Question 515 Marks
Find A, if $\begin{bmatrix}4\\1\\3\end{bmatrix}\text{A}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}.$
AnswerWe have, $\begin{bmatrix}4\\1\\3\end{bmatrix}_{3\times1}\text{A}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}_{3\times3}$Let $\text{A}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\end{bmatrix}$
$\therefore\ \begin{bmatrix}4\\1\\3\end{bmatrix}_{3\times1}\begin{bmatrix}\text{x}&\text{y}&\text{z}\end{bmatrix}_{1\times3}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}_{3\times3}$
$\Rightarrow\ \begin{bmatrix}4\text{x}&4\text{y}&4\text{z}\\\text{x}&\text{y}&\text{z}\\3\text{x}&3\text{y}&3\text{z}\end{bmatrix}=\begin{bmatrix}-4&8&4\\-1&2&1\\-3&6&3\end{bmatrix}$
$\Rightarrow\ 4\text{x}=-4\Rightarrow\ \text{x}=-1,\ 4\text{y}=8$
$\Rightarrow\ \text{y}=2\ \text{and }4\text{z}=4$
$\Rightarrow\ \text{z}=1$
$\therefore\ \text{A}=\begin{bmatrix}-1&2&1\end{bmatrix}$
View full question & answer→Question 525 Marks
$\text{if} \ \text{A}'=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix},\text{and}\ \text{B}=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix},\text{then verify that}$
- $\text{(A+B)}'=\text{A}'+\text{B}'$
- $\text{(A}-\text{B)}'=\text{A}'-\text{B}'$
Answer
- It is know that $ \text{A} = (\text{A}')'$
Therefore, we have:
$\text{A}=\begin{bmatrix}3&-1&0\\4&2&1\end{bmatrix} $
$ \text{B}'=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}$
$\text{A + B}=\begin{bmatrix}3&-1&0\\4&2&1\end{bmatrix}+\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}=\begin{bmatrix}2&1&1\\5&4&4\end{bmatrix}$
$\therefore\text{(A + B)}'=\begin{bmatrix}2&5\\1&4\\1&4\end{bmatrix}$
$\text{A}'+\text{B}'=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}+\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}=\begin{bmatrix}2&5\\1&4\\1&4\end{bmatrix}$
Thus, we have verify that $ (\text{A} + \text{B})' = \text{A}' + \text{B}'$
- $\text{A} -\text{B}=\begin{bmatrix}3&-1&0\\4&2&1\end{bmatrix}-\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}=\begin{bmatrix}4&-3&-1\\3&0&-2\end{bmatrix}$
$\therefore\text{(A} - \text{B})'=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
$\text{A}' - \text{B}'=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
Thus, we have verify that $ (\text{A} - \text{B})' = \text{A}' - \text{B}'$ View full question & answer→Question 535 Marks
If $\text{A}=\begin{bmatrix}1&2\\-2&1\end{bmatrix},\ \text{B}=\begin{bmatrix}2&3\\3&-4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&0\\-1&0\end{bmatrix},$ verify $(\text{AB})\text{C}=\text{A}(\text{BC}).$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-2&1\end{bmatrix},\ \text{B}=\begin{bmatrix}2&3\\3&-4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&0\\-1&0\end{bmatrix}$
$\text{AB}=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}2&3\\3&-4\end{bmatrix}$
$=\begin{bmatrix}2+6&3-8\\-4+3&-6-4\end{bmatrix}=\begin{bmatrix}8&-5\\-1&-10\end{bmatrix}$
and $(\text{AB})\text{C}=\begin{bmatrix}8&-5\\-1&-10\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}$
$\begin{bmatrix}8+5&0\\-1+10&0\end{bmatrix}=\begin{bmatrix}13&0\\9&0\end{bmatrix}\ ....(\text{i})$
Again, $(\text{BC})=\begin{bmatrix}2&3\\3&-4\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}$
$=\begin{bmatrix}2-3&0\\3+4&0\end{bmatrix}=\begin{bmatrix}-1&0\\7&0\end{bmatrix}$
And $\text{A}(\text{BC})=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}-1&0\\7&0\end{bmatrix}$
$=\begin{bmatrix}-1+14&0\\2+7&0\end{bmatrix}=\begin{bmatrix}13&0\\9&0\end{bmatrix}\ ....(\text{ii})$
From (i) and (ii), we get
$\therefore\ (\text{AB})\text{C}=\text{A}(\text{BC})$
View full question & answer→Question 545 Marks
Find non-zero values of x satisfying the matrix equation:$\text{x}\begin{bmatrix}2\text{x}&2\\3&\text{x}\end{bmatrix}+2\begin{bmatrix}8&5\text{x}\\4&4\text{x}\end{bmatrix}=2\begin{bmatrix}(\text{x}^2+8)&24\$10)&6\text{x}\end{bmatrix}$
AnswerConsider, $\text{x}\begin{bmatrix}2\text{x}&2\\3&\text{x}\end{bmatrix}+2\begin{bmatrix}8&5\text{x}\\4&4\text{x}\end{bmatrix}=2\begin{bmatrix}(\text{x}^2+8)&24\$10)&6\text{x}\end{bmatrix}$$\Rightarrow\ \begin{bmatrix}2\text{x}^2&2\text{x}\\3\text{x}&\text{x}^2\end{bmatrix}+\begin{bmatrix}16&10\text{x}\\8&8\text{x}\end{bmatrix}=\begin{bmatrix}2\text{x}^2+16&48\\20&12\text{x}\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2\text{x}^2+16&2\text{x}+10\text{x}\\3\text{x}+8&\text{x}^2+8\end{bmatrix}=\begin{bmatrix}2\text{x}^2+16&48\\20&12\text{x}\end{bmatrix}$
$\Rightarrow\ 2\text{x}+10\text{x}=48$
$\Rightarrow\ 12\text{x}=48$
$\Rightarrow\ \text{x}=\frac{48}{12}=4$
View full question & answer→Question 555 Marks
If $\text{A}=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix},$ then prove that $A^2 - 4A - 5I = 0.$
AnswerGiven: $\text{A}=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}1+4+4&2+2+4&2+4+2\\2+2+4&4+1+4&4+2+2\\2+4+2&4+2+2&4+4+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}$
$\text{A}^2-4\text{A}-5\text{I}$
$\Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-4\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}-\begin{bmatrix}4&8&8\\8&4&8\\8&8&4\end{bmatrix}-\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}$
$ \Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}9-4-5&8-8-0&8-8-0\\8-8-0&9-4-5&8-8-0\\8-8-0&8-8-0&9-4-5\end{bmatrix}$
$ \Rightarrow\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 565 Marks
If $\text{A} = \begin{bmatrix}3&-4\\1&-1\end{bmatrix},$then prove that $\text{A}'' = \begin{bmatrix}1 + 2n & -4n \\n & 1 - 2n \end{bmatrix}$ where n is any positive integer.
AnswerGiven: $\text{A}^{"}=\begin{bmatrix}1+2n&-4n\\n&1-2n\end{bmatrix}\ \ \ \therefore\ \text{A}^{n}=\begin{bmatrix}1+2n&-4n\\n&1-2n\end{bmatrix}$
$\Rightarrow \text{A}^{1}=\begin{bmatrix}1+2&-4\\1&1-2\end{bmatrix}=\begin{bmatrix}3&-4\\1&-1\end{bmatrix}$which is true for n =1.
Now, $\text{A}^{k}=\begin{bmatrix}1+2k&-4k\\k&1-2k\end{bmatrix}...\text{(i)}$
Again $\text{A}^{k + 1}=\begin{bmatrix}1 + 2(k + 1)&-4(k + 1)\$k + 1)&1-2(k + 1)\end{bmatrix}...\text{(ii)}$
$\Rightarrow\text{A}^{k}.\text{A}=\begin{bmatrix}1+2(k+1)&-4(k+1)\$k+1)&1-2(k+1)\end{bmatrix}\begin{bmatrix}3&-4\\1&1\end{bmatrix}$ [From eq.(i)]
$\Rightarrow\text{ A}^{k}.\text{A}=\begin{bmatrix}3+6k-4k&-4-8k+4k\\3k+1-2k&-4k-1+2k\end{bmatrix}=\begin{bmatrix}3+2k&-4-4k\\1+k&-1-2k\end{bmatrix}$
$\Rightarrow \text{A}^{k}.\text{A=}\begin{bmatrix}1+2(k+1)&-4(k+1)\$k+1)&1-2(k+1)\end{bmatrix}$
Therefore, the result is true for n = k + 1.
Hence, by the principal of mathematical induction, the result is true for all positive integers n.
