Question 1015 Marks
Compute the elements $a_{43}$ and $a_{22}$ of the matrix:
$\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}2&-1\\-3&2\\4&3\end{bmatrix}\begin{bmatrix}0&1&-1&2&-2\\3&-3&4&-4&0\end{bmatrix}$
AnswerWe have,
Given: $\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}2&-1\\-3&2\\4&3\end{bmatrix}\begin{bmatrix}0&1&-1&2&-2\\3&-3&4&-4&0\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0&3&2\\4&0&4\end{bmatrix}\begin{bmatrix}0-3&2+3&-2-4&4+4&-4-0\\0+6&-3-6&3+8&-6-8&6+0\\0+9&4-9&-4+12&8-12&-8+0\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}0&1&0\\2&0&2\\0 &3&2\\4&0&4\end{bmatrix}\begin{bmatrix}-3&5&-6&8&-4\\6&-9&11&-14&6\\9&-5&8&-4&-8\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}0+6+0&0-9-0&0+11+0&0-14-0&0+6-0\\-6+0+18&10-0-10&-12+0+16&16-0-8&-8+0-16\\0+18+18&0-27-10&0+33+16&0-42-8&0+18-16\\-12+0+36&20-0-20&-24+0+32&32-0-16&-16+0-32\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}6&-9&11&-14&6\\12&0&4&8&-24\\36&-37&49&-50&2\\24&0&8&16&-48\end{bmatrix}$
$\therefore\ \text{a}_{43}=8\text{ and a}_{22}=0$
View full question & answer→Question 1025 Marks
If possible, find BA and AB, where $\text{A}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}.$
AnswerWe have, $\text{A}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}$ and $\text{B}=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}_{3\times2}$
Since, in both AB and BA, the number of columns of first is equal to the number of rows of second.
So, AB and BA both are possible.
$\therefore\ \text{AB}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}8+2+2&2+3+4\\4+4+4&1+6+8\end{bmatrix}=\begin{bmatrix}12&9\\12&15\end{bmatrix}$
Also, $\text{BA}=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}$
$=\begin{bmatrix}8+1&4+2&8+4\\4+3&2+6&4+12\\2+2&1+4&2+8\end{bmatrix}=\begin{bmatrix}9&6&12\\7&8&16\\4&5&10\end{bmatrix}$
View full question & answer→Question 1035 Marks
If $\text{A}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix},$ then show that $A$ is a root of the polynomial $f(x) = x^3 - 6x^2 + 7x + 2.$
AnswerGiven: $f(x) = x^3 - 6x^2 + 7x + 2$
$f(A) = A^3 - 6A^2 + 7A + 2I_3$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$ \Rightarrow\text{A}^3=\begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}$
$\Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$ \Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2\end{bmatrix}$
$ \Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Since $f(A) = 0, A$ is the root of $f(x) = x^3 - 6x^2 + 7x + 2.$
View full question & answer→Question 1045 Marks
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ show that $\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $\text{A}^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}.$
AnswerGiven, $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1 \end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0&1+1\\0+0&0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$
$ \text{A}^3=\text{A}^2\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1&2\\0&1 \end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1+0&1+2\\0+0&0+1 \end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1&3\\0&1 \end{bmatrix}$
Hence proved.
View full question & answer→Question 1055 Marks
Show that the matrix $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ satisfies the equation $A^3 - 4A^2 + A = 0.$
AnswerGiven, $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^3=\text{A}^2.\text{A}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}14+12&21+24\\8+7&12+14\end{bmatrix}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}$
Hence, $\text{A}^3-4\text{A}^2+\text{A}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}-4\begin{bmatrix}7&12\\4&7\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}26-28+2&45-48+3\\15-16+1&26-28+2\end{bmatrix}$
$ =\begin{bmatrix}0&0\\0&0\end{bmatrix}$
So, $ \text{A}^3-4\text{A}+\text{A}=0$
View full question & answer→Question 1065 Marks
If $\text{A}=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix},$ show that $A^2 - 7A + 10I_3 = 0.$
AnswerGiven,
$\text{A}=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}$
$ \text{A}^2-7\text{A}+10\text{I}_3$
$ =\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}-7\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}+10\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$ =\begin{bmatrix}9+2+0&6+8+0&0+0+0\\3+4+0&2+16+0&0+0+0\\0+0+0&0+0+0&0+0+25\end{bmatrix}-\begin{bmatrix}21&14&0\\7&28&0\\0&0&35\end{bmatrix}+\begin{bmatrix}10&0&0\\0&10&0\\0&0&10\end{bmatrix} $
$=\begin{bmatrix}11&14&0\\7&18&0\\0&0&25\end{bmatrix}-\begin{bmatrix}21&14&0\\7&28&0\\0&0&35\end{bmatrix}+\begin{bmatrix}10&0&0\\0&10&0\\0&0&10\end{bmatrix}$
$=\begin{bmatrix}11-21+10&14-14+0&0-0+0\\7-7+0&18-28+10&0-0+0\\0-0+0&0-0+0&25-35+10\end{bmatrix}$
$ =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Hence,
$ \text{A}^2-7\text{A}+10\text{I}_3=0$
View full question & answer→Question 1075 Marks
If $\text{A}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix},$ find $A^T - B^T.$
AnswerGiven: $\text{A}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}$
Now,
$\text{A}^\text{T}-\text{B}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}-\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}$
$=\begin{bmatrix}3+1&4-1\\-1-2&2-2\\0-1&1-3\end{bmatrix}$
$=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
Therefore,
$\text{A}^\text{T}-\text{B}^\text{T}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
View full question & answer→Question 1085 Marks
If $f(x) = x^2 - 2x,$ find $f(A),$ where $\text{A}=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
AnswerGiven: $f(x) = x^2 - 2x$
$f(A) = A^2 - 2A$
Now,
$\text{A}^2=\text{AA}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}0+4+0&0+5+4&0+0+6\\0+20+0&4+25+0&8+0+0\\0+8+0&0+10+6&0+0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}$
$\text{f(A)}=\text{A}^2-2\text{A}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-2\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-\begin{bmatrix}0&2&4\\8&10&0\\0&4&6\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4-0&9-2&6-4\\20-8&29-10&8-0\\8-0&16-4&9-6\end{bmatrix}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&7&2\\12&19&8\\8&12&3\end{bmatrix}$
View full question & answer→Question 1095 Marks
If $\text{A} = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix},$ prove that $\text{A}'' = \begin{bmatrix}3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\end{bmatrix}n \in \text{N}.$
AnswerGiven: $\text{A}=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
Let p(n): $\text A^{n}=\begin{bmatrix}3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\end{bmatrix}\ $
p(1): $\Rightarrow \ \ \ \text A=\begin{bmatrix}3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0}\end{bmatrix}\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
$\therefore$ p(1) is true for n = 1.
Now p(k): $\text A^{k}=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}...\text{(ii)}$
Multiplying eq. (ii) by eq. (i) $\text{A}^{k}\text{A}=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
$\Rightarrow \text{ A}^ \text{k+1}=\begin{bmatrix}3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k}\end{bmatrix}=\text{p(k+1)}$
Therefore, p(n) is true for all natural numbers by P.M.I.
