&96&60\end{bmatrix}$ Matrix representation of selling price (Rs.) of each book is given by, $\text{Y}=\begin{bmatrix}8.30&\text{Physics}\\3.45&\text{Chemistry}\\4.50&\text{Mathematics}\end{bmatrix}$ So, total amount recieved by the store from selling all the items is given by, $\text{XY}=\begin{bmatrix}120&96&60\end{bmatrix}\begin{bmatrix}8.30\\3.45\\4.50\end{bmatrix}$ $\big[(120)(8.30)+(96)(3.45)+(60)(4.50)\big]$ $=\big[996+331.20+270\big]$ $=\big[1597.20\big]$Required amount = Rs. 1597.20
View full question & answer→Question 862 Marks
If $\text{A}=\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix},$ find the matrix C such that A + B + C is zeor matrix.
AnswerGiven,
$\text{A}=\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix},\text{B}=\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix},$
And
$\text{A}+\text{B}+\text{C}=0$
$\Rightarrow\text{C}=-\text{A}-\text{B}+0$
$\Rightarrow\text{C}=-\text{A}-\text{B}$
$\Rightarrow\text{C}=-\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix}-\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-1&3&-2\\-2&0&-2\end{bmatrix}-\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-1-2&3+1&-2+1\\-2-1&0-0&-2+1\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix}$
Hence,
$\text{C}=\begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix}$
View full question & answer→Question 872 Marks
If $\text{A}=\begin{bmatrix}-3&0\\0&-3\end{bmatrix}$, find $A^4.$
AnswerHere,
$\text{A}^2 = \text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-3&0\\0&-3\end{bmatrix}\begin{bmatrix}-3&0\\0&-3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9+0&0+0\\0+0&0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9&0\\0&9\end{bmatrix}$
Now,
$\text{A}^4 = \text{A}^2\cdot\text{A}^2$
$\Rightarrow\text{A}^4=\begin{bmatrix}9&0\\0&9\end{bmatrix}\begin{bmatrix}9&0\\0&9\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}81+0&0+0\\0+0&0+81\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}81&0\\0&81\end{bmatrix}$
View full question & answer→Question 882 Marks
What is the total number of $2\times 2$ matrices with each entry $0$ or $1$?
AnswerIn a $2\times 2$ matrix
Total number of elements are $4$ and each entry can be writte in $2$ ways.
So, Number of ways in which $4$ entries can be written
$= 4^2$
$= 16$
So, Total number of $2 \times 2$ matrices with each entry $0$ or $1 = 16$
View full question & answer→Question 892 Marks
If matrix $A = [1 2 3],$ write $AA^T.$
AnswerGiven: $A = [1 2 3]$
$\text{A}^{\text{T}}=\begin{bmatrix}1\\2\\3 \end{bmatrix}$
$\text{AA}^{\text{T}}=\begin{bmatrix}1&2&3 \end{bmatrix}\begin{bmatrix}1\\2\\3 \end{bmatrix}$
$\Rightarrow AA^{T }= 1 + 4 + 9$
$\Rightarrow AA^{T }= 14$
View full question & answer→Question 902 Marks
Simplify: $$$\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} + \sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\ \cos\theta&\sin\theta\end{bmatrix}$
AnswerGiven: $\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} + \sin\theta\begin{bmatrix}\sin\theta&-\cos\theta \\ \cos\theta&\ \sin\theta\end{bmatrix}$
$=\begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{bmatrix} + \begin{bmatrix}\sin^2\theta&-\cos\theta\sin\theta\\ \cos\theta\sin\theta&\ \sin^2\theta\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→Question 912 Marks
If $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix},$ find $A + A^T.$
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 922 Marks
If $A = [a_{ij}]$ is a skew-symmetric matrix, then write the value of $\sum\limits_\text{i}\sum\limits_\text{j}\text{a}_\text{ij}.$
AnswerGiven: $A = [a_{ij}]$ is a skew-symmetric matrix.