View full question & answer→Question 575 Marks
If $A$ is a square matrix, using mathematical induction prove that $(A^T)^n = (A^n)^T$ for all $n \in N.$
AnswerLet the given statement $P(n),$ be given as
$P(n): (A^T)^n = (A^n)^T$ for all $n \in N.$
We observe that
$P(1): (A^T)^1 = A^T = (A^1)^T$
Thus, $P(n)$ is true for $n = 1.$
Assume that $P(n)$ is true for $n = k \in N.$
i.e., $P(k): (A^T)^k = (A^k)^T$
To prove that $P(k + 1)$ is true, we have
$(A^T)^{k + 1} = (A^T)^k.(A^T)^1$
$= (A^k)^T.(A^1)^T$
$= (A^{k + 1})^T$
Thus, $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the Principle of mathematical induction, $P(n)$ is true for all $n \in N.$
View full question & answer→Question 585 Marks
If $\text{A}=\begin{bmatrix}1&2\\-2&1\end{bmatrix},\ \text{B}=\begin{bmatrix}2&3\\3&-4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&0\\-1&0\end{bmatrix},$ verify $\text{A}(\text{B}+\text{C})=\text{AB}+\text{AC}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-2&1\end{bmatrix},\ \text{B}=\begin{bmatrix}2&3\\3&-4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&0\\-1&0\end{bmatrix}$
$\text{B}+\text{C}=\begin{bmatrix}2&3\\3&-4\end{bmatrix}+\begin{bmatrix}1&0\\-1&0\end{bmatrix}=\begin{bmatrix}3&3\\2&-4\end{bmatrix}$
$\Rightarrow\ \text{A}.(\text{B}+\text{C})=\begin{bmatrix}1&2\\-2&1\end{bmatrix}.\begin{bmatrix}3&3\\2&-4\end{bmatrix}$
$=\begin{bmatrix}3+4&3-8\\-6+2&-6-4\end{bmatrix}$
$=\begin{bmatrix}7&-5\\-4&-10\end{bmatrix}\ ....(\text{i})$
$\text{AB}=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}2&3\\3&-4\end{bmatrix}$
$=\begin{bmatrix}2+6&3-8\\-4+3&-6-4\end{bmatrix}=\begin{bmatrix}8&-5\\-1&-10\end{bmatrix}$
and $\text{AC}=\begin{bmatrix}1&2\\-2&1\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}$
$=\begin{bmatrix}1-2&0\\-2-1&0\end{bmatrix}=\begin{bmatrix}-1&0\\-3&0\end{bmatrix}$
$\therefore\ \text{AB}+\text{AC}=\begin{bmatrix}8&-5\\-1&-10\end{bmatrix}+\begin{bmatrix}-1&0\\-3&0\end{bmatrix}$
$\Rightarrow\ \text{AB}+\text{AC}=\begin{bmatrix}7&-5\\-4&-10\end{bmatrix}\ ....(\text{ii})$
From Eq. (i) and (ii),
$\text{A}(\text{B}+\text{C})=\text{AB}+\text{AC}.$
View full question & answer→Question 595 Marks
If $\text{A}=\begin{bmatrix}1&5\\7&12\end{bmatrix}$ and $\text{B}=\begin{bmatrix}9&1\\7&8\end{bmatrix},$ find a matrix C such that 3A + 5B + 2C is a null matrix.
AnswerWe have, $\text{A}=\begin{bmatrix}1&5\\7&12\end{bmatrix}$ and $\text{B}=\begin{bmatrix}9&1\\7&8\end{bmatrix}$ Let $\text{C}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}$ $\therefore\ 3\text{A}+5\text{B}+2\text{C}=0$ $\Rightarrow\ \begin{bmatrix}3&15\\21&36\end{bmatrix}+\begin{bmatrix}45&5\\35&40\end{bmatrix}+\begin{bmatrix}2\text{a}&2\text{b}\\2\text{c}&2\text{d}\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ $\Rightarrow\ \begin{bmatrix}48+2\text{a}&20+2\text{b}\\56+2\text{c}&76+2\text{d}\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ $\Rightarrow\ 2\text{a}+48=0\Rightarrow\ \text{a}=-24$ Also, $20+2\text{b}=0\Rightarrow\ \text{b}=-10$ $56+2\text{c}=0\Rightarrow\ \text{c}=-28$And $76+2\text{d}=0\Rightarrow\ \text{d}=-38$
$\therefore\ \text{C}=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}$
View full question & answer→Question 605 Marks
If $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},$ then find $A^2 - 5A - 14I.$ Hence, obtain $A^3.$
AnswerWe have, $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\ ....(\text{i})$
$\therefore\ \text{A}^2=\text{A}.\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
$=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}$
$\therefore\ \text{A}^2-5\text{A}-14\text{I}$
$=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-\begin{bmatrix}15&-25\\-20&10\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
Now, $\text{A}^2-5\text{A}-14\text{I}=0$
$\Rightarrow\ \text{A}.\text{A}^2-5\text{A}.\text{A}=14\text{AI}=0$
$\Rightarrow\ \text{A}^3-5\text{A}^2-14\text{A}=0$
$\Rightarrow\ \text{A}^3=5\text{A}^2+14\text{A}$
$=5\begin{bmatrix}29&-25\\-20&24\end{bmatrix}+14\begin{bmatrix}3&-5\\-4&-2\end{bmatrix}$
$=\begin{bmatrix}145&-125\\-100&120\end{bmatrix}+\begin{bmatrix}42&-70\\-56&28\end{bmatrix}$
$=\begin{bmatrix}187&-195\\-156&148\end{bmatrix}$
View full question & answer→Question 615 Marks
If $\text{A}=\begin{bmatrix}1&0\\-1&7\end{bmatrix},$ find $k$ such that $A^2 - 8A + kI = 0.$
AnswerGiven: $\text{A}=\begin{bmatrix}1&0\\-1&7\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\-1&7\end{bmatrix}\begin{bmatrix}1&0\\-1&7\end{bmatrix}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}1-0&0+0\\-1-7&0+49\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\-8&49\end{bmatrix}$
$ \text{A}^2-8\text{A}+\text{kI}=0$
$ \Rightarrow\begin{bmatrix}1&0\\-8&49\end{bmatrix}-8\begin{bmatrix}1&0\\-1&7\end{bmatrix}+\text{k}\begin{bmatrix}1&0\\0&1\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}1&0\\-8&49\end{bmatrix}-\begin{bmatrix}8&0\\-8&56\end{bmatrix}+\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}1-8+\text{k}&0-0+0\\-8+8+0&49-56+\text{k}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}-7+\text{k}&0\\0&-7+\text{k}\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore-7+\text{k}=0$
$\Rightarrow\text{k}=7$
View full question & answer→Question 625 Marks
If $B, C$ are $n$ rowed square matrices and if $A = B + C, BC = CB, C^2 = O,$ then show that for every $n \in N, A^{n+1} = B^n(B + (n + 1)C).$
AnswerLet $P(n)$ be the statement given by $P(n) : A^{n+1} = B^n(B + (n + 1)C)$ For $n = 1,$ We have
$P(1) : A^2 = B(B + 2C)$
Here,
$\text{LHS} = A^2$
$= (B + C)(B + C)$
$= B(B + C)+ C(B + C)$
$= B^2+ BC + CB + C^2$
$= B^2 + 2BC [\because BC = CB$ and $C^2 = O]$
$= B(B + 2C) = \text{RHS}$
$\text{RHS}$ Hence, the statement is true for $n=1.$
If the statement is true for $n = k,$ then
$P(k) : A^{k+1} = B^k(B + (k + 2)C) ...(1)$
For $P(k + 1)$ to be true, we must have
$P(k + 1) : A^{k+2} = B^{k+1}(B + (k + 2)C)$
Now,
$A^{k+2} =A^{k+1}A$
$= [B^k(B + (k + 1)C)] (B + C) [$From eqn. $(1)]$
$= [B^{k+1} + (k + 1)B^kC] (B + C)$
$= B^{k+1}(B + C) + (k + 1)B^kC(B + C)$
$= B^{k+2} + B^{k+1}C + (k + 1)B^kCB + (k + 1)B^kC^2$
$= B^{k+2} + B^{k+1}C + (k + 1)B^kBC [\because BC = CB$ and $C^2 = 0]$
$= B^{k+2} + B^{k+1}C + (k + 1)B^{k+1}C$
$= B^{k+2} + (k + 2)B^{k+1}C$
$= B^{k+1 }[B + (k + 2)C]$
So, the statement is true for $n = k + 1.$
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N.$
View full question & answer→Question 635 Marks
Prove by Mathematical Induction that $(A′)^n = (A^n)′,$ where $\text{n}\in\text{N}$ for any square matrix $A.$
AnswerLet $P(n) : (A')^n = (A^n) '$
$\therefore P(1) : (A') = (A)'$
$\Rightarrow A' = A'$
$\Rightarrow P(1)$ is true.
Now, let $P(k) = (A')^k = (A^k)',$
where $\text{k}\in\text{N}$
and $P(k + 1) : (A')^{K+1} = (A')^kA'$
$= (A^k)'A'$
$= (AA^k)' (as (AB)' = B'A')$
$= (A^{k+1})'$
Thus $P(1)$ is true and whenever $P(k)$ is true $P(k + 1)$ is true.
So, $P(n)$ is true for all $\text{n}\in\text{N}.$
View full question & answer→Question 645 Marks
If $\text{P}(\text{x})=\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).