View full question & answer→Question 1105 Marks
Find the value of x if $\begin{bmatrix}1&\text{x}&1\end{bmatrix}\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}=0$
AnswerWe have, $\begin{bmatrix}1&\text{x}&1\end{bmatrix}_{1\times3}\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix}_{3\times3}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}_{3\times1}=0$
$\Rightarrow\ \begin{bmatrix}1+2\text{x}+15&3+5\text{x}+3&2+\text{x}+2\end{bmatrix}_{1\times3}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}_{3\times1}=0$
$\Rightarrow\ \begin{bmatrix}16+2\text{x}&5\text{x}+6&\text{x}+4\end{bmatrix}_{1\times3}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}_{3\times1}=0$
$\Rightarrow\ \begin{bmatrix}16+2\text{x}+10\text{x}+12+\text{x}^2+4\text{x}\end{bmatrix}=0$
$\Rightarrow\ [\text{x}^2+16\text{x}+28]=0$
$\Rightarrow\ \text{x}^2+16\text{x}+28=0$
$\Rightarrow\ (\text{x}+2)(\text{x}+4)=0$
$\Rightarrow\ \text{x}=-2,-14$
View full question & answer→Question 1115 Marks
Let $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix},$ Find $A^T, B^T$ and verify that.$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}+\text{A}^\text{T}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$ $\text{A}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ and $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$ $\text{AB}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$
$\Rightarrow\ \text{AB}=\begin{bmatrix}1-2+0&2-1+0&3-3+0\\2+2+0&4+1+3&6+3+3\\1+4+0&2+2+1&3+6+1\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-1&1&0\\4&8&12\\5&5&10\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(1)$
Now, $\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$
$\Rightarrow\text{B}^\text{A}\text{A}^\text{T}=\begin{bmatrix}1-2+0&2+2+0&1+4+0\\2-1+0&4+1+3&2+2+1\\3-3+0&6+3+3&3+6+1\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(2)$
$\Rightarrow(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T} [$from eqs. $(1)$ and $(2)]$
View full question & answer→Question 1125 Marks
Without using the concept of inverse of a matrix, find the matrix $\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}$ such that $\begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}5\text{x}-7\text{z}&5\text{y}-7\text{u}\\-2\text{x}+3\text{z}&-2\text{y}+3\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$ \therefore\ 5\text{x}-7\text{z}=-16\ \dots(1)$
$5\text{y}-7\text{u}=-6\ \dots(2)$
$-2\text{y}+3\text{u}=2$
$ \Rightarrow3\text{u}=2+2\text{y}$
$\Rightarrow\text{u}=\frac{2+2\text{y}}{3}\ \dots(3)$
$-2\text{x}+3\text{z}=7$
$\Rightarrow3\text{z}=7+2\text{x}$
$ \Rightarrow\text{z}=\frac{7+2\text{x}}{3}\ \dots(4)$
Putting the value of z in eq. (1), we get
$5\text{x}-7\Big(\frac{7+2\text{x}}{3}\Big)=-16$
$\Rightarrow5\text{x}-\frac{49+14\text{x}}{3}=-16$
$ \Rightarrow\frac{15\text{x}-49+14\text{x}}{3}=-16$
$\Rightarrow\text{x}-49=-48$
$\Rightarrow\text{x}=-48+49$
$\therefore\ \text{x}=1$
Putting the value of x in eq. (4), we get
$\text{z}=\frac{7+2(1)}{3}$
$\text{z}=\frac{9}{3}=3$
Putting the value of u in eq. (2), we get
$5\text{y}-7\Big(\frac{2+2\text{y}}{3}\Big)=-6$
$\Rightarrow5\text{y}-\frac{14+14\text{y}}{3}=-6$
$\Rightarrow\frac{15\text{y}-14+14\text{y}}{3}=-6$
$ \Rightarrow\text{y}-14=-18$
$\Rightarrow\text{y}=-18+14$
$ \Rightarrow\text{y}=-4$
Putting the value of y in eq. (3), we get
$ \text{u}=\frac{2+2(-4)}{3}$
$ \Rightarrow\text{u}=-2$
$\therefore\ \begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
View full question & answer→Question 1135 Marks
If $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -l,$ then show that $(A + B)^2 = A^2 + B^2.$
AnswerWe have, $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -1$
$\therefore\ (\text{A}+\text{B})=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$
$\therefore\ (\text{A}+\text{B})^2=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$
$=\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ ....(\text{i})$
Also, $\text{A}^2=\text{A}.\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}$
$=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}$
And $\text{B}^2=\text{B}.\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\therefore\ \text{A}^2+\text{B}^2=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}+\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ ....(\text{ii})$
From eq. $(i)$ and $(ii),$ we have $(\text{A}+\text{B})^2=\text{A}^2+\text{B}^2$
View full question & answer→Question 1145 Marks
If the matrix $\begin{bmatrix}0&\text{a}&3\\2&\text{b}&-1\\\text{c}&1&0\end{bmatrix}$ is a skew-symmetric matrix, then find the values of a, b and c.
AnswerLet $\text{A}=\begin{bmatrix}0&\text{a}&3\\2&\text{b}&-1\\\text{c}&1&0\end{bmatrix}$ Since, A is skew-symmetric matrix. $\therefore\ \text{A}'=-\text{A}$$\Rightarrow\ \begin{bmatrix}0&2&\text{c}\\\text{a}&\text{b}&1\\3&-1&0\end{bmatrix}=\begin{bmatrix}0&\text{a}&3\\2&\text{b}&-1\\\text{c}&1&0\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}0&2&\text{c}\\\text{a}&\text{b}&1\\3&-1&0\end{bmatrix}=\begin{bmatrix}0&-\text{a}&-3\\-2&-\text{b}&1\\-\text{c}&-1&0\end{bmatrix}$ By equality of matrices, we get a = -2, c = -3 and b = -b ⇒ b = 0 $\therefore$ a = -2, b = 0 and c = -3
View full question & answer→Question 1155 Marks
If $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then find $\lambda,\mu$ so that $\text{A}^2=\lambda\text{A}+\mu\text{I}$
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^2=\lambda\text{A}=\mu\text{I}$
$\Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\lambda\begin{bmatrix}2&3\\1&2\end{bmatrix}+\mu\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda&3\lambda\\1\lambda&2\lambda\end{bmatrix}+\begin{bmatrix}\mu&0\\0&\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda+0\\\lambda+0&2\lambda+\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda\\\lambda&2\lambda+\mu\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$ \therefore\ 7=2\lambda+\mu\ \dots(1)$
$ 12=3\lambda$
$\Rightarrow\lambda=\frac{12}{3}=4$
Putting the value of $\lambda$ in eq. (1), we get
$7=2(4)+\mu$
$\Rightarrow7-8=\mu$
$\therefore\ \mu=-1$
View full question & answer→Question 1165 Marks
Find a 2 × 2 matrix A such that.
$\text{A}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
AnswerLet $\text{A}=\begin{bmatrix}\text{w}&\text{x}\\\text{y}&\text{z}\end{bmatrix}$
Now,
$\begin{bmatrix}\text{w}&\text{x}\\\text{y}&\text{z}\end{bmatrix}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
$ \Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=6\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=\begin{bmatrix}6&0\\0&6\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ w + x = 6
⇒ w = 6 - x ...(1)
-2w + 4x = 0 ...(2)
Putting the value of w in eq. (2), we get
-2(6 - x) + 4x = 0
⇒ -12 + 2x + 4x = 0
⇒ -12 + 6x = 0
⇒ 6x = 12
⇒ x = 2
Putting the value of x in eq. (1), we get
w = 6 - 2
⇒ w = 4
Now,
y + z = 0
⇒ y = -z ...(3)
-2y + 4z = 6 ...(4)
Putting the value of y in eq. (4), we get
-2(-z) + 4z = 6
⇒ 2z + 4z = 6
⇒ 6z = 6
⇒ z = 1
Putting the value of z in eq. (3), we get
y = -1
$ \therefore\ \text{A}=\begin{bmatrix}4&2\\-1&1\end{bmatrix}$
View full question & answer→Question 1175 Marks
$ \text{Find}\ \frac{1}{2}(\text{A}+\text{A}')\ \text{and}\frac{1}{2}(\text{A}-\text{A}'),\ \text{when}\ \text{A}=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}$
AnswerThe given matrix is $\text{A}=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix} $
$\therefore \text{A}'=\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}$
$\text{A + A}'=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}+\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\therefore\ \frac{1}{2}\text{(A + A}')=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
Now, $\text{A} - \text{A}'=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}-\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}$$=\begin{bmatrix}0&2\text{a}&2\text{b}\\-2\text{a}&0&2\text{c}\\-2\text{b}&-2\text{c}&0\end{bmatrix}$
$\therefore\ \frac{1}{2}\text{(A} -\text{A}')=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}$
View full question & answer→Question 1185 Marks
If $\text{A}=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix},$ then verify $A^2 + A = A(A + I),$ where $I$ is the identity matrix.
AnswerTo verify: $A^2 + A = A(A + I).$
Given: $\text{A}=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$ =\begin{bmatrix}1+0+0&0+0-3&-3+0-3\\2+2+0&0+1+3&-6+3+3\\0+2+0&0+1+1&0+3+1\end{bmatrix}$
$ =\begin{bmatrix}1&-3&-6\\4&4&0\\2&21&4\end{bmatrix}$
$\ce{LHS}:$
$ \text{A}^2+\text{A}=\begin{bmatrix}1&-3&-6\\4&4&0\\2&21&4\end{bmatrix}+\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$=\begin{bmatrix}1+1&-3+0&-6-3\\4+2&4+1&0+3\\2+0&2+1&4+1\end{bmatrix}$
$=\begin{bmatrix}2&-3&-9\\6&5&3\\2&3&5\end{bmatrix}$
$\ce{RHS}:$
$ \text{A}+\text{I}=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}+\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}1+1&0+0&-3+0\\2+0&1+0&3+0\\0+0&1+0&1+1\end{bmatrix}$
$ =\begin{bmatrix}2&0&-3\\2&2&3\\0&1&2\end{bmatrix}$
$ \text{A}(\text{A}+\text{I})=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}2&0&-3\\2&2&3\\0&1&2\end{bmatrix}$
$ =\begin{bmatrix}2+0+0&0+0-3&-3+0-6\\4+2+0&0+2+3&-6+3+6\\0+2+0&0+2+1&0+3+2\end{bmatrix}$
$=\begin{bmatrix}2&-3&-9\\6&5&3\\2&3&5\end{bmatrix}$
Therefore, $\ce{LHS = RHS}$
Hence, $A^2 + A = A(A + I)$ is verified.