$\Rightarrow a_{ij} = -a_{ij} [$For all values of $i, j]$
$\Rightarrow a_{ij} = -a_{ij} [$For all values of $i]$
$\Rightarrow a_{ij} = 0$
Now,
$\sum\limits_{\text{i}}\sum\limits_{\text{j}} \text{a}_{\text{ij}} = \text{a}_{11} +\text{a}_{12} +\text{a}_{13 } + ... + \text{a}_{21} +\text{a}_{22} +\text{a}_{23} + ... + \text{a}_{31} +\text{a}_{32} +\text{a}_{33} +...$
$= 0 + \text{a}_{12} +\text{a}_{13} + ... -\text{a}_{12} + 0 + \text{a}_{23} +...-\text{a}_{13} - \text{a}_{23} +0 + ...$
$=0$
Thus,
$\sum\limits_{\text{i}}\sum\limits_{\text{j}} \text{a}_{\text{ij}} =0$
View full question & answer→Question 932 Marks
If $\begin{bmatrix}2&1&3 \end{bmatrix}$ $\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}\begin{pmatrix}1\\0\\-1 \end{pmatrix}=\text{A},$ then write the order of matrix A.
AnswerConsider, $\begin{pmatrix}2&1&3 \end{pmatrix}\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}\begin{pmatrix}1\\0\\-1 \end{pmatrix}=\text{A}$
Order of matrix $\begin{pmatrix}2&1&3 \end{pmatrix}$ is 1 × 3.
Order of matrix $\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}$ is 3 × 3
Order of matrix $\begin{pmatrix}1\\0\\-1 \end{pmatrix}$ is 3 × 1
Therefore, order of $\begin{pmatrix}2&1&3 \end{pmatrix}\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}\begin{pmatrix}1\\0\\-1 \end{pmatrix}$ is 1 × 1.
View full question & answer→Question 942 Marks
Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ 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 952 Marks
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ 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 962 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 972 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,
$2x + 1 = x + 3 ...(i)$
$2y = y^2 + 2 ...(ii)$
$y^2 - 5y = -6 ...(iii)$
Solving equation $(i),$
$2x + 1 = x + 3$
$2x - x = 3 - 1$
$x = 2$
Solving equation $(ii),$
$2y = y^2 + 2$
$y^2 - 2y + 2 = 0$
$D = b^2 - 4ac$
$= (-2)^2 - 4$
$= 4 - 8$
$= -2$
So, There is no real value of y from equation $(ii),$
Solving equation $(iii),$
$y^2 - 5y = -6$
$y^2 - 5y + 6 = 0$
$y^2 - 3y - 2y + 6 = 0$
$y(y - 3) - 2(y - 3) = 0$
$(y - 3)(y - 2) = 0$
$y = 3$ 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 982 Marks
If $A$ and $B$ are square matrices of the same order such that $AB = BA,$ then show that $(A + B)^2 = A^2 + 2AB + B^2.$
AnswerGiven,
$A$ and $B$ two square matrices of same order such that $AB = BA$
To prove: $(A + B)^2 = A^2 + 2AB + B^2$
Now, solving $LHS$ gives,
$(A + B)^2 = (A + B)(A + B)$
$= A(A + B) + B(A + B) [$by dist, of matrix multiplication over addition$]$
$= A^2 + AB + BA + B^2 [$by dist, of matrix multiplication over addition$]$
$= A^2 + 2AB + B^2 [$As$, AB = BA]$
$= RHS$
Hence proved.
View full question & answer→Question 992 Marks
If matrix $\text{A}=\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$ and $A^{2 }= pA,$ then write the value of $p.$
AnswerGiven: $\text{A}=\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$
$\text{A}^{2}=\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$
$=\begin{bmatrix}4+4&-4-4\\-4-4&4+4 \end{bmatrix}$
$=\begin{bmatrix}8&-8\\-8&8 \end{bmatrix}$
$=4\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$
$=4\text{A}$
Hence, $p = 4.$
View full question & answer→Question 1002 Marks
The sales figure of two car dealers during January 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 standard cars. Total sales over the 2 month period of January-February revealed that dealer A sold 8 deluxe 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars. Write 2 × 3 matrices summarizing sales data for January and 2-month period for each dealer.
AnswerAccording to the data, dealer A sold 5 deluxe cars, 3 premium cars and 4 standard cars in January.
Also, dealer B sold 7 deluxe cars, 2 premium cars and 3 standard cars in January.
The above information can be given by,
Deluxe Premium Standard $\begin{matrix}\text{Dealer A} \\\text{Dealer B} \end{matrix}\begin{bmatrix}5&3&4\\7&2&3\end{bmatrix}$
Total sales over the period of January-February reveal that dealer A sold 8 deluxe cars,7 premium cars and 6 standard cars, while dealer B sold 10 deluxe cars, 5 premium cars and 7 standard cars.
This information can be given by,
Deluxe Premium Standard $\begin{matrix}\text{Dealer A} \\\text{Dealer B} \end{matrix}\begin{bmatrix}8&7&6\\10&5&7\end{bmatrix}$
View full question & answer→