AnswerWe have, $\text{P}(\text{x})=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$ $\Rightarrow\ \text{P}(\text{y})=\begin{bmatrix}\cos\text{y}&\sin\text{y}\\-\sin\text{y}&\cos\text{y}\end{bmatrix}$ Consider, P(x).P(y) = P(x + y)LHS = P(x).P(y)
$=\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}.\sin\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}$
$\begin{bmatrix}\because\ \cos(\text{x}+\text{y})=\cos\text{x}.\cos\text{y}-\sin\text{x}.\sin\text{y}\\\text{and }\sin(\text{x}+\text{y})=\sin\text{x}.\cos\text{y}+\cos\text{x}.\sin\text{y}\end{bmatrix}$
$=\text{P}(\text{x}+\text{y})$
= RHS Hence proved.
View full question & answer→Question 655 Marks
If $\text{A}=\begin{bmatrix}3&-4\\1&1\\2&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix},$ then verify $(\text{BA})^2\neq\text{B}^2\text{A}^2.$
AnswerWe have, $\text{A}=\begin{bmatrix}3&-4\\1&1\\2&0\end{bmatrix}_{3\times2}$ and $\text{B}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}$
$\therefore\ \text{BA}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}\begin{bmatrix}3&-4\\1&1\\2&0\end{bmatrix}_{3\times2}$
$=\begin{bmatrix}6+1+4&-8+1+0\\3+2+8&-4+2+0\end{bmatrix}=\begin{bmatrix}11&-7\\13&-2\end{bmatrix}$
And $(\text{BA}).(\text{BA})=\begin{bmatrix}11&-7\\13&-2\end{bmatrix}.\begin{bmatrix}11&-7\\13&-2\end{bmatrix}$
$\Rightarrow\ (\text{BA})^2=\begin{bmatrix}121-91&-77+14\\143-26&-91+4\end{bmatrix}$ $=\begin{bmatrix}30&-63\\117&-87\end{bmatrix}\ ....(\text{i})$
Also,$\text{B}^2=\text{B}.\text{B}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}.\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}$
So $B^2$ is not possible, since the $B$ is not a square matrix.
Hence, $(\text{BA})^2\neq\text{B}^2\text{A}^2.$
View full question & answer→Question 665 Marks
If $\text{X}=\begin{bmatrix}3&1&-1\\5&-2&-3\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}2&1&-1\\7&2&4\end{bmatrix},$ then find:
- X + Y
- 2X - 3Y
- A matrix Z such that X + Y + Z is a zero matrix.
AnswerWe have, $\text{X}=\begin{bmatrix}3&1&-1\\5&-2&-3\end{bmatrix}_{2\times3}$ and $\text{Y}=\begin{bmatrix}2&1&-1\\7&2&4\end{bmatrix}_{2\times3}$
- $\text{X}+\text{Y}=\begin{bmatrix}3+2&1+1&-1-1\\5+7&-2+2&-3+4\end{bmatrix}$
$=\begin{bmatrix}5&2&-2\\12&0&1\end{bmatrix}$
- $\because\ 2\text{X}-3\text{Y}=2\begin{bmatrix}3&1&-1\\5&-2&-3\end{bmatrix}-3\begin{bmatrix}2&1&-1\\7&2&4\end{bmatrix}$
$=\begin{bmatrix}6&2&-2\\10&-4&-6\end{bmatrix}-\begin{bmatrix}6&3&-3\\21&6&12\end{bmatrix}$
$=\begin{bmatrix}6-6&2-3&-2+3\\10-21&-4-6&-6-12\end{bmatrix}$
$=\begin{bmatrix}0&-1&1\\-11&-10&-18\end{bmatrix}$
- $\text{X}+\text{Y}=\begin{bmatrix}5&2&-2\\12&0&1\end{bmatrix}$
Also, $\text{X}+\text{Y}+\text{Z}=\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}$
So, Z is the additive inverse of (X + Y) or negative of (X + Y).
$\therefore\ \text{Z}=-(\text{X}+\text{Y})=\begin{bmatrix}-5&-2&2\\-12&0&-1\end{bmatrix}$ View full question & answer→Question 675 Marks
If $\text{A}=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix},$ then verify that $\text{A}^{2}+\text{A}=\text{A}(\text{A}+\text{I}),$ where I is 3 × 3 unit matrix.
AnswerWe have, $\text{A}=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}$$\therefore\ \text{A}^2=\text{A}.\text{A}$
$=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}.\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}$
$=\begin{bmatrix}1+0+0&0+0-1&-1+0-1\\2+2+0&0+1+3&-2+3+3\\0+2+0&0+1+1&0+3+1\end{bmatrix}$
$=\begin{bmatrix}1&-1&-2\\4&4&4\\2&2&4\end{bmatrix}$
$\therefore\ \text{A}^2+\text{A}=\begin{bmatrix}1&-1&-2\\4&4&4\\2&2&4\end{bmatrix}+\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}$
$=\begin{bmatrix}2&-1&-3\\6&5&7\\2&3&5\end{bmatrix}\ ....(\text{i})$
Now, $\text{A}+\text{I}=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}+\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}2&0&-1\\2&2&3\\0&1&2\end{bmatrix}$
So, $\text{A}(\text{A}+\text{I})=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}2&0&-1\\2&2&3\\0&1&2\end{bmatrix}$
$=\begin{bmatrix}2+0+0&0+0-1&-1+0-2\\4+2+0&0+2+3&-2+3+6\\0+2+0&0+2+1&0+3+2\end{bmatrix}$
$=\begin{bmatrix}2&-1&-3\\6&5&7\\2&3&5\end{bmatrix}\ ....(\text{ii})$
From (i) and (ii), we get $\text{A}^{2}+\text{A}=\text{A}(\text{A}+\text{I}).$
View full question & answer→Question 685 Marks
Find the matrix A satisfying the matrix equation:$\begin{bmatrix}2&1\\3&2\end{bmatrix}\text{A}\begin{bmatrix}-3&2\\5&-3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$
AnswerWe have, $\begin{bmatrix}2&1\\3&2\end{bmatrix}\text{A}\begin{bmatrix}-3&2\\5&-3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ Or PAQ = I, where $\text{P}=\begin{bmatrix}2&1\\3&2\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}-3&2\\5&-3\end{bmatrix}$ $\therefore\ \text{P}^{-1}\text{PAQ}=\text{P}^{-1}\text{I}$ $\Rightarrow\ \text{IAQ}=\text{P}^{-1}$ $\Rightarrow\ \text{AQ}=\text{P}^{-1}$ $\Rightarrow\ \text{AQQ}^{-1}=\text{P}^{-1}\text{Q}^{-1}$ $\Rightarrow\ \text{AI}=\text{P}^{-1}\text{Q}^{-1}$ $\Rightarrow\ \text{A}=\text{P}^{-1}\text{Q}^{-1}$ Now adj. $\text{P}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}$ and $|\text{P}|=1$ $\therefore\ \text{P}^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}$ Also adj. $\text{Q}=\begin{bmatrix}-3&-2\\-5&-3\end{bmatrix}$ and $|\text{Q}|=-1$ $\therefore\ \text{Q}^{-1}=\begin{bmatrix}3&2\\5&3\end{bmatrix}$$\Rightarrow\ \text{A}=\text{P}^{-1}\text{Q}^{-1}$
$=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}\begin{bmatrix}3&2\\5&3\end{bmatrix}$
$=\begin{bmatrix}6-5&4-3\\-9+10&-6+6\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}$
View full question & answer→Question 695 Marks
If $\text{A}=\begin{bmatrix}3&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}7&3\end{bmatrix},$ then find a non-zero matrix C such that AC = BC.
AnswerWe have, $\text{A}=\begin{bmatrix}3&5\end{bmatrix}_{2\times1}$ and $\text{B}=\begin{bmatrix}7&3\end{bmatrix}_{1\times2}$ Let $\text{C}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}_{2\times1}$ is a non-zero matrix of order 2 × 1. $\therefore\ \text{AC}=\begin{bmatrix}3&5\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}3\text{x}+5\text{y}\end{bmatrix}$ And $\text{BC}=\begin{bmatrix}7&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}7\text{x}+3\text{y}\end{bmatrix}$ For AC = BC,$\begin{bmatrix}3\text{x}+5\text{y}\end{bmatrix}=\begin{bmatrix}7\text{x}+3\text{y}\end{bmatrix}$
On using equality of matrix, we get $\Rightarrow\ 3\text{x}+5\text{y}=7\text{x}+3\text{y}$ $\Rightarrow\ 4\text{x}=2\text{y}$ $\Rightarrow\ \text{x}=\frac{1}{2}\text{y}$ $\Rightarrow\ \text{y}=2\text{x}$ $\therefore\ \text{C}=\begin{bmatrix}\text{x}\\2\text{x}\end{bmatrix}$ We see that on taking C of order 2 × 1, 2 × 2, 2 × 3, .....we get $\text{C}=\begin{bmatrix}\text{x}\\2\text{x}\end{bmatrix},\begin{bmatrix}\text{x}&\text{x}\\2\text{x}&2\text{x}\end{bmatrix},\begin{bmatrix}\text{x}&\text{x}&\text{x}\\2\text{x}&2\text{x}&2\text{x}\end{bmatrix}....$ In general, $\text{C}=\begin{bmatrix}\text{k}\\2\text{k}\end{bmatrix},\begin{bmatrix}\text{k}&\text{k}\\2\text{k}&2\text{k}\end{bmatrix}\ \text{etc }.....$Where, k is any real number.