View full question & answer→Question 1195 Marks
Solve the matrix equations:
$\begin{bmatrix}\text{x}&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
AnswerHere,
$\begin{bmatrix}\text{x}&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}\text{x}-2&0-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}(\text{x}-2)\text{x}-15\end{bmatrix} =0$
$\Rightarrow\text{x}^2-2\text{x}-15=0$
$ \Rightarrow\text{x}^2-5\text{x}+3\text{x}-15=0$
$ \Rightarrow\text{x}(\text{x}-5)+3(\text{x}-5)=0$
$\Rightarrow(\text{x}-5)(\text{x}+3)=0$
$ \Rightarrow\text{x}-5=0\ \text{or}\ \text{x}+3=0$
$ \Rightarrow\text{x}=5\ \text{or}\ \text{x}=-3$
So,
$\text{x}=5\text{ or }-3$
View full question & answer→Question 1205 Marks
If $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix},$ prove that $\text{A}^\text{n}=\begin{bmatrix}\text{a}^\text{n}&\text{b}\Big(\frac{\text{a}^\text{n}-1}{\text{a}-1}\Big)\\0&1\end{bmatrix}$ for every positive integer n.
AnswerWe shall prove the result by the principle of mathematical induction on n. Step 1: If n = 1, by definition of integral power of a matrix, we have $\text{A}^1=\begin{bmatrix}\text{a}^1&\text{b}\frac{(\text{a}^1-1)}{\text{a}-1}\\0&1\end{bmatrix}=\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix}=\text{A}$So, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
$\text{A}^{\text{m}}=\begin{bmatrix}\text{a}^{\text{m}}&\text{b}\frac{(\text{a}^{\text{m}}-1)}{\text{a}-1}\\0&1\end{bmatrix}\ \dots(1)$
Now, we shall show that the result is true for n = m + 1.
Here,
$\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\text{b}\frac{(\text{a}^{\text{m}+1}-1)}{\text{a}-1}\\0&1\end{bmatrix}$
By definition of integral power of a matrix, we have
$\text{A}^{\text{m}+1}=\text{A}^\text{m}\text{A}$
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^\text{m}&\text{b}\frac{(\text{a}^\text{m}-1)}{\text{a}-1}\\0&1&\end{bmatrix}\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix}$ [From eq. (1)]
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^\text{m}\text{a}+0&\frac{\big\{\text{a}^\text{m}\text{b}+\text{b}(\text{a}^\text{m}-1)\big\}}{\text{a}-1}\\0+0&0+1\end{bmatrix}$
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\frac{(\text{a}^{\text{m}+1}\text{b}-\text{a}^\text{m}\text{b}+\text{a}^\text{m}\text{b}-\text{b})}{\text{a}-1}\\0&1\end{bmatrix}$
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\text{b}\frac{(\text{a}^{\text{m}+1}-1)}{\text{a}-1}\\0&1\end{bmatrix}$
This shows that when the result is true for n = m, it is also true for n = m + 1.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.
View full question & answer→Question 1215 Marks
If $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}-1&3&5\\1&-3&-5\\-1&3&5\end{bmatrix},$ show that $AB = BA = O_{3\times 3}$
AnswerHere,
$\text{AB}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\begin{bmatrix}-1&3&5\\1&-3&-5\\-1&3&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-2-3+5&6+9-15&10+15-25\\1+4-5&-3-12+15&-5-20 +25\\-1-3+4&3+9-12&5+15-20\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{AB}=0_{3\times3}\ \dots(1)$
$\text{BA}=\begin{bmatrix}-1&3&5\\1&-3&-5\\-1&3&5\end{bmatrix}\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$ \Rightarrow\text{BA}=\begin{bmatrix}-2-3+5&3+12-15&5+15-20\\2+3-5&-3-12+15&-5-15+20\\-2-3+5&3+12-15&5+15-20\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{BA}=0_{3\times3}\ \dots(2)$
$\Rightarrow\text{AB}=\text{BA}=0_{3\times3} [$From $eqs. (1)$ and $(2)]$
View full question & answer→Question 1225 Marks
If $\text{A}=\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix},$ show that $A^2 = 0$
AnswerGiven: $\text{A}=\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix}\begin{bmatrix}\text{ab}&\text{b}^2\\-\text{a}^2&-\text{ab}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{a}^2\text{b}^2-\text{a}^2\text{b}^2&\text{ab}^3-\text{ab}^3\\-\text{a}^3\text{b}+\text{a}^3\text{b}&-\text{a}^2\text{b}^2+\text{a}^2\text{b}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2=0.$
Hence proved.
View full question & answer→Question 1235 Marks
For the matrices $A$ and $B,$ verify that $(AB)^T = B^TA^T,$ where $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix},\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}1+6&4+15\\2+8&8+20\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}7&19\\10&28\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(1)$
Also,
$\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1+6&2+8\\4+15&8+20\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(2)$
$\therefore\ (\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T} [$From eqs. $(1)$ and $(2)]$
View full question & answer→Question 1245 Marks
If $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix},$ find $k$ such that $A^2 = kA - 2I_2.$
AnswerGiven: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\text{A}^2=\text{kA}-2\text{I}_2$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\text{k}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3\text{k}&-2\text{k}\\4\text{k}&-2\text{k}\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&-2\\4&-4\end{bmatrix}=\begin{bmatrix}3\text{k}-2&-2\text{k}-0\\4\text{k}-0&-2\text{k}-2\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore\ 1=3\text{k}-2$
$\Rightarrow1+2=3\text{k}$
$\Rightarrow3=3\text{k}$
$\Rightarrow\text{k}=1$
View full question & answer→Question 1255 Marks
If $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}, f(x) = x^2 - 2x - 3,$ show that $f(A) = 0$
AnswerGiven: $\text{A}=\begin{bmatrix}1&2\\2&1\end{bmatrix}$ and $f(x) = x^2 - 2x - 3$
$\text{f(A)}=\text{A}^2-2\text{A}-3\text{I}$
$=\begin{bmatrix}1&2\\2&1\end{bmatrix}\begin{bmatrix}1&2\\2&1\end{bmatrix}-2\begin{bmatrix}1&2\\2&1\end{bmatrix}-3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}1+4&2+2\\2+2&4+1\end{bmatrix}-\begin{bmatrix}2&4\\4&2\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$ =\begin{bmatrix}5&4\\4&5\end{bmatrix}-\begin{bmatrix}2&4\\4&2\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$=\begin{bmatrix}5-2-3&4-4-0\\4-4-0&5-2-3\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
So,
$\text{f(A)}=0$
View full question & answer→Question 1265 Marks
If $A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$, prove that $A^3 - 6A^2 + 7A + 2I = 0.$
Answer$\ce{L.H.S = A^3 - 6A^2 + 7A + 2I}$
$A^3 =\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$A^3 =\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\text{A}^3=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$A^3=\begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}$$=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}...\text{(i)}$
$6A^2 =-6\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$$= 6\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}$
$6A^2 =\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}...\text{(ii)}$
$7A + 2I =7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$7A + 2I =\begin{bmatrix}7+2&0+0&14+0\\0+0&14+2&7+0\\14+0&0+0&21+2\end{bmatrix}$$ =\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}...\text{(iii)}$
Now from $(i), (ii),$ and $(iii)$ equation, we get
$A^3 - 6A^2 + 7A + 2I$
$=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}$
$=\begin{bmatrix}21-30&0-0&34-48\\12-12&8-24&23-30\\34-48&0-0&55-78\end{bmatrix}\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}$
$=\begin{bmatrix}-9&0&-14\\0&-16&-7\\-14&0&-23\end{bmatrix}+\begin{bmatrix}9&0&14\\0&16&7\\14&0&23\end{bmatrix}$
$=\begin{bmatrix}-9+9&0+0&-14+14\\0+0&-16+16&-7+7\\-14+14&0+0&-23+23\end{bmatrix}$
$=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}= 0 ($Zero matrix$) = \text{R.H.S.}$
View full question & answer→Question 1275 Marks
If $f(x) = x^3 + 4x^2 - x,$ find $f(A),$ where $\text{A}=\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
AnswerGiven,
$\text{A}=\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
And $\text{f(x)}=\text{x}^3+4\text{x}^2-\text{x}$
$\Rightarrow\text{f(x)}=\text{A}^3+4\text{A}^2-\text{A}\ \dots(\text{i})$
$\text{A}^2=\text{AA}$
$=\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
$=\begin{bmatrix}0+2+2&0-3-2&0+0+0\\0-6+0&2+9+0&4+0+0\\0-2+0&1+3+0&2+0+0\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}4&-5&0\\-6&11&4\\-2&4&2\end{bmatrix}$
$\text{A}^3=\text{A}^2\times\text{A}$
$ =\begin{bmatrix}4&-5&0\\-6&11&4\\-2&4&2\end{bmatrix}\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
$=\begin{bmatrix}0-10+0&4+15+0&8+0+0\\0+22+4&-6-33-4&-12+0+0\\0+8+2&-2-12-2&-4+0+0\end{bmatrix}$
$=\begin{bmatrix}-10&19&8\\26&-43&-12\\10&-16&-4\end{bmatrix}$
Put the value of $A, A^2, A^3$ in equation $(i)$
$ \text{f(A)}=\text{A}^3+4\text{A}^2-\text{A}$
$=\begin{bmatrix}-10&19&8\\26&-43&-12\\10&-16&-4\end{bmatrix}+4\begin{bmatrix}4&-5&0\\-6&11&4\\-2&4&2\end{bmatrix}-\begin{bmatrix}0&1&2\\2&-3&0\\1&-1&0\end{bmatrix}$
$ =\begin{bmatrix}-10+16-0&19-20-1&8+0-2\\26-24-2&-43+44+3&-12+16+0\\10-8-1&-16+16+1&-4+8-0\end{bmatrix}$
$ =\begin{bmatrix}6&-2&6\\0&4&4\\1&1&4\end{bmatrix}$
Hence,
$\text{f(A)} =\begin{bmatrix}6&-2&6\\0&4&4\\1&1&4\end{bmatrix}$
View full question & answer→Question 1285 Marks
If $\text{P}\big(\text{x}\big)=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix},$ then show that P(x)P(y) = P(x + y) = P(y)P(x).
