Question 512 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 522 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 532 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 542 Marks
If A and B are square matrices of the same order, explain, why in general:
$(A + B)(A - B) \neq A^2 - B^2.$
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,
$(A + B)(A - B) \neq A^2 - B^2$
View full question & answer→Question 552 Marks
If F(x) =$\begin{bmatrix}\cos x& -\sin x& 0\\ \sin x& \cos x&0\\0&0&1\end{bmatrix}, $ show that F(x) F(y) = F(x + y).
AnswerGiven: F(x) = $\text{F(x)}=\begin{bmatrix} \cos x &-\sin x&0\\ \sin x&\cos x&0\\0&0&1\end{bmatrix}.....(1)$
Changing x to y in eq. (i), $\text{F(y)}=\begin{bmatrix} \cos y &-\sin y&0\\ \sin y&\cos y&0\\0&0&1\end{bmatrix} $
$\text{L.H.S}=\begin{bmatrix} \cos x &-\sin x&0\\ \sin x&\cos x&0\\0&0&1\end{bmatrix}.\begin{bmatrix} \cos y &-\sin y&0\\ \sin y&\cos y&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}\cos x \cos y- \sin x \sin y+0&- \cos x \sin y- \sin x \cos y+0&0-0+0\ &\\\sin x \cos y+ \cos x \sin y +0&- \sin x \sin y+ \cos x \cos y+0&0+0+0\\0+0+0&0+0+0&0+0+1\end{bmatrix}$
$=\begin{bmatrix} \cos(x+y)&-\sin(x+y)&0\\ \sin(x+y)&\ \cos(x+y)&0\\0&0&1\end{bmatrix}=\text F(x+y)= \text{R.H.S}$ [changing x to to (x + y) in eq. (i)]
View full question & answer→Question 562 Marks
If $A$ is a square matrix such that $A^2=A$, then write the value of $7 A-(I+A)^3$, where $I$ is the identity matrix.
Answer$A^2=A$
$A^3=A^2=A$
$7 A-(I+A)^3$
$=7 A-\left(I^3+A^3+3 A^2 I+3 A I^2\right)$
$=7 A-(I+A+3 A+3 A)$
$=7 A-(I+7 A)$
$=-1$
View full question & answer→Question 572 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, $ABA^T$ is a skew-symmetric matrix.
View full question & answer→Question 582 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 592 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 602 Marks
Show by an example that for $\text{A}\neq0,\ \text{B}\neq0,\ \text{AB}=0.$
AnswerLet $\text{A}=\begin{bmatrix}0&1\\0&2\end{bmatrix}\neq0$ and $\text{B}=\begin{bmatrix}-1&1\\0&0\end{bmatrix}\neq0$
$\therefore\ \text{AB}=\begin{bmatrix}0&1\\0&2\end{bmatrix}\begin{bmatrix}-1&1\\0&0\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 612 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 622 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 632 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 642 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 652 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 662 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 672 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 682 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 692 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 702 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 712 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 722 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 732 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 742 Marks
If a matrix has 24 elements, what are the possible orders it can have? What, if has 13 elements?
AnswerWe know that a matrix of order m × n has mn elements. Therefore, for finding all possible orders of a matrix with 24 elements, we will find all ordered pairs with products of elements as 24.$\therefore$ all possible ordered pairs are
(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)
$\therefore$ possible orders are
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4
if number of elements = 13, then possible orders are 1 × 13, 13 × 1.
View full question & answer→Question 752 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 762 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 $A B=B A$
To prove: $(A+B)^2=A^2+2 A B+B^2$
Now, solving LHS gives,
$(A+B)^2=(A+B)(A+B)$
$=A(A+B)+B(A+B)$[ by dist, of matrix multiplication over addition]
$=A^2+A B+B A+B^2$[ by dist, of matrix multiplication over addition ]
$=A^2+2 A B+B^2[A s, A B=B A]$
$=R H S$
Hence proved.
View full question & answer→Question 772 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 782 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→Question 792 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$(A-B)^2-(A-B)(A-B)$
$=A(A-B)-B(A-B)\{\text { using distributive property }\}$
$=A \times A-A B-B A+B \times B$
$=A^2-A B-B A+B^2$
$\neq A^2-2 A B+B^2$
Since, in general matrix multiplication is not commutative $(A B \neq B A)$,
So, $(A-B)^2 \neq A^2-2 A B+B^2$
View full question & answer→Question 802 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 812 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 822 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 832 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 842 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 852 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 862 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, $A^{\top}=A$ and $B^{\top}=B$.
Here,
$(A B-B A)^{\top}=(A B)^{\top}-(B A)^{\top}$
$\Rightarrow(A B-B A)^{\top}=B^{\top} A^{\top}-A^{\top} B^{\top}\left[\because(A B)^{\top}=B^{\top} A^{\top}\right]$
$\Rightarrow(A B-B A)^{\top}=B A-A B\left[\because B^{\top}=B \text { and } A^{\top}=A\right]$
$\Rightarrow(A B-B A)^{\top}=-(A B-B A)$
Therefore, $A B$ - $B A$ is skew - symmetric.
View full question & answer→Question 872 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 882 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 892 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 902 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 912 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 922 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 932 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 942 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 952 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:$a_{ij} = 2i - j$
AnswerHere,
$a_{11} = 2(1) -1 = 1, a_{12} = 2(1) -2 = 0, a_{13} = 2(1) -3 = -1$
$a_{21} = 2(2) -1 = 3, a_{22} = 2(2) -2 = 2, a_{23} = 2(2) -3 = 1$
Using equation (i)
$\text{A}=\begin{bmatrix}1 &0&-1\\3&2&1\end{bmatrix}$
View full question & answer→Question 962 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 972 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 982 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 992 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 1002 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
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