View full question & answer→Question 705 Marks
Show that $\text{A}=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}$ satisfies the equation $A^2 - 3A - 7I = 0$ and hence find $A^{-1}.$
AnswerWe have, $\text{A}=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}$
$\therefore\ \text{A}^2=\text{A.A}=\begin{bmatrix}5&3\\-1&-2\end{bmatrix}.\begin{bmatrix}5&3\\-1&-2\end{bmatrix}$
$=\begin{bmatrix}25-3&15-6\\-5+2&-3+4\end{bmatrix}=\begin{bmatrix}22&9\\-3&1\end{bmatrix}$
$3\text{A}=3\begin{bmatrix}5&3\\-1&-2\end{bmatrix}=\begin{bmatrix}15&9\\-3&-6\end{bmatrix}$
And $7\text{I}=7\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\therefore\ \text{A}^2-3\text{A}-7\text{I}$
$=\begin{bmatrix}22&9\\-3&1\end{bmatrix}-\begin{bmatrix}15&9\\-3&-6\end{bmatrix}-\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}22-15-7&9-9-0\\-3+3-0&1+6-7\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
Hence proved.
Since, $\text{A}^2-3\text{A}-7\text{I}=0$
$\Rightarrow\ \text{A}^{-1}[(\text{A}^2)-3\text{A}-7\text{I}]=\text{A}^{-1}0$
$\Rightarrow\ \text{A}^{-1}\text{A}.\text{A}-3\text{A}^{-1}\text{A}-7\text{A}^{-1}\text{I}=0$
$[\because\ \text{A}^{-1}0=0]$
$\Rightarrow\ \text{IA}-3\text{I}-7\text{A}^{-1}=0$
$[\because\ \text{A}^{-1}\text{A}=\text{I}]$
$\Rightarrow\ \text{A}-3\text{I}-7\text{A}^{-1}=0$
$[\because\ \text{A}^{-1}\text{I}=\text{A}^{-1}]$
$\Rightarrow\ -7\text{A}^{-1}=-\text{A}+3\text{I}$
$=\begin{bmatrix}-5&-3\\1&2\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}=\begin{bmatrix}-2&-3\\1&5\end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{-1}{7}\begin{bmatrix}-2&-3\\1&5\end{bmatrix}$
View full question & answer→Question 715 Marks
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
AnswerGiven: $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
Now,
$\text{P}\text{Q}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{y}\text{b}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{zc}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(4)$
Also,
$\text{Q}\text{P}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$=\begin{bmatrix}\text{ax}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{by}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{cz}\end{bmatrix}$
$=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(5)$
From (4) and (5), we get
$\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
View full question & answer→Question 725 Marks
If $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then prove that $A^2 - A + 2I = O.$
AnswerGiven: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}+2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1-3&-2+2\\4-4&-4+2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2+2&0+0\\0+0&-2+2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=0$
Hence proved.
View full question & answer→Question 735 Marks
Show that A′A and AA′ are both symmetric matrices for any matrix A.
AnswerLet P = A'A
$\therefore$ P' = (AA')'
= A'(A')' $[\because$ (AB')' = BA'$]$
= A'A = P
So, A’A is symmetric matrix for any matrix A.
Similarly, let Q = AA’
$\therefore$ Q' = (AA')' = (A')'(A)'
= A(A')' = Q
So, AA’ is symmetric matrix for any matrix A.
View full question & answer→Question 745 Marks
If $\text{A}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix},$ show that $AB = BA = O_{3 \times 3}$
AnswerHere,
$\text{AB}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0+\text{abc}-\text{abc}&0+\text{b}^2\text{c}-\text{b}^2\text{c}&0+\text{bc}^2-\text{bc}^2\\-\text{a}^2\text{c}+0+\text{a}^2\text{c}&-\text{abc}+0+\text{abc}&-\text{ac}^2+0+\text{ac}^2\\\text{a}^2\text{b}-\text{a}^2\text{b}+0&\text{ab}^2-\text{ab}^2+0&\text{abc}-\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{AB}=\text{O}_{3\times3}\ \dots(1)$
$\text{BA}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0-\text{abc}+\text{abc}&\text{a}^2\text{c}+0-\text{a}^2\text{c}&-\text{a}^2\text{b}+\text{a}^2\text{b}+0\\0-\text{b}^2\text{c}+\text{b}^2\text{c}&\text{abc}+0-\text{abc}&-\text{ab}^2+\text{ab}^2+0\\0-\text{bc}^2+\text{bc}^2&\text{ac}^2+0-\text{ac}^2&-\text{abc}+\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{BA}=\text{O}_{3\times3}\ \dots(2)$
$ \Rightarrow\text{AB}=\text{BA}=0_{3\times3} [$From eqs. $(1)$ and $(2)]$
View full question & answer→Question 755 Marks
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ prove that $\text{A}^\text{n}=\begin{bmatrix}1&\text{n}\\0&1\end{bmatrix}$ for all positive integers $n.$
AnswerGiven, $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ To prove $\text{A}^\text{n}=\begin{bmatrix}1&\text{n}\\0&1\end{bmatrix}$ we will use the principle of mathematical induction.
Step $1:$ Put $n - 1 \text{A}^1=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
So, $A^n$ is true for $n = 1$
Step $2:$ Let, $A^n$ be true for $n = k,$ then $\text{A}^\text{k}=\begin{bmatrix}1&\text{k}\\0&1\end{bmatrix}\ \dots(\text{i})$
Step $3:$ We have to show that $ \text{A}^\text{k+1}=\begin{bmatrix}1&\text{k}+1\\0&1\end{bmatrix}$
So,
$\text{A}^\text{k+1}=\text{A}^\text{k}\times\text{A}$
$=\begin{bmatrix}1&\text{k}\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix} ($using equation $(i)$ and given$)$
$ =\begin{bmatrix}1+0&1+\text{k}\\0+0&0+1\end{bmatrix}$
$\text{A}^{\text{k}+1}=\begin{bmatrix}1&1+\text{k}\\0&1\end{bmatrix}$
This shows that $A^n$ is true for $n = k + 1$ whenever it is true for $n = k$
Hence, by the principle of mathematical induction $A^n$ is true for all positive integer.
View full question & answer→Question 765 Marks
Find the value of x for which the matrix product
$\begin{bmatrix}2&0&7\\0&1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}-\text{x}&14\text{x}&7\text{x}\\0&1&0\\\text{x}&-4\text{x}&-2\text{x}\end{bmatrix}$ equal an identity matrix.
AnswerHere,
$\begin{bmatrix}2&0&7\\0&1&0\\1&-2&1\end{bmatrix}\begin{bmatrix}-\text{x}&14\text{x}&7\text{x}\\0&1&0\\\text{x}&-4\text{x}&-2\text{x}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}-2\text{x}+0+7\text{x}&28\text{x}+0-28\text{x}&14\text{x}+0-14\text{x}\\0+0+0&0+1-0&0+0-0\\-\text{x}-0+\text{x}&14\text{x}-2-4\text{x}&7\text{x}-0-2\text{x}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}5\text{x}&0&0\\0&1&0\\0&10\text{x}-2&5\text{x}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore\ 5\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{5}$
View full question & answer→Question 775 Marks
Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result:
$\text{A}=\begin{bmatrix}3 & -2 &-4\\3 & -2&-5\\-1&-1& 2\end{bmatrix}$
AnswerLet, $\text{A}=\begin{bmatrix}3 & -2 &-4\\3 & -2&-5\\-1&-1& 2\end{bmatrix}$
$\Rightarrow\text{A}^{\text{T}}=\begin{bmatrix}3 & 3 &-1\\-2&-2 & 1\\-4&-5&2 \end{bmatrix}$
Let, $\text{X}=\frac{1}{2}(\text{A}+\text{A}^{\text{T}})=\frac{1}{2}\begin{bmatrix}3&-2 & -4 \\3&-2 & -5\\-1&1&2 \end{bmatrix}+\begin{bmatrix}3&3&-1 \\-2&-2&1\\-4&-5&2 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}3+3 & -2+3&-4-1 \\3-2& -2-2&-5+1\\-1-4&1-5&2+2 \end{bmatrix}=\frac{1}{2}\begin{bmatrix}6 & 1&-5 \\1 & -4&-4\\-5&-4&4 \end{bmatrix}$
$=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}$
Now, $\text{X}^{\text{T}}=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&2\\\frac{-5}{2}&-2&2 \end{bmatrix}^{\text{T}}=\begin{bmatrix}3& \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}=\text{X}$
⇒ X is a symmatric matrix.