AnswerGiven: $\text{P}\big(\text{x}\big)=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
Then, $\text{P}\big(\text{y}\big)=\begin{bmatrix}\cos\text{y}&\sin\text{y}\\-\sin\text{y}&\cos\text{y}\end{bmatrix}$
Now,
$\text{P}\big(\text{x}\big)\text{P}\big(\text{y}\big)=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}\begin{bmatrix}\cos\text{y}&\sin\text{y}\\-\sin\text{y}&\cos\text{y}\end{bmatrix}$
$=\begin{bmatrix}\cos\text{x}\cos\text{y}-\sin\text{x}\sin\text{y}&\cos\text{x}\sin\text{y}+\sin\text{x}\cos\text{y}\\-\sin\text{x}\cos\text{y}-\cos\text{x}\sin\text{y}&-\sin\text{x}\sin\text{y}+\cos\text{x}\cos\text{y}\end{bmatrix}$
$=\begin{bmatrix}\cos(\text{x}+\text{y})&\sin(\text{x}+\text{y})\\-\sin(\text{x}+\text{y})&\cos(\text{x}+\text{y})\end{bmatrix}\ \dots(1)$
Also, $\text{P}\big(\text{x}+\text{y}\big)=\begin{bmatrix}\cos(\text{x}+\text{y})&\sin(\text{x}+\text{y})\\-\sin(\text{x}+\text{y})&\cos(\text{x}+\text{y})\end{bmatrix}\ \dots(2)$
Now,
$=\text{P}\big(\text{y}\big)\text{P}\big(\text{x}\big)=\begin{bmatrix}\cos\text{y}&\sin\text{y}\\-\sin\text{y}&\cos\text{y}\end{bmatrix}\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$=\begin{bmatrix}\cos\text{y}\cos\text{x}-\sin\text{y}\sin\text{x}&\cos\text{y}\sin\text{x}+\sin\text{y}\cos\text{x}\\-\sin\text{y}\cos\text{x}\cos\text{y}\sin\text{x}&\sin\text{y}\sin\text{x}+\cos\text{y}\cos\text{x}\end{bmatrix}$
$=\begin{bmatrix}\cos(\text{x}+\text{y})&\sin(\text{x}+\text{y})\\-\sin(\text{x}+\text{y})&\cos(\text{x}+\text{y})\end{bmatrix}\ \dots(3)$
From (1), (2) and (3), we get
P(x)P(y) = P(x + y) = P(y)P(x)
View full question & answer→Question 1295 Marks
If $\text{A}=\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix},$ prove that
$ \text{A}^2=\begin{bmatrix}\cos\text{n}\alpha+\sin\text{n}\alpha&\sqrt{2}\sin\text{n}\alpha\\-\sqrt{2}\sin\text{n}\alpha&\cos\text{n}\alpha-\sin\text{n}\alpha\end{bmatrix}$ for all $\text{n}\in\text{N}.$
AnswerGiven,
$\text{A}=\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix}$
To prove $P(n): \text{A}^2=\begin{bmatrix}\cos\text{n}\alpha+\sin\text{n}\alpha&\sqrt{2}\sin\text{n}\alpha\\-\sqrt{2}\sin\text{n}\alpha&\cos\text{n}\alpha-\sin\text{n}\alpha\end{bmatrix}$ we use mathematical induction.
Step $1:$ To show $P(1)$ is true.
$A^n$ is true for $n = 1$
Step $2:$ Let $P(k)$ be true, So
$\text{A}^\text{k}=\begin{bmatrix}\cos\text{k}\alpha+\sin\text{k}\alpha&\sqrt{2}\sin\text{k}\alpha\\-\sqrt{2}\sin\text{k}\alpha&\cos\text{k}\alpha-\sin\text{k}\alpha\end{bmatrix}$
Step $3:$ Let $P(k)$ is true.
Now, we have to show that
$ \text{A}^\text{k+1}=\begin{bmatrix}\cos(\text{k}+1)\alpha+\sin(\text{k}+1)\alpha&\sqrt{2}\sin(\text{k}+1)\alpha\\-\sqrt{2}\sin(\text{k}+1)\alpha&\cos(\text{k}++1)\alpha-\sin(\text{k}+1)\alpha\end{bmatrix}$
Now,
$\text{A}^{\text{k}+1}=\text{A}^\text{k}\times\text{A}$
$=\begin{bmatrix}\cos\text{k}\alpha+\sin\text{k}\alpha&\sqrt{2}\sin\text{k}\alpha\\-\sqrt{2}\sin\text{k}\alpha&\cos\text{k}\alpha\sin\text{k}\alpha\end{bmatrix}\begin{bmatrix}\cos\alpha+\sin\alpha&\sqrt{2}\sin\alpha\\-\sqrt{2}\sin\alpha&\cos\alpha-\sin\alpha\end{bmatrix}$
$ =\begin{bmatrix}(\cos\text{k}\alpha+\sin\text{k}\alpha)(\cos\alpha+\sin\alpha)-2\sin\alpha\sin\text{k}\alpha&(\cos\text{k}\alpha+\sin\text{k}\alpha)\sqrt{2}\sin\alpha+\sqrt{2}\sin\text{k}\alpha(\sin\alpha-\cos\alpha)\$\cos\alpha+\sin\alpha)(-\sqrt{2}\sin\text{k}\alpha)-\sqrt{2}\sin\alpha(\cos\text{k}\alpha-\sin\text{k}\alpha)&-2\sin\text{k}\alpha\sin\alpha+(\cos\text{k}\alpha-\sin\text{k}\alpha)(\cos\alpha-\sin\alpha)\end{bmatrix}$
$ =\begin{bmatrix}\cos\text{k}\alpha\cos\alpha+\sin\text{k}\alpha\cos\alpha+\cos\text{k}\alpha\sin\alpha+\sin\alpha\sin\text{k}\alpha-2\sin\alpha\sin\text{k}\alpha&\sqrt{2}\cos\text{k}\alpha\sin\alpha+\sqrt{2}\sin\alpha\sin\text{k}\alpha+\sqrt{2}\sin\text{k}\alpha\cos\alpha-\sqrt{2}\sin\text{k}\alpha\sin\alpha\\-\sqrt{2}\cos\alpha\sin\alpha-\sqrt{2}\sin\alpha\sin\text{k}\alpha-\sqrt{2}\sin\alpha\cos\text{k}\alpha+\sqrt{2}\sin\alpha\sin\text{k}\alpha&-2\sin\text{k}\alpha\sin\alpha+\cos\text{k}\alpha\cos\alpha-\cos\alpha\sin\text{k}\alpha-\sin\alpha\cos\text{k}\alpha\sin\alpha\sin\text{k}\alpha\end{bmatrix}$
$ =\begin{bmatrix}\cos\alpha\cos\text{k}\alpha+\sin\alpha\sin\text{k}\alpha+\sin\alpha\cos\text{k}\alpha+\sin\text{k}\alpha\cos\alpha&\sqrt{2}(\sin\text{k}\alpha\cos\alpha+\cos\text{k}\alpha\sin\alpha)\\-\sqrt{2}(\sin\text{k}\alpha\cos\alpha+\cos\text{k}\alpha\sin\alpha)&\cos\text{k}\alpha\cos\alpha-\sin\text{k}\alpha\sin\alpha-(\sin\text{k}\alpha\cos\alpha+\sin\alpha\cos\text{k}\alpha)\end{bmatrix}$
$ =\begin{bmatrix}\cos(\text{k}+1)\alpha+\sin(\text{k}+1)\alpha&\sqrt{2}\sin(\text{k}+1)\alpha\\-\sqrt{2}\sin(\text{k}+1)\alpha&\cos(\text{k}+1)\alpha-\sin(\text{k}+1)\alpha\end{bmatrix}$
So, $P(k + 1)$ is true whenever $P(k)$ is true.
Hence, by principle of mathematical induction $P(n)$ is true for all positive in teger.
View full question & answer→Question 1305 Marks
Solve the matrix equations:
$\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\\text{x}\end{bmatrix}=0$
Answer$\begin{bmatrix}1&2&1\end{bmatrix}\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}\begin{bmatrix}0\\2\\\text{x}\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}1+4+1&2+0+0&0+2+2\end{bmatrix}\begin{bmatrix}0\\2\\\text{x}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}6&2&4\end{bmatrix}\begin{bmatrix}0\\2\\\text{x}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}0+4+4\text{x}\end{bmatrix}=0$
$\Rightarrow4+4\text{x}=0$
$\Rightarrow4\text{x}=-4$
$\therefore\ \text{x}=\frac{-4}{4}=-1$
View full question & answer→Question 1315 Marks
If w is a complex cube root of unity, show that.