Let, $\text{y}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}3&-2&-4 \\3&-2&-5 \\-1&1&2 \end{bmatrix}-\begin{bmatrix}3&3&-1 \\-2&-2&1 \\-4&-5&2 \end{bmatrix}$ $=\frac{1}{2}\begin{bmatrix}3-3&-2-3&-4+1 \\3+2 & -2+2&-5-1\\-1+4&1+5&2-2 \end{bmatrix} $
$=\frac{1}{2}\begin{bmatrix}0&-5&-3 \\5&0&6\\3&6&0 \end{bmatrix}=\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\\frac{5}{2} &0&-3\\\frac{3}{2}&3&0 \end{bmatrix}$
$-\text{Y}^{\text{T}}=-\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\5&0&6\\3&6&0 \end{bmatrix}=\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\\frac{5}{2} &0&-3\\\frac{3}{2}&3&0 \end{bmatrix}=\text{Y}$
⇒ Y is a skew symmetric matrix
$\text{X}+\text{Y}=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2} & -2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}+\begin{bmatrix}0&\frac{-5}{2}&\frac{-3}{2} \\\frac{5}{2} & 0&-3\\\frac{3}{2}&3&0 \end{bmatrix}=\begin{bmatrix}3+0&{\frac{1}{2}-\frac{5}{2}}&{\frac{-5}{-2}-\frac{3}{2}} \\{\frac{1}{2}+\frac{5}{2}} & -2+0&-2-3&\\{\frac{-5}{2}+\frac{3}{2}}&-2+3&2+0 \end{bmatrix}$
$=\begin{bmatrix}3 & -2&-4 \\3&-2&-5\\-1&1&2 \end{bmatrix}=\text{A}$
Hence, symmetric matrix $\text{X}=\begin{bmatrix}3 & \frac{1}{2}&\frac{-5}{2} \\\frac{1}{2}&-2&-2\\\frac{-5}{2}&-2&2 \end{bmatrix}$
View full question & answer→Question 785 Marks
A manufacturer produces three types of bolts, $x, y$ and $z$ which he sells in two markets. Annual sales $($in $₹)$ are indicated below:
| Markets |
Products |
| $X$ |
$Y$ |
$Z$ |
| $I$ |
$10000$ |
$2000$ |
$18000$ |
| $II$ |
$6000$ |
$20000$ |
$8000$ |
If unit sales prices of $x, y$ and z are $₹\ 2.50, ₹\ 1.50$ and $₹\ 1.00$ respectively, then answer the following questions using the concept of matrices.
- Find the total revenue collected from the Market$-I.$
- $₹\ 44000$
- $₹\ 48000$
- $₹\ 46000$
- $₹\ 53000$
- Find the total revenue collected from the Market$-II.$
- $₹\ 51000$
- $₹\ 53000$
- $₹\ 46000$
- $₹\ 49000$
- If the unit costs of the above three commodities are $₹\ 2.00, ₹\ 1.00$ and $50$ paise respectively, then find the gross profit from both the markets.
- $₹\ 53000$
- $₹\ 46000$
- $₹\ 34000$
- $₹\ 32000$
- If matrix $\text{A}=[\text{a}_\text{ij}]_{2\times2},$ where $\text{a}_\text{ij}=1,$ if $\text{i}\neq\text{j}$ and $\text{a}_\text{ij}=0,$ if $\text{i}=\text{j}$ then $A^2$ is equal to:
- $I$
- $A$
- $OR$
- None of these
- If $A$ and $B$ are matrices of same order, then $(AB' - BA')$ is a.
- Skew$-$synunetric matrix.
- Null matrix.
- Symmetric matrix.
- Unit matrix.
AnswerLet A be the $2 \times 3$ matrix representing the annual sales of products in two markets. $\begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{X}&\ \ \ \ \ \ \ \text{Y}&\ \ \ \ \ \ \ \text{Z}\end{matrix}\\\therefore\ \ \text{A}=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{matrix}\text{Market I}\\\text{Market II}\end{matrix}$
Let $B$ be the column matrix representing the sale price of each unit of products $x, y, z.$
$\therefore\ \text{B}=\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}$
Now, revenue $=$ sale price$ \times $ number of items sold $=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}$ $=\begin{bmatrix}25000+3000+18000\\15000+30000+8000\end{bmatrix}=\begin{bmatrix}46000\\53000\end{bmatrix}$
Therefore, the revenue collected from Market $I = ₹\ 46000$ and the revenue collected from Market $II = ₹\ 53000.$
- $(c) ₹\ 46000$
- $(b) ₹\ 53000$
- $(d) ₹\ 32000$
Let $C$ be the column matrix representing cost price of each unit of products $x, y, z.$
Then, $\text{C}=\begin{bmatrix}2\\1\\0.5\end{bmatrix}$
$\therefore$ Total cost in each market is given by
$\text{AC}=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2\\1\\0.5\end{bmatrix}$
$=\begin{bmatrix}20000+2000+9000\\12000+20000+4000\end{bmatrix}=\begin{bmatrix}31000\\36000\end{bmatrix}$
Now, Profit matrix $=$ Revenue matrix $-$ Cost matrix
$=\begin{bmatrix}46000\\53000\end{bmatrix}-\begin{bmatrix}31000\\36000\end{bmatrix}=\begin{bmatrix}15000\\17000\end{bmatrix}$
Therefore, the gross profit from both the markets $= ₹\ 15000 + ₹\ 17000 = ₹\ 32000$
- $(a) I$
We have, $\text{A}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\therefore\ \ \text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
- $(a)$ Skew$-$synunetric matrix.
We have, $(AB' - BA')' = (B')' A' - (A')' B' = BA' - AB'= -(AB' - BA')$
Thus$, AB' - BA'$ is a skew$-$symmetric matrix. View full question & answer→Question 795 Marks
If $\text{A}=\begin{bmatrix}2&3\\-1&0\end{bmatrix},$ show that $A^2 - 2A + 3I_2 = 0.$
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\-1&0\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&3\\-1&0\end{bmatrix}\begin{bmatrix}2&3\\-1&0\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4-3&6+0\\-2+0&-3+0\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}$
$ \text{A}^2-2\text{A}+3\text{I}_2$
$\Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}-2\begin{bmatrix}2&3\\-1&0\end{bmatrix}+3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}-\begin{bmatrix}4&6\\-2&0\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$ \Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1-4+3&6-6+0\\-2+2+0&-3+0+3\end{bmatrix}$
$\Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 805 Marks
If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that $A^2 - 5A + 7I_2 = 0.$
AnswerGiven: $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
Now,
$ \text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{A}^2-5\text{A}+7\text{I}_2$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}_2=0$
Hence proved.
View full question & answer→Question 815 Marks
A trust fund has $₹\ 30, 000$ that must be inverted in two different types of bond. the first bond pays $5\%$ interest per year and the second bond pays $7\%$ interest per year. using matrix multiplication, determine how to divide $₹\ 30,000$ in two types of bonds, if the trust fund must obtain an annual interest of $(a) ₹\ 1800, (b) ₹\ 2000.$
AnswerLet the investment in first bond be $₹\ x ,$ then the investment in the second bond $= ₹\ (30000 - x)$ Interest paid by first bond $= 5\% = \frac{5}{100}$ per rupee and interest paid by second bond $= 5\% = \frac{7}{100}$ per rupee. Matrix of investment is $A = [x 30000 - x]_{1 \times 2}$ Matrix of annual interest per rupee $B = \begin{bmatrix}\frac5{100}\\\frac7{100}\end{bmatrix}_{2 \times 1} $ Matrix of total annual interest is $AB =\left[ x\ 30000- x\right]\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}\frac{5x}{100}+\frac{7(30000-x)}{100}\end{bmatrix}$ $= \begin{bmatrix}\frac{5x + 210000 - 7x}{100}\end{bmatrix}=\begin{bmatrix}\frac{210000 - 2x}{100}\end{bmatrix}$ $\therefore$ Total annual interest $= ₹\ \frac{210000 - 2x}{100}$
- According to question, $\frac{210000-2x}{100}=1800 $
$₹\ 210000 - 2x = 180000 ₹\ x = 15, 000$
Therefore, Investment in first bond $= ₹\ 15, 000$
And Investment in second bond $= ₹\ (30000 - 15000) = ₹\ 15, 000$
- According to question, $\frac{210000-2x}{100} = 2000$
$\Rightarrow 210000 - 2x = 200000\ ₹\ x = 5, 000$
Therefore, Investment in first bond $= ₹\ 5, 000$
And Investment in second bond $= ₹\ (30000 - 5000) = ₹\ 25, 000$ View full question & answer→Question 825 Marks
If $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},$ find $A^2 - 5A - 14.$
AnswerGiven: $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9+20&-15-10\\-12-8&20+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}$
$\text{A}^2-5\text{A}-14\text{I}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-5\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-\begin{bmatrix}15&-25\\-20&10\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29-15-14&-25+25+0\\-20+20+0&24-10-14\end{bmatrix}$
$ \Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
View full question & answer→Question 835 Marks
Evaluate the following:
$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2 \end{bmatrix}\end{pmatrix}$
Answer$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2 \end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1-0&0-1&2-2\\2-1&0-0&1-2 \end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{bmatrix}1&-1&0\\1&0&1\end{bmatrix}$
$=\begin{bmatrix}1-1&-1+0&0+1\\0+2&0+0&0-2\\2+3&-2+0&0-3\end{bmatrix}$
$=\begin{bmatrix}0&-1&1\\2&0&-2\\5&-2&-3\end{bmatrix}$
Hence,
$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix} 0&1&2\\1&0&2\end{bmatrix} \end{pmatrix}\begin{bmatrix}0&-1&1\\2&0&-2\\5&-2&-3\end{bmatrix}$
View full question & answer→Question 845 Marks
If $\text{A}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\text{B}=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix},$ verify that A(B - C) = AB - AC.