$\begin{pmatrix}\begin{bmatrix}1&w&w^2\\w&w^2&1\\w^2&1&w\end{bmatrix} +\begin{bmatrix}w&w^2&1\\w^2&1&w\\w&w^2&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
AnswerHere,
$\text{LHS}=\begin{pmatrix}\begin{bmatrix}1&w&w^2\\w&w^2&1\\w^2&1&w\end{bmatrix} +\begin{bmatrix}w&w^2&1\\w^2&1&w\\w&w^2&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}$
$=\begin{bmatrix}1+w&w+w^2&w^2+1\\w+w^2&w^2+1&1+w\\w^2+w&1+w^2&w+1\end{bmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}$
$=\begin{bmatrix}-w^2&-1&-w\\-1&-w&-w^2\\-1&-w&-w^2\end{bmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}$ $\big(\because1+\text{w}+\text{w}^2=0\text{ and w}^3=1\big)$
$=\begin{bmatrix}-w^2-w-w^3\\-1-w^2-w^4\\-1-w^2-w^4\end{bmatrix}$
$=\begin{bmatrix}-w(1+w+w^2)\\-1-w^2-w^3w\\-1-w^2-w^3w\end{bmatrix}$
$=\begin{bmatrix}-w\times0\\-1-w^2-w\\-1-w^2-w\end{bmatrix}$ $\big(\because1+\text{w}+\text{w}^2=0\text{ and w}^3=1\big)$
$=\begin{bmatrix}0\\-0\\-0\end{bmatrix}$
$=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$\therefore\ \begin{pmatrix}\begin{bmatrix}1&w&w^2\\w&w^2&1\\w^2&1&w\end{bmatrix} +\begin{bmatrix}w&w^2&1\\w^2&1&w\\w&w^2&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1\\w\\w^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
View full question & answer→Question 1325 Marks
If $\text{A}=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix},$ compute $A^2 - 4A + 3I_3.$
AnswerGiven: $\text{A}=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+6+0&2-8-0&0+10+0\\3-12+0&6+16-5&0-20+15\\0-3+0&0+4-3&0-5+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}$
$\text{A}^2-4\text{A}+3\text{I}_3$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}-4\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}-\begin{bmatrix}4&8&0\\12&-16&20\\0&-4&12\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7-4+3&-6-8+0&10-0+0\\-9-12+0&17+16+3&-5-20+0\\-3-0+0&1+4+0&4-12+3\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}6&-14&10\\-21&36&-25\\-3&5&-5\end{bmatrix}$
View full question & answer→Question 1335 Marks
Find the matrix A such that
$\text{A}=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$
AnswerIt is given that:
$\text{A}=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$
The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on the L.H.S. of the equation is a 2 × 3 matrix. Therefore, X has to be a 2 × 2 matrix.
$ \text{X}=\begin{bmatrix}\text{a}&\text{c}\\\text{b}&\text{d}\end{bmatrix}$
Therefore, we have:
$ \begin{bmatrix}\text{a}&\text{c}\\\text{b}&\text{d}\end{bmatrix}\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$
$\begin{bmatrix}\text{a}+4\text{c}&2\text{a}+5\text{c}&3\text{a}+6\text{c}\\\text{b}+4\text{d}&2\text{b}+5\text{d}&3\text{b}+6\text{d}\end{bmatrix}=\begin{bmatrix}-7&-8&-9\\2&4&6\end{bmatrix}$
Equating the corresponding elements of the two matrices, we have:
a + 4c = -7, 2a + 5c = -8, 3a + 6c = -9
b + 4d = 2, 2b + 5d = 4, 3b + 6d = 6
Now, a + 4c = -7 ⇒ a = -7 - 4c
$\therefore$ 2a + 5c = -8 ⇒ -14 - 8c + 5c = -8
⇒ -3c = 6
⇒ c = -2
$\therefore$ a = -7 - 4(-2) = -7 + 8 = 1
Now, b + 4d = 2 ⇒ b = 2 - 4d
$\therefore$ 2b + 5d = 4 ⇒ 4 - 8d + 5d = 4
⇒ -3d = 0
⇒ d = 0
$\therefore$ b = 2 - 4(0) = 2
Thus, a = 1, b = 2, c = -2, d = 0
Hence, the required matrix X is $\begin{bmatrix}1&-2\\2&0\end{bmatrix}.$
View full question & answer→Question 1345 Marks
Evaluate the following:
$\begin{bmatrix}1&2&3\end{bmatrix}\begin{bmatrix}1&0&2\\2&0&1\\0&1&2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
Answer$\begin{bmatrix}1&2&3\end{bmatrix}\begin{bmatrix}1&0&2\\2&0&1\\0&1&2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+4+0&0+0+3&2+2+6\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
$\Rightarrow\begin{bmatrix}5&3&10\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}$
$\Rightarrow[10+12+60]$
$\Rightarrow[82]$
View full question & answer→Question 1355 Marks
If $\text{A}=\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix},$ and $I$ is the identity matrix of order $3,$ show that $A^3 = pI + qA + rA^2.$
AnswerGiven,
$\text{A}=\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix},$
$\text{A}^2=\text{A}\times\text{A}$
$=\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix}$
$\begin{bmatrix}0+0+0&0+0+0&0+1+0\\0+0+\text{p}&0+0+\text{q}&0+0+\text{r}\\0+0+\text{pr}&\text{p}+0+\text{qr}&0+\text{q}+\text{r}^2\end{bmatrix}$
$\text{A}^3=\text{A}^2\times\text{A}$
$=\begin{bmatrix}0&0&1\\\text{p}&\text{q}&\text{r}\\\text{pr}&\text{p}+\text{qr}&\text{q}+\text{r}^2\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix}$
$=\begin{bmatrix}0+0+\text{p}&0 +0+\text{q}&0+0+\text{r}\\0+0+\text{pr}&\text{p}+0+\text{qr}&0+\text{q}+\text{r}^2\\0+0+\text{pq}+\text{pr}^2&\text{pr}+0+\text{q}^2+\text{qr}^2&0+\text{p}+\text{qr}+\text{qr}+\text{r}^2\end{bmatrix}$
$\text{A}^3= \begin{bmatrix}\text{p}&\text{q}&\text{r}\\\text{pr}&\text{p}+\text{qr}&\text{q}+\text{r}^2\\\text{pq}+\text{pr}^2&\text{pr}+\text{q}^2+\text{qr}^2&\text{p}+2\text{qr}+\text{r}^2\end{bmatrix}$
$\text{pI}+\text{qA}+\text{rA}^2$
$=\text{p} \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+\text{q}\begin{bmatrix}0&1&0\\0&0&1\\\text{p}&\text{q}&\text{r}\end{bmatrix}+\text{r}\begin{bmatrix}0&0&1\\\text{p}&\text{q}&\text{r}\\\text{pr}&\text{p}+\text{qr}&\text{q}+\text{r}^2\end{bmatrix}$
$= \begin{bmatrix}\text{p}+0+0&0+\text{q}+0&0+0+\text{r}\\0+0+\text{pr}&\text{p}+0+\text{qr}&0+\text{q}+\text{r}^2\\0+\text{pq}+\text{pr}^2&0+\text{q}^2+\text{pr}+\text{qr}^2&\text{p}+\text{qr}+\text{qr}+\text{r}^2\end{bmatrix}$
$\text{pI}+\text{qA}+\text{rA}^2$
$= \begin{bmatrix}\text{p}&\text{q}&\text{r}\\\text{pr}&\text{p}+\text{pr}&\text{q}+\text{r}^2\\\text{pq}+\text{pr}^2&\text{pr}+\text{q}^2+\text{qr}^2&\text{p}+2\text{qr}+\text{r}^2\end{bmatrix}$
View full question & answer→Question 1365 Marks
Let $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix},$ compute $A^2 - B^2.$
AnswerGiven: $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+3-5&-1-3-5&1+3-5\\-3-9+15&3+9+15&-3-9+15\\-5+15+25&5-15+25&-5+15+25\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&-9&-1\\3&27&3\\35&15&35\end{bmatrix}$
$\text{B}^2=\text{BB}$
$\Rightarrow\text{B}^2=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}$
$ \Rightarrow\text{B}^2=\begin{bmatrix}0+4-3&0-12+12&0-12+12\\0-3+3&4+9-12&3+9-12\\0+4-4&-4-12+16&-3-12+16\end{bmatrix}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A}^2-\text{B}^2$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-1&-9&-1\\3&27&3\\35&15&35\end{bmatrix}-\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-1-1&-9-0&-1-0\\3-0&27-1&3-0\\35-0&15-0&35-1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-2&-1&-9\\3&26&3\\35&15&34\end{bmatrix}$
View full question & answer→Question 1375 Marks
If $\text{A}=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix},$ find $A^2.$
AnswerGiven: $\text{A}=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}$
Now,
$\text{A}^2=\text{A.A}$
$=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}$
$=\begin{bmatrix}\cos^22\theta-\sin^22\theta&\cos2\theta\sin^2+\cos2\theta\sin^2\theta\\-\cos2\theta\sin^2\theta-\sin^2\theta\cos^2\theta&-\sin^22\theta+\cos^22\theta\end{bmatrix}$
$=\begin{bmatrix}\cos4\theta&2\sin^2\theta\cos^2\theta\\-2\sin^2\cos2\theta&\cos4\theta\end{bmatrix}$
$\begin{Bmatrix}\text{ since }\cos^2\theta-\sin^2\theta=\cos2\theta\end{Bmatrix}$
$=\begin{bmatrix}\cos4\theta&\sin4\theta\\-\sin4\theta&\cos4\theta\end{bmatrix}$
$\begin{Bmatrix}\text{ since }\sin^2\theta=2\sin\theta\cos\theta\end{Bmatrix}$
Hence,
$\text{A}^2=\begin{bmatrix}\cos4\theta&\sin4\theta\\-\sin4\theta&\cos4\theta\end{bmatrix}$
View full question & answer→Question 1385 Marks
Find inverse, by elementary row operations (if possible), of the following matrices.