AnswerGiven, $\text{A}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix},\text{B}=\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}$ $\text{C}=\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ $ \text{A}(\text{B}-\text{C})=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}\end{bmatrix}$ $=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}0-1&5-5&-4-2\\-2+1&1-1&3-0\\-1-0&0+1&2-1\end{bmatrix}$ $=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}-1&0&-6\\-1&0&3\\-1&1&1\end{bmatrix}$ $=\begin{bmatrix}-1+0+2&0+0-2&-6+0-2\\-3+1+0&0+0+0&-18-3+0\\2-1-1&0+0+1&12+3+1\end{bmatrix}$ $\text{A}(\text{B}-\text{C})=\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}\ \dots(\text{i})$ $ \text{AB}-\text{AC}=\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}0&5&-4\\-2&1&3\\-1&0&2\end{bmatrix}-\begin{bmatrix}1&0&-2\\3&-1&0\\-2&1&1\end{bmatrix}\begin{bmatrix}1&5&2\\-1&1&0\\0&-1&1\end{bmatrix}$ $=\begin{bmatrix}0+0+2&5+0+0&-4+0-4\\0+2+0&15-1+0&-12-3+0\\0-2-1&-10+1+0&8+3+2\end{bmatrix}-\begin{bmatrix}1+0+0&5+0+2&2+0-2\\3+1+0&15-1+0&6+0+0\\0-2-1&-10+1+1&-4+0+1\end{bmatrix}$ $=\begin{bmatrix}2&5&-8\\2&14&-15\\-3&-9&13\end{bmatrix}-\begin{bmatrix}1&7&0\\4&14&6\\-3&-10&-3\end{bmatrix}$ $=\begin{bmatrix}2-1&5-7&-8-0\\2-4&14-14&-14-6\\-3+3&-9+10&3+3\end{bmatrix}$ $\text{AB}-\text{AC}=\begin{bmatrix}1&-2&-8\\-2&0&-21\\0&1&16\end{bmatrix}\ \dots(\text{ii})$ From equation (i) and (ii),$\text{A}(\text{B}-\text{C})=\text{AB}-\text{AC}$
View full question & answer→Question 855 Marks
Express the matrix $\begin{bmatrix}2&3&1\\1&-1&2\\4&1&2\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
AnswerWe have, $\text{A}=\begin{bmatrix}2&3&1\\1&-1&2\\4&1&2\end{bmatrix}$
$\therefore\ \text{A}'=\begin{bmatrix}2&1&4\\3&-1&1\\1&2&2\end{bmatrix}$
Now, $\Rightarrow\ \frac{\text{A}+\text{A}'}{2}=\frac{1}{2}\begin{bmatrix}4&4&5\\4&-2&3\\5&3&4\end{bmatrix}$
$=\begin{bmatrix}2&2&\frac{5}{2}\\2&-1&\frac{3}{2}\\\frac{5}{2}&\frac{3}{2}&2\end{bmatrix}$
And $\frac{\text{A}-\text{A}'}{2}=\frac{1}{2}\begin{bmatrix}0&2&-3\\-2&0&1\\3&-1&0\end{bmatrix}$
$=\begin{bmatrix}0&1&\frac{-3}{2}\\-1&0&\frac{1}{2}\\\frac{3}{2}&\frac{-1}{2}&0\end{bmatrix}$
$\Rightarrow\ \text{A}=\frac{\text{A}+\text{A}'}{2}+\frac{\text{A}-\text{A}'}{2}$
$=\begin{bmatrix}2&2&\frac{5}{2}\\2&-1&\frac{3}{2}\\\frac{5}{2}&\frac{3}{2}&2\end{bmatrix}+\begin{bmatrix}0&1&\frac{-3}{2}\\-1&0&\frac{1}{2}\\\frac{3}{2}&\frac{-1}{2}&0\end{bmatrix}$
Which is the required expression.
View full question & answer→Question 865 Marks
Evaluate the following:
$\begin{pmatrix}\begin{bmatrix}1&3\\-1&-4 \end{bmatrix}+\begin{bmatrix}3&-2\\-1&1 \end{bmatrix}\end{pmatrix}\begin{bmatrix}1&3&5\\2&4&6 \end{bmatrix}$
Answer$\begin{pmatrix}\begin{bmatrix}1&3\\-1&-4 \end{bmatrix}+\begin{bmatrix}3&-2\\-1&1 \end{bmatrix}\end{pmatrix}\begin{bmatrix}1&3&5\\2&4&6 \end{bmatrix}$
$\Rightarrow\begin{pmatrix}\begin{bmatrix}1+3&3-2\\-1-1&-4+1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1&3&5\\2&4&6 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}4&1\\-2&-3\end{bmatrix}\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}$
$\Rightarrow\begin{bmatrix}4+2&12+4&20+6\\-2-6&-6-12&-10-18\end{bmatrix}$
$\Rightarrow\begin{bmatrix}6&16&26\\-8&-18&-28\end{bmatrix}$
View full question & answer→Question 875 Marks
Find the matrix A such that $\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\text{A}=\begin{bmatrix}-1&-8&-10\\1&-2&-10\\9&22&15\end{bmatrix}.$
AnswerWe have, $\begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}_{3\times2}\text{A}=\begin{bmatrix}-1&-8&-10\\1&-2&-10\\9&22&15\end{bmatrix}_{3\times3}$ From the given equation, it is clear that order of A should be 2 × 3. Let $\text{A}=\begin{bmatrix}\text{a}&\text{b}&\text{c}\\\text{d}&\text{e}&\text{f}\end{bmatrix}$ $\therefore\ \begin{bmatrix}2&-1\\1&0\\-3&4\end{bmatrix}\begin{bmatrix}\text{a}&\text{b}&\text{c}\\\text{d}&\text{e}&\text{f}\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-10\\9&22&15\end{bmatrix}$ $\Rightarrow\ \begin{bmatrix}2\text{a}-\text{d}&2\text{b}-\text{e}&2\text{c}-\text{f}\\\text{a}&\text{b}&\text{c}\\-3\text{a}+4\text{d}&-3\text{b}+4\text{e}&-3\text{c}+4\text{f}\end{bmatrix}=\begin{bmatrix}-1&-8&-10\\1&-2&-5\\9&22&15\end{bmatrix}$ By equality of matrices, we geta = 1, b = -2, c = -5
and 2a - d = -1 ⇒ d = 2a + 1 =3;
2b - e = -8 ⇒ e = 2(-2) + 8 = 4
2c - f = -10 ⇒ f = 2c + 10 = 0
$\therefore\ \text{A}=\begin{bmatrix}1&-2&-5\\3&4&0\end{bmatrix}$
View full question & answer→Question 885 Marks
If $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix},\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix},$ verify that $(AB)^T = B^TA^T$
AnswerGiven: $\text{A}=\begin{bmatrix}-2\\4\\5\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}-2&4&5\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&3&-6\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1\\3\\-6\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}-2\\4\\5\end{bmatrix}\begin{bmatrix}1&3&-6\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-2&-6&12\\4&12&-24\\5&15&-30\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\ \dots(1)$
$\text{B}^\text{T}\text{A}^\text{A}=\begin{bmatrix}1\\3\\-6\end{bmatrix}\begin{bmatrix}-2&4&5\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}-2&4&5\\-6&12&15\\12&-24&-30\end{bmatrix}\ \dots(2)$
$\therefore\ (\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T} [$From eqs. $(1)$ and $(2)]$
View full question & answer→Question 895 Marks
If $l_i, m_i, n_i; i = 1, 2, 3$ denotes the direction cosines of three mutually perpendicular vectors in space, prove that $AA^T = I,$ where
$\text{A}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}.$
AnswerGiven,
$l_i, m_i, n_i$ are direction cosines of three mutually perpendicular vectors
$\begin{matrix}\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2=0 \\\text{l}_2\text{l}_3+\text{m}_2\text{m}_3+\text{n}_2\text{n}_3=0\\\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3=0\end{matrix}\Bigg\}\ \dots(\text{A})$
And,
$\begin{matrix}\text{l}_1{^2}+\text{m}_1{^2}+\text{n}_1{^2}=1\\\text{l}_2{^2}+\text{m}_2{^2}+\text{n}_2{^2}=1\\\text{l}_3{^2}+\text{m}_3{^2}+\text{n}_3{^2}=1\end{matrix}\Bigg\}\ \dots(\text{B})$
Given
$\text{A}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}$
$\text{AA}^{T}=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}^\text{T}$
$=\begin{bmatrix}\text{l}_1&\text{m}_1&\text{n}_1\\\text{l}_2&\text{m}_2&\text{n}_2\\\text{l}_3&\text{m}_3&\text{n}_3 \end{bmatrix}\begin{bmatrix}\text{l}_1&\text{l}_1&\text{l}_3\\\text{m}_1&\text{m}_2&\text{m}_3\\\text{n}_1&\text{n}_2&\text{n}_3 \end{bmatrix}$
$=\begin{bmatrix}\text{l}_1{^2}+\text{m}_1{^2}+\text{n}_1{^2}&\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2&\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3\\\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2&\text{l}_2{^2}+\text{m}_2{^2}+\text{n}_2{^2}&\text{l}_2\text{l}_3+\text{m}_2\text{m}_3+\text{n}_2\text{n}_3\\\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3&\text{l}_3\text{l}_2+\text{m}_3\text{m}_2+\text{n}_3\text{n}_2&\text{l}_3{^2}+\text{m}_3{^2}+\text{n}_3{^2}\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \\ \end{bmatrix} \{$using $(A)$ and $(B)\}$
$=\text{l}$
Hence,
$\text{AA}^{\text{T}}=\text{l}$
View full question & answer→Question 905 Marks
Find the values of x, y, z if the matrix A = $\begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}\\\text{x}&-\text{y}&\text{z}\end{bmatrix}$satisfies the equation A’A = I.