$\begin{bmatrix}1&3\\-5&7\end{bmatrix}.$
AnswerLet $\text{A}=\begin{bmatrix}1&3\\-5&7\end{bmatrix}$ Consider, $\text{A}=\text{IA}$ $\Rightarrow\ \begin{bmatrix}1&3\\-5&7\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\text{A}$ $\Rightarrow\ \begin{bmatrix}1&3\\0&22\end{bmatrix}=\begin{bmatrix}1&0\\5&1\end{bmatrix}\text{A}$ $[\because\ \text{R}_2\rightarrow\ \text{R}_2+5\text{R}_1]$ $\Rightarrow\ \begin{bmatrix}1&3\\0&1\end{bmatrix}=\begin{bmatrix}1&0\\\frac{5}{22}&\frac{1}{22}\end{bmatrix}\text{A}$$[\because\ \text{R}_2\rightarrow\ \frac{1}{22}\text{R}_2]$
$\Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}\frac{7}{22}&\frac{-3}{22}\\\frac{5}{22}&\frac{1}{22}\end{bmatrix}\text{A}$ $[\because\ \text{R}_2\rightarrow\ \text{R}_1+3\text{R}_2]$ $\Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac{1}{22}\begin{bmatrix}7&-3\\5&1\end{bmatrix}\text{A}$ $\Rightarrow\ \text{A}^{-1}=\frac{1}{22}\begin{bmatrix}7&-3\\5&1\end{bmatrix}$
View full question & answer→Question 1395 Marks
Express the matrix $\text{A}=\begin{bmatrix}4&2&-1 \\3 & 5&7\\1&-2&1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
AnswerGiven,
$\text{A}=\begin{bmatrix}4&2&-1 \\3 & 5&7\\1&-2&1 \end{bmatrix}\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}4&3&1 \\2&5&-2\\-1&7&1 \end{bmatrix}$
Let $\text{x}=\frac{1}{2}(\text{A}+\text{A})^{\text{T}}$
$=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}4&2&-1 \\3&5&7\\1&-2&1 \end{bmatrix}+\begin{bmatrix}4&3&1 \\2&5&-2\\-1&7&1 \end{bmatrix} \end{pmatrix}$
$ =\frac{1}{2}\begin{bmatrix}4+4&2+3&-1+1 \\3+2&5+5&7-2\\1-1&-2+7&1+1 \end{bmatrix}$ $=\frac{1}{2}\begin{bmatrix}8&5&0 \\5&10&2\\0&5&2\end{bmatrix}=\begin{bmatrix}4&\frac{5}{2}&0 \\\frac{5}{2}&5&\frac{5}{2}\\0&\frac{5}{2}&1 \end{bmatrix}$
$\text{x}^{\text{T}}=\begin{bmatrix}4&\frac{5}{2}&0 \\\frac{5}{2} & 5&\frac{5}{2}\\0&\frac{5}{2}&1 \end{bmatrix}^{\text{T}}=\begin{bmatrix}4&\frac{5}{2}&0 \\\frac{5}{2} & 5&\frac{5}{2}\\0&\frac{5}{2}&1 \end{bmatrix}^{\text{T}}=\text{X}$
$\therefore$ x is symmetric matrix
Now,
$\text{y}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})$
$=\frac{1}{2}\begin{pmatrix} \begin{bmatrix}4&2&-1 \\3&5&7\\1&-2&1 \end{bmatrix}-\begin{bmatrix}4&3&1 \\2&5&-2\\-1&7&1 \end{bmatrix}\end{pmatrix} $
$=\frac{1}{2}\begin{bmatrix}4-4&2-3&-1-1 \\3-2&5-5&7+2\\1+1&-2-7&1-1 \end{bmatrix}=\frac{1}{2}\begin{bmatrix}0&-1&-2 \\1&0&9\\2&-9&0 \end{bmatrix}$
$\therefore\ \text{y}=\begin{bmatrix}0&\frac{-1}{2}&-1 \\\frac{1}{2}&0&\frac{9}{2}\\1&\frac{-9}{2}&0 \end{bmatrix}$
$\Rightarrow-\text{Y}^{\text{T}}=-\begin{bmatrix}0&\frac{-1}{2}&-1 \\\frac{1}{2} & 0&\frac{9}{2}\\1&\frac{-9}{2}&0 \end{bmatrix}=\begin{bmatrix}0&\frac{-1}{2}&-1 \\\frac{1}{2} &0 &\frac{9}{2}\\1&\frac{-9}{2}&0 \end{bmatrix}=\text{Y}$
⇒ y is a skew symmetric matrix.
Now,
$\text{X}+\text{Y}=\begin{bmatrix}4&\frac{5}{2}&0 \\\frac{5}{2} &5&\frac{5}{2}\\0&\frac{5}{2}&1 \end{bmatrix}+\begin{bmatrix}0&\frac{-1}{2}&-1 \\\frac{1}{2} &0 &\frac{9}{2}\\1&\frac{-9}{2}&0 \end{bmatrix}$
$=\begin{bmatrix}4+0&{\frac{5}{2}-\frac{1}{2}}&0-1 \\\frac{5}{2}+{\frac{1}{2}} &5+0&{\frac{5}{2}+\frac{9}{2}}\\0+1&{\frac{5}{2}-\frac{9}{2}}&1+0 \end{bmatrix}$
$=\begin{bmatrix}4&2&-1 \\3&5 & 7\\1&-2&1 \end{bmatrix}=\text{A}$
View full question & answer→Question 1405 Marks
Show that the matrix $\text{A}=\begin{bmatrix}5&3\\12&7\end{bmatrix}$ is root of the equation $A^2 - 12A - I = 0.$
AnswerGive: $\text{A}=\begin{bmatrix}5&3\\12&7\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}5&3\\12&7\end{bmatrix}\begin{bmatrix}5&3\\12&7\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}25+36&15+21\\60+84&36+49\end{bmatrix}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}61&36\\144&85\end{bmatrix}$
$\text{A}^2-12\text{A}-\text{I}$
$\Rightarrow\text{A}^2-12\text{A}-\text{I}=\begin{bmatrix}61&36\\144&85\end{bmatrix}-12\begin{bmatrix}5&3\\12&7\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-12\text{A}-\text{I}=\begin{bmatrix}61&36\\144&85\end{bmatrix}-\begin{bmatrix}60&36\\144&84\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-12\text{A}-\text{I}=\begin{bmatrix}61-60-1&36-36+0\\144-144+0&85-84-1\end{bmatrix}$
$\Rightarrow\text{A}^2-12\text{A}-\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
Since $A$ is satisfying the equation $A^2 - 12A - I, A$ is the root of the equation $A^2 - 12A - I.$
View full question & answer→Question 1415 Marks
If $A =$ diag $(a, b, c),$ show that $A^n =$ diag$(a^n, b^n, c^n)$ for all positive integer $n.$
AnswerGiven, $A =$ diag$(a, b, c)$ Show that, $A^n =$ diag$(a^n, b^n, c^n)$
Step $1:$ Put $n = 1 A^1 =$ diag$(a^1, b^1, c^1) A =$ diag$(a, b, c)$
So, $A^n$ is true for $n = 1$
Step $2:$ Let, An be true for $n = k,$
so, $A^k =$ diag$(a^k, b^k, c^k) ...(i)$
Step $3:$ Now, we have to show that, $A^{k+1} =$ diag$(a^{k+1}, b^{k+1}, c^{k+1})$
Now, $A^{k+1} = A^k \times k^3 =$ diag$(a^k, b^k, c^k) \times$ diag$(a, b, c)$ {using equation $(i)$ and given} $\text{A}^{\text{k}+1}=\begin{bmatrix}\text{a}^\text{k}&0&0\\0&\text{b}^\text{k}&0\\0&0&\text{c}^\text{k}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
$ =\begin{bmatrix}\text{a}^\text{k}\times\text{a}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{b}^\text{k}\times\text{b}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{c}^\text{k}\times\text{c}\end{bmatrix}$
$=\begin{bmatrix}\text{a}^\text{k+1}&0&0\\0&\text{b}^\text{k+1}&0\\0&0&\text{c}^\text{k+1}\end{bmatrix}$
$\text{A}^{\text{k}+1}=\text{diag}\big(\text{a}^{\text{k}+1},\text{b}^{\text{k}+1},\text{c}^{\text{k}+1}\big)$
So,$ P(n)$ is true for $n = k + 1$ whenever $P(n)$ is true for $n = k$
Hence, by principle of mathematical induction $A^n$ is true for all positive integer.
View full question & answer→Question 1425 Marks
Using elementary transformation, find the inverse of each of the matrices, $\begin{bmatrix}2& 0&-1 \\5 & 1&0\\0&1&3 \end{bmatrix}$
AnswerLet $\text{A}=\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix},$ Since $\text{A = IA}\ \Rightarrow\ \ \begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\text{A}$
$\Rightarrow \ \begin{bmatrix}2&0&-1\\1&1&2\\0&1&3\end{bmatrix}=\begin{bmatrix}1&0&0\\-2&1&0\\0&0&1\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \ \left[\text R_2\rightarrow \text R_2-2\text R_1\right]$
$\Rightarrow \ \begin{bmatrix}1&1&2\\2&0&-1\\0&1&3\end{bmatrix}=\begin{bmatrix}-2&1&0\\1&0&0\\0&0&1\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \ \ \ \ \ \left[\text R_1\leftrightarrow\text R_2\right]$
$\Rightarrow \ \begin{bmatrix}1&1&2\\0&-2&-5\\0&1&3\end{bmatrix}=\begin{bmatrix}-2&1&0\\5&-2&0\\0&0&1\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \ \left[\text R_2\rightarrow\text R_2-2\text R_1\right]$
$\Rightarrow\ \begin{bmatrix}1&1&2\\0&1&3\\0&-2&-5\end{bmatrix}=\begin{bmatrix}-2&1&0\\5&-2&0\\0&0&1\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \left[\text R_2\leftrightarrow \text R_3\right]$
$\Rightarrow\ \begin{bmatrix}1&0&-1\\0&1&3\\0&0&1\end{bmatrix}=\begin{bmatrix}-2&1&-1\\0&0&1\\5&-2&2\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \left[\text R_1\rightarrow \text R_1-\text R_2\ \text{and}\ \ \text R_3\rightarrow\text R_3+2 \text R_2\right]$
$\Rightarrow \ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\text{A}\ \ \ \ \ \ \ \ \ \left[\text R_1\rightarrow\text R_1+\text R_3 \ \text{and}\ \text R_2\rightarrow\text R_2-3\text R_3\right]$
$\therefore \ \text{ A}^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
View full question & answer→Question 1435 Marks
$\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&\text{x}\end{bmatrix}$ find x.
AnswerGiven: $\begin{bmatrix}2&3\\5&7\end{bmatrix}\begin{bmatrix}1&-3\\-2&4\end{bmatrix}=\begin{bmatrix}-4&6\\-9&\text{x}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2-6&-6+12\\5-14&-15+28\end{bmatrix}=\begin{bmatrix}-4&6\\-9&\text{x} \end{bmatrix}$
$\Rightarrow\begin{bmatrix}-4&6\\-9&\text{x} \end{bmatrix}=\begin{bmatrix}-4&6\\-9&\text{x} \end{bmatrix}$
$\Rightarrow\text{x}=3$
$\therefore\ \text{x}=13$
View full question & answer→Question 1445 Marks
Let $\text{A}=\begin{bmatrix}3 & 2&7 \\1 & 4&3\\-2&5&8 \end{bmatrix}.$ Find matrices X and Y such that X + Y = A, where X is a symmetric and Y is a skew-symmetric matrix.
AnswerGiven: $\text{A}=\begin{bmatrix}3 & 2&7 \\1 & 4&3\\-2&5&8 \end{bmatrix}$ $\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}3 & 1&-2 \\2 & 4&5\\7&3&8 \end{bmatrix}$Let $\text{x}=\frac{1}{2}(\text{A}+\text{A})^\text{T}$
$=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3&2&7 \\1&4&3\\-2&5&8 \end{bmatrix}+\begin{bmatrix}3&1 & -2 \\2 & 4&5\\7&3&8 \end{bmatrix}\ \end{pmatrix}=\begin{bmatrix}3&\frac{3}{2}&\frac{5}{2}\\\frac{3}{2}&4&4\\\frac{5}{2}&4&8 \end{bmatrix}$
Let $\text{y}=\frac{1}{2}(\text{A}+\text{A}^{\text{T}})$ $=\frac{1}{2}\begin{pmatrix}\begin{bmatrix}3&2&7 & \\1&4 & 3\\-2&5&8 \end{bmatrix}+\begin{bmatrix}3 & 1&-2 \\2 & 4&5\\7&3&8 \end{bmatrix}\end{pmatrix} =\begin{bmatrix}0&\frac{1}{2} & \frac{9}{2} \\\frac{-1}{2}& 0&-1\\\frac{-9}{2}&0&0 \end{bmatrix}$ $\text{x}^{\text{T}}=\begin{bmatrix}3 & \frac{3}{2}&\frac{5}{2} \\\frac{3}{2} & 4&4\\\frac{5}{2}&4&8 \end{bmatrix}^{\text{T}}=\begin{bmatrix}3&\frac{3}{2} & \frac{5}{2} \\\frac{3}{2} & 4&4\\\frac{5}{2}&4&8 \end{bmatrix}^{\text{T}}=\text{x}$ $\text{y}^{\text{T}}=\begin{bmatrix}0 & \frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} & 0&-1\\\frac{-9}{2}&1&0 \end{bmatrix}=\begin{bmatrix}0 & \frac{-1}{2}&\frac{-9}{2} \\\frac{1}{2} & 0&1\\\frac{9}{2} &-1&0\end{bmatrix}=-\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} & 0&-1\\\frac{-9}{2}&1&0 \end{bmatrix}=\text{y}$ Thus, x is a symmetric matrix and y is skew- symmrteic matrix. Now, $\text{x}+{\text{y}}=\begin{bmatrix}3 & \frac{3}{2}&\frac{5}{2} \\\frac{-3}{2} & 4&4\\\frac{5}{2}&4&8 \end{bmatrix}=\begin{bmatrix}0 & \frac{-1}{2}&\frac{-9}{2} \\\frac{1}{2} & 0&-1\\\frac{9}{2} &1&0\end{bmatrix}$ $=\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} & 0&-1\\\frac{-9}{2}&1&0 \end{bmatrix}=\begin{bmatrix}3&2&7 \\1&4 & 0\\-2&5&8 \end{bmatrix}=\text{A}$ $\therefore\ \text{x}=\begin{bmatrix}3&\frac{3}{2}&\frac{5}{2} \\\frac{3}{2} & 4&4\\\frac{5}{2}&4&8\end{bmatrix}$ and $\text{y}=\begin{bmatrix}0&\frac{1}{2}&\frac{9}{2} \\\frac{-1}{2} &0&-1\\\frac{-9}{3}&1&0\end{bmatrix}$
View full question & answer→Question 1455 Marks
Find $x, y, z$ if $\text{A}=\begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}\\\text{x}&-\text{y}&\text{z}\end{bmatrix}$ satisfies $A′ = A^{-1}.$
AnswerMatrix A is such that $A' = A^{-1}$
$\Rightarrow AA' = I$
$\Rightarrow\ \begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}\\\text{x}&-\text{y}&\text{z}\end{bmatrix}\begin{bmatrix}0&\text{x}&\text{x}\\2\text{y}&\text{y}&-\text{y}\\\text{z}&-\text{z}&\text{z}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}4\text{y}^2+\text{z}^2&2\text{y}^2-\text{z}^2&-2\text{y}^2+\text{z}^2\\2\text{y}^2-\text{z}^2&\text{x}^2+\text{y}^2+\text{z}^2&\text{x}^2-\text{y}^2-\text{z}^2\\-2\text{y}^2+\text{z}^2&\text{x}^2-\text{y}^2+\text{z}^2&\text{x}^2+\text{y}^2+\text{z}^2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$ $\Rightarrow2\text{y}^2-\text{z}^2=0$
$\Rightarrow2\text{y}^2=\text{x}^2$
$\Rightarrow4\text{y}^2+\text{z}^2=1$
$\Rightarrow2\text{x}^2+\text{z}^2=1$ $\text{z}=\pm\frac{1}{\sqrt{3}}$
$\therefore\text{y}^2=\frac{\text{z}^2}{2}$
$\Rightarrow\text{y}=\pm\frac{1}{\sqrt{6}}$ Also $\text{x}^2+\text{y}^2+\text{z}^2=1$
$\Rightarrow\text{x}^2=1-\text{y}^2-\text{z}^2=1-\frac{1}{6}-\frac{1}{3}$ $=1-\frac{3}{6}=\frac{1}{2}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt{2}}$
$\therefore\text{x}=\pm,\frac{1}{\sqrt{2}},\text{y}=\pm\frac{1}{\sqrt{6}}$ and $\text{z}=\pm\frac{1}{\sqrt{3}}$
View full question & answer→Question 1465 Marks
If $AB = BA$ for any two sqaure matrices, prove by mathematical induction that $(AB)^n = A^nB^n.$
AnswerLet $P(n) : (AB)^n = A^nB^n$
$\therefore P(1) : (AB)^1 = A^1B^1 $
$\Rightarrow AB = ABSo, P(1)$ is true.