AnswerGiven: $\text{A} = \begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}\\\text{x}&-\text{y}&\text{z}\end{bmatrix}$ $\Rightarrow\text{A}'= \begin{bmatrix}0&\text{x}&\text{x}\\2\text{y}&\text{y}&-\text{y}\\\text{z}&-\text{z}&\text{z}\end{bmatrix}$
Now $\text{A}'\text{A} =\text{ I}\Rightarrow\begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}&\\\text{x}&-\text{y}&\text{z}\end{bmatrix} \begin{bmatrix}0&\text{x}&\text{x}\\2\text{y}&\text{y}&-\text{y}\\\text{z}&-\text{z}&\text{z}\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}0+\text{x}^{2}+\text{x}^{2}&0+\text{xy}-\text{xy}&0-\text{xz}+\text{xz}\\0+\text{xy}-\text{xy}&4\text{y}^{\text{y}}+\text{y}^{2}+\text{y}^{2}&2\text{yz}-\text{yz}-\text{yz}\\0-\text{zx}+\text{zx}&2\text{yz}-\text{yz}-\text{yz}&\text{z}^{2}+\text{z}^{2}+\text{z}^{2}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2\text{x}^{2}&0&0\\0&6\text{y}^{2}&0\\0&0&3\text{z}^{2}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Equating corresponding entries, we have
$2\text{x}^{2}=1\ \Rightarrow \ \text{x}^{2}=\frac{1}{2} \ \Rightarrow\ \text{x}=\pm\frac{1}{\sqrt{2}}$
And $6\text{y}^{2}=1\ \Rightarrow \ \text{y}^{2}=\frac{1}{6} \ \Rightarrow\ \text{y}=\pm\frac{1}{\sqrt{6}}$
And $3\text{z}^{2}=1\ \Rightarrow \ \text{z}^{2}=\frac{1}{3} \ \Rightarrow\ \text{z}=\pm\frac{1}{\sqrt{3}}$
$\therefore \ \text{x}=\pm \frac{1}{\sqrt{2}},\text{y}=\pm\frac{1}{\sqrt{6}},\text{z}=\pm\frac{1}{\sqrt{3}}$
View full question & answer→Question 915 Marks
If $\text{A}=\begin{bmatrix}4&2\\-1&-1 \end{bmatrix},$ prove that (A - 2I)(A - 3I) = 0
AnswerGiven, $\text{A}=\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}$
$(\text{A}-2\text{I})(\text{A}-3\text{I})$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-2\begin{bmatrix}1&0\\0&1 \end{bmatrix}\end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-3\begin{bmatrix}1&0\\0&1 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}2&0\\0&2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}3&0\\0&3 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4-2&2-0\\-1-0&1-2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4-3&2-0\\-1-0&1-3 \end{bmatrix} \end{pmatrix}$
$=\begin{bmatrix}2&2\\-1&-1 \end{bmatrix}\begin{bmatrix}1&2\\-1&-2 \end{bmatrix}$
$=\begin{bmatrix}2-2&4-4\\-1+1&-2+2 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
$=0$
Hence,
$(\text{A}-2\text{I})(\text{A}-3\text{I})=0$
View full question & answer→Question 925 Marks
If $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix},$ show that AB = A and BA = B.
AnswerGiven, $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix},\text{B}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$\text{AB}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$=\begin{bmatrix}4+3-5&-4-9+10&-8-12+15\\-2-4+5&2+12-10&4+16-15\\2+3-4&-2-9+18&-4-12+12\end{bmatrix}$
$=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$\text{AB}=\text{A}$
$\text{BA}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$
$=\begin{bmatrix}4+2-4&-6-8+12&-10-10+16\\-2-3+4&3+12-12&5+15-16\\2+2-3&-3-8+9&-5-10+12\end{bmatrix}$
$=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}$
$\text{BA}=\text{B}$
View full question & answer→Question 935 Marks
If $\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix},\text{B}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}0&1\\1&0\end{bmatrix},$ then show that $A^2 = B^2 = C^2 = l_2.$
AnswerHere,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0&0+0\\0+0&0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(1)$
$\text{B}^2=\text{BB}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1+0&0-0\\0-0&0+1\end{bmatrix}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(2)$
$\text{C}^2=\text{CC}$
$\Rightarrow\text{C}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\Rightarrow\text{C}^2=\begin{bmatrix}0+1&0+0\\0+0&1+0\end{bmatrix}$
$\Rightarrow\text{C}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(3)$
We know,
$\text{I}_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\ \dots(4)$
$\Rightarrow\text{A}^2=\text{B}^2=\text{C}^2=\text{I}^2 [$From eqs. $(1), (2), (3)$ and $(4)]$
View full question & answer→Question 945 Marks
For the following matrices verify the associativity of multiplication i.e., (AB) C = A(BC):
$\text{A}=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
AnswerGiven,
$\text{A}=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix},\text{C}=\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$(\text{AB})\text{C}=\begin{pmatrix}\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}\end{pmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}4+0+6&-4+2-3&4+4+3\\1+0+4&-1+1-2&1+2+2\\3+0+2&-3+0-1&3+0+1\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}10&-5&11\\5&-2&5\\5&-4&4\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}10-15+0&20+0+0&-10+5+11\\5-6+0&10+0+0&-5-2+5\\5-12+0&10+0+0&-5-4+4\end{bmatrix}$
$(\text{AB})\text{C}=\begin{bmatrix}-5&20&-4\\-1&10&-2\\-7&10&-5\end{bmatrix}\ \dots(\text{i})$
$\text{A}(\text{BC})=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}\end{pmatrix}$
$=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}1-3+0&2+0+0&-1-1+1\\0+3+0&0+0+0&0+1+2\\2-3+0&4+0+0&-2-1+1\end{bmatrix}$
$=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix}\begin{bmatrix}-2&2&-1\\3&0&3\\-1&4&-2\end{bmatrix}$
$=\begin{bmatrix}-8+6-3&8+0+12&-4+6-6\\-2+3-2&2+0+8&-1+3-4\\-6+0-1&6+0+4&-3+0-2\end{bmatrix}$
$\text{A}(\text{BC})=\begin{bmatrix}-5&20&-4\\-1&10&-2\\-7&10&-5\end{bmatrix}\ \dots\text{(ii)}$
From equation (i) and (ii),
$(\text{AB})\text{C}=\text{A}(\text{BC})$
View full question & answer→Question 955 Marks
If $\text{A}=\begin{bmatrix}2&-1\\3&2 \end{bmatrix} $ and $\text{B}=\begin{bmatrix}0&4\\-1&7\end{bmatrix} ,$ find $3A^2 - 2B + l$
AnswerGiven: $\text{A}=\begin{bmatrix}2&-1\\3&2 \end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&-1\\3&2 \end{bmatrix}\begin{bmatrix}2&-1\\3&2 \end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4-3&-2-2\\6+6&-3+4 \end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-4\\12&1 \end{bmatrix}$
$3\text{A}^2-2\text{B}+\text{I}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=3\begin{bmatrix}1&-4\\12&1 \end{bmatrix}-2\begin{bmatrix}0&4\\-1&7 \end{bmatrix}+\begin{bmatrix}1&0\\0&1 \end{bmatrix}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=\begin{bmatrix}3&-12\\36&3 \end{bmatrix}-\begin{bmatrix}0&8\\-2&14 \end{bmatrix}+\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=\begin{bmatrix}3-0+1&-12-8+0\\36+2+0&3-14+1\end{bmatrix}$
$\Rightarrow3\text{A}^2-2\text{B}+\text{I}=\begin{bmatrix}4&-20\\38&-10 \end{bmatrix}$
View full question & answer→Question 965 Marks
If $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix},$ then find the value of $x, y, z$ and $w.$
AnswerWe have, $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix}$
By comparing elements we get, $x + y = 6, xy = 8, z + 6 = 0$ and $w = 4$
From first two equations
$(6 - y).y = 8$
$\Rightarrow y^2 - 6y + 8 = 0$
$\Rightarrow (y - 2)(y - 4) = 0$
$\Rightarrow y = 2$ or $y = 4$
$\therefore x = 4$ or $x = 2$
Also, $z + 6 = 0$
$\Rightarrow z = -6$ and $w = 4$
$\therefore x = 2, y = 4$ or $x = 4, y = 2, z = -6$ and $w = 4$
View full question & answer→Question 975 Marks
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:
| Market |
|
Products |
|
| I |
10, 000 |
2, 000 |
18, 000 |
| II |
6, 000 |
20, 000 |
8, 000 |
- If unit sales prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00 respectively, find the total revenue in each market with the help of matrix algebra.