Now, $P(k) : (AB)^k = A^kB^k \text{k}\in\text{N}$
So, $P(k)$ is true, whenever $P(k + 1)$ is true. $\therefore P(k + 1) $
$\Rightarrow (AB)^{k+1} = A^{k+1}.B^{k+1}$
$\therefore P(k + 1 : AB)^{k+1} = A^{k+1}B^{k+1} $
$\Rightarrow A^k.B^k.AB $
$\Rightarrow A^k.B^kBA $
$\Rightarrow A^kB^{k+1}A $
$\Rightarrow A^kA.B^{k+1} $
$\Rightarrow A^{k+1}B^{k+1} $
$\Rightarrow (A.B)^{k+1} = A^{k+1}B^{k+1} So, P(k+1)$ is true for all $\text{n}\in\text{N},$
whenever $P(k)$ is true. By mathematical induction $(AB) = A^nB^n$ is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 1475 Marks
If A = $\begin{bmatrix}0&-\tan\frac{\alpha}{2}\\ \tan\frac{\alpha}{2}&0\end{bmatrix}$ and I is the identity matrix of order 2, show that I + A = ( I - A) $\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin{\alpha}&\cos\alpha\end{bmatrix}.$
Answer$\text{L.H.S}=\text{I}+\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}+\begin{bmatrix}0&-\tan\frac\alpha{2}\\ \tan\frac\alpha{2}&0\end{bmatrix}=\begin{bmatrix}1&-\tan\frac\alpha{2}\\ \tan\frac\alpha{2}&1\end{bmatrix}$
Now, $\text{I}-\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}0&-\tan\frac\alpha{2}\\ \tan\frac\alpha{2}&0\end{bmatrix}=\begin{bmatrix}1&\tan\frac\alpha{2}\\ -\tan\frac\alpha{2}&1\end{bmatrix}$
R.HS. $=(\text{I - A})\begin{bmatrix}\cos \alpha&-\sin\alpha\\ \sin \alpha&\cos \alpha\end{bmatrix}=\begin{bmatrix}1&\tan\frac\alpha{2}\\-\tan\frac\alpha{2}&1\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix}$
$=\begin{bmatrix}\cos\alpha+\sin\alpha\tan\frac\alpha{2}&-\sin\alpha+\cos\alpha\tan\frac\alpha{2}\\-\cos\alpha\tan\frac\alpha{2}+\sin\alpha&\sin\alpha\tan\frac\alpha{2}+\cos\alpha\end{bmatrix} $
$=\begin{bmatrix}\cos\alpha+\sin\alpha\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}& -\sin\alpha+\cos\alpha\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}\\- \cos\alpha\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}+\sin\alpha&\sin\alpha\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}+\cos\alpha\end{bmatrix}$
$=\begin{bmatrix}\frac{\cos\alpha\cos\frac{\alpha}{2}+\sin\alpha\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}&\frac{-\sin\alpha\cos\frac{\alpha}{2}+\cos\alpha\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}\\\frac{-\cos\alpha\sin\frac{\alpha}{2}+\sin\alpha\cos\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}&\frac{\sin\alpha\sin\frac{\alpha}{2}+\cos\alpha\cos\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}\end{bmatrix}$
$=\begin{bmatrix}\frac{\cos\left(\alpha-\frac{\alpha}{2}\right)}{\cos\frac{\alpha}{2}}&\frac{-\sin\left(\alpha-\frac{\alpha}{2}\right)}{\cos\frac{\alpha}{2}}\\ \frac{\sin\left(\alpha-\frac{\alpha}{2}\right)}{\cos\frac{\alpha}{2}}&\frac{\cos\left(\alpha-\frac{\alpha}{2}\right)}{\cos\frac{\alpha}{2}}\end{bmatrix}=\begin{bmatrix}\frac{\cos\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}&\frac{-\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}\\ \frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}&\frac{\cos\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}\end{bmatrix}$
$= \begin{bmatrix}1&-\tan\frac{\alpha}{2}\\ \tan\frac{\alpha}{2}&1\end{bmatrix}$
$\therefore $ L.H.S = R.H.S
Proved
View full question & answer→Question 1485 Marks
If $\text{A}=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix},\text{ B}=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix},$ find matrix X such that 2A + 3X = 5B.
AnswerGiven, $2\text{A}+3\text{X}=5\text{B}$
$\Rightarrow2\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}+3\text{X}=5\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}$
$\Rightarrow\begin{bmatrix}4&-4\\8&4\\-10&2\end{bmatrix}+3\text{X}=\begin{bmatrix}40&0\\20&-10\\15&30\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}40&0\\20&-10\\15&30\end{bmatrix}-\begin{bmatrix}4&-4\\8&4\\-10&2\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}40-4&0+4\\20-8&-10-4\\15+10&30-2\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}36&4\\12&-14\\25&28\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{3}\begin{bmatrix}36&4\\12&-14\\25&28\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}12&\frac{4}{3}\\4&\frac{-14}{3}\\\frac{25}{3}&\frac{28}{3}\end{bmatrix}$
View full question & answer→Question 1495 Marks
If possible, using elementary row transformations, find the inverse of the following matrices:$\begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}.$
AnswerFor getting the inverse of the given matrix A by row elementary operations we may write the given matrix as A = IA$\therefore\ \begin{bmatrix}2&0&-1\\5&1&0\\0&1&3\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\text{A}$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\3&1&1\\0&1&3\end{bmatrix}\begin{bmatrix}1&0&0\\-1&1&0\\0&0&1\end{bmatrix}\text{A}\ [\because\ \text{R}_2\rightarrow\ \text{R}_2+\text{R}_1]$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\1&1&2\\2&1&2\end{bmatrix}\begin{bmatrix}1&0&0\\-2&1&0\\1&0&1\end{bmatrix}\text{A}\begin{bmatrix}\because\ \text{R}_2\rightarrow\ \text{R}_2+\text{R}_1\\\text{and }\text{R}_3\rightarrow\ \text{R}_3+\text{R}_1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\0&1&\frac{5}{2}\\4&1&1\end{bmatrix}\begin{bmatrix}1&0&0\\\frac{-5}{2}&1&0\\2&0&1\end{bmatrix}\text{A}\ \begin{bmatrix}\because\ \text{R}_3\rightarrow\ \text{R}_3+\text{R}_1\\\text{and }\text{R}_2\rightarrow\ \text{R}_2\frac{1}{2}\text{R}_1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\0&1&\frac{5}{2}\\0&1&3\end{bmatrix}\begin{bmatrix}1&0&0\\\frac{-5}{2}&1&0\\0&0&1\end{bmatrix}\text{A}\ [\because\ \text{R}_3\rightarrow\ \text{R}_3-2\text{R}_1]$
$\Rightarrow\ \begin{bmatrix}2&0&-1\\0&1&\frac{5}{2}\\0&0&\frac{1}{2}\end{bmatrix}\begin{bmatrix}1&0&0\\\frac{-5}{2}&1&0\\\frac{5}{2}&-1&1\end{bmatrix}\text{A}\ [\because\ \text{R}_3\rightarrow\ \text{R}_3-\text{R}_2]$
$\Rightarrow\ \begin{bmatrix}1&0&\frac{-1}{2}\\0&1&\frac{5}{2}\\0&0&1\end{bmatrix}\begin{bmatrix}\frac{1}{2}&0&0\\\frac{-5}{2}&1&0\\5&-2&2\end{bmatrix}\text{A}\ \begin{bmatrix}\because\ \text{R}_1\rightarrow\ \frac{1}{2}\text{R}_1\\\text{and }\text{R}_ 3\rightarrow\ 2\text{R}_3\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\text{A}\ \begin{bmatrix}\because\ \text{R}_1\rightarrow\ \text{R}_1\frac{1}{2}\text{R}_3\\\text{and }\text{R}_2\rightarrow\ \text{R}_2-\frac{5}{2}\text{R}_3\end{bmatrix}$
Hence, $\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$ is the inverse of given matrix.
View full question & answer→Question 1505 Marks
If $\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix},$ find $A^{16}$.
AnswerGiven,
$\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix}$
$ \text{A}^2=\text{A}\times\text{A}$
$ =\begin{bmatrix}0&0\\4&0\end{bmatrix}\begin{bmatrix}0&0\\4&0\end{bmatrix}$
$ =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
$ \text{A}^4=\text{A}^2\times\text{A}^2$
$=0\times0$
$=0$
$ \text{A}^{16}=\text{A}^4\times\text{A}^4$
$=0\times0$
$=0$
So,
$A^{16}$ is a null matrix
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