- If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
AnswerAccording to question, the matrix $\text{A}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\\10,000&2,000&18,000\\6,000&20,000&8,000\end{bmatrix}$
- Let B be the column matrix representing sale price of each unit of products x, y, z.
Then $\text{B}=\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}_{3\times1}$
Now Revenue = Sale price c Number of items sold$\Rightarrow\ \begin{bmatrix}10,000&2,000&18,000\\6,000&20,000&8,000\end{bmatrix}\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}=\begin{bmatrix}25,000+3,000+18,000\\15,000+30,000+8,000\end{bmatrix}=\begin{bmatrix}46,000\\53,000\end{bmatrix}$
Therefore, the revenue collected by sale of all items in Market I = ₹ 46, 000 and the revenue collected by sale of all items in Market II = ₹ 53, 000.
- Let C be the column matrix representing cost price of each unit of products x, y, z.
Then $\text{C} = \begin{bmatrix}2\\1\\0.5\end{bmatrix}_{3\times1}$
$\therefore$ Total cost = AC = $\begin{bmatrix}10,000&2,000&18,000\\6,000&20,000&8,000\end{bmatrix}\begin{bmatrix}2\\1\\0.5\end{bmatrix}$
$= \begin{bmatrix}20,000+2,000+9,000\\12,000+20,000+4,000\end{bmatrix} = \begin{bmatrix}46,000\\53,000\end{bmatrix}\begin{bmatrix}31,000\\36,000\end{bmatrix} $
$\therefore$ The profit collected in two markets is given in matrix form as
Profit matrix = Revenue matrix – Cost matrix
$\Rightarrow\begin{bmatrix}46,000\\53,000\end{bmatrix}-\begin{bmatrix}31,000\\36,000\end{bmatrix} = \begin{bmatrix}15,000\\17,000\end{bmatrix}$
Therefore, the gross profit in both the market = ₹15000 + ₹ 17000 = ₹ 32,000. View full question & answer→Question 985 Marks
If $\text{A}=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix},$ find $A^2 - 5A + 4I$ and hence find a matrix $X$ such that $A^2 - 5A + 4I + X = 0.$
AnswerGiven: $\text{A}=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}$
$ \text{A}^2=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}$
$=\begin{bmatrix}4+0+1&0+0-1&2+0+0\\4+2+3&0+1-3&2+3+0\\2-2+0&0-1-0&1-3+0\end{bmatrix}$
$ =\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}$
Now,
$ \text{A}^2-5\text{A}+4\text{I}=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\end{bmatrix}+4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$ =\begin{bmatrix}5-10+4&-1-0+0&2-5+0\\9-10+0&-2-5+4&5-15+0\\0-5+0&-1+5+0&-2-0+4\end{bmatrix}$
$ =\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}$
Now, $ \text{A}^2-5\text{A}+4\text{I}+\text{X}=0$
$ =\text{X}=-(\text{A}^2-5\text{A}+4\text{I})$
$ \therefore\ \text{X}=-\begin{bmatrix}-1&-1&-3\\-1&-3&-10\\-5&4&2\end{bmatrix}=\begin{bmatrix}1&1&3\\1&3&10\\5&-4&-2\end{bmatrix}$
View full question & answer→Question 995 Marks
If $\text{A}=\begin{bmatrix}\cos\theta&\text{i}\sin\theta\\\text{i}\sin\theta&\cos\theta\end{bmatrix},$ then prove by principle of mathematical induction that $\text{A}^\text{n}=\begin{bmatrix}\cos\text{n}\theta&\text{i}\sin\text{n}\theta\\\text{i}\sin\text{n}\theta&\cos\text{n}\theta\end{bmatrix}$ for all $\text{n}\in\text{N}.$
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 $\text{A}^1=\begin{bmatrix}\cos1\theta&\text{i}\sin1\theta\\\text{i}\sin1\theta&\cos1\theta\end{bmatrix}=\begin{bmatrix}\cos\theta&\text{i}\sin\theta\\\text{i}\sin\theta&\cos\theta\end{bmatrix}=\text{A}$ Thus, the result is true for n = 1. Step 2: Let the result be true for n = m. Then, $\text{A}^\text{m}=\begin{bmatrix}\cos\text{m}\theta&\text{i}\sin\text{m}\theta\\\text{i}\sin\text{m}\theta&\cos\text{m}\theta\end{bmatrix}$ Now we shall show that the result is true for n = m + 1. Here, $ \text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}+1\theta&\text{i}\sin\text{m}+1\theta\\\text{i}\sin\text{m}+1\theta&\cos\text{m}+1\theta\end{bmatrix}$ By definition of integral power of matrix, we have $\text{A}^\text{m+1}=\text{A}^\text{m}\text{A}$ $\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}\theta&\text{i}\sin\text{m}\theta\\\text{i}\sin\text{m}\theta&\cos\text{m}\theta\end{bmatrix}\begin{bmatrix}\cos\theta&\text{i}\sin\theta\\\text{i}\sin\theta&\cos\theta\end{bmatrix}$ [From eq. (1)] $\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}\theta.\cos\theta+\text{i}\sin\text{m}\theta.\text{i}\sin\theta&\cos\text{m}\theta.\text{i}\sin\theta+\text{i}\sin\text{m}\theta.\cos\theta\\\text{i}\sin\text{m}\theta.\cos\theta+\cos\text{m}\theta.\text{i}\sin\theta&\text{i}\sin\text{m}\theta.\text{i}\sin\theta+\cos\text{m}\theta.\cos\theta\end{bmatrix}$ $ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\cos\text{m}\theta.\cos\theta-\sin\text{m}\theta.\sin\theta&\text{i}(\cos\text{m}\theta.\sin\theta+\sin\text{m}\theta.\cos\theta)\\\text{i}(\sin\text{m}\theta.\cos\theta+\cos\text{m}\theta.\sin\theta)&-\sin\text{m}\theta.\sin\theta+\cos\text{m}\theta.\cos\theta\end{bmatrix}$ $\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos\text{m}\theta.\cos\theta-\sin\text{m}\theta.\sin\theta&\text{i}(\cos\text{m}\theta.\sin\theta+\sin\text{m}\theta.\cos\theta)\\\text{i}(\sin\text{m}\theta.\cos\theta+\cos\text{m}\theta.\sin\theta)&\cos\text{m}\theta.\cos\theta-\sin\text{m}\theta.\sin\theta\end{bmatrix}$ $\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos(\text{m}\theta+\theta)&\text{i}\sin(\text{m}\theta+\theta)\\\text{i}\sin(\text{m}\theta+\theta)&\cos(\text{m}\theta+\theta)\end{bmatrix}$ $ \Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}\cos(\text{m}+1)\theta&\text{i}\sin(\text{m}+1)\theta\\\text{i}\sin(\text{m}+1)\theta&\cos(\text{m}+1)\theta\end{bmatrix}$This shows that when the result is true for n = m, it is true for n = m + 1.
Hence, by the principle of mathematical induction, the result is valid for all $\text{n}\in\text{N}.$
View full question & answer→Question 1005 Marks
If $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that $A^2 - 5A + 7I = 0$ use this to find $A^4.$
AnswerGiven: $\text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{A}^2-5\text{A}+7\text{I}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8-15+7&5-5+0\\-5+5+0&3-10+7\end{bmatrix}$
$ \Rightarrow\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
Given: $\text{A}^2-5\text{A}+7\text{I}=0$
$\Rightarrow\text{A}^2=5\text{A}-7\text{I}\ \dots(1)$
$\Rightarrow\text{A}^3=\text{A}(5\text{A}-7\text{I}) ($Multipying by $A$ on both sides$)$
$\Rightarrow\text{A}^3=5\text{A}^2-7\text{AI}$
$\Rightarrow\text{A}^3=5(5\text{A}-7\text{I})-7\text{A} [$From $eq. (1)]$
$\Rightarrow\text{A}^3=25\text{A}-35\text{I}-7\text{A}$
$\Rightarrow\text{A}^3=18\text{A}-35\text{I}$
$\Rightarrow\text{A}^4=(18\text{A}-35\text{I})\text{A} ($Multipying by $A$ on both sides$)$
View full question & answer→