Question 1015 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 1025 Marks
Construct $a_{2 \times 2}$ matrix, where,
- $\text{a}_\text{ij}=\frac{(\text{i}-2\text{j})^2}{2}$
- $\text{a}_\text{ij}=|-2\text{i}+3\text{j}|$
Answer
- We have, $\text{A}=[\text{a}_\text{ij}]_{2\times2}$
Such that, $\text{a}_\text{ij}=\frac{(\text{i}-2\text{j})^2}{2};$ where $1\leq\text{i}\leq2;1\leq\text{j}\leq2\ ....(\text{i})$
$\therefore\ \text{a}_{11}=\frac{(1-2)^2}{2}=\frac{1}{2}$
$\text{a}_{12}=\frac{(1-2\times2)^2}{2}=\frac{9}{2}$
$\text{a}_{21}=\frac{(2-2\times1)^2}{2}=0$
$\text{a}_{22}=\frac{(2-2\times2)^2}{2}=2$
So, $\text{A}=\begin{bmatrix}\frac{1}{2}&\frac{9}{2}\\0&2\end{bmatrix}$
- We have, $\text{A}=[\text{a}_\text{ij}]_{2\times2}$
Such that, $\text{a}_\text{ij}=|-2\text{i}+3\text{j}|;$ where $1\leq\text{i}\leq2;1\leq\text{j}\leq2$
$\therefore\ \text{a}_{11}=|-2\times1+3\times1|=1$
$\text{a}_{12}=|-2\times1+3\times2|=4$
$\text{a}_{21}=|-2\times2+3\times1|=-1$
$\text{a}_{22}=|-2\times2+3\times2|=2$
$\therefore\ \text{A}=\begin{bmatrix}1&4\\-1&2\end{bmatrix}$ View full question & answer→Question 1035 Marks
If A is a square matrix, using mathematical induction prove that $(A^T)^n = (A^n)^T$ for all $n ∈ N$.
AnswerLet the given statement $P(n)$, be given as
$P(n):\left(A^{\top}\right)^n=\left(A^n\right)^{\top} \text { for all } n \in N \text {. }$
We observe that
$P(1):\left(A^{\top}\right)^1=A^{\top}=\left(A^1\right)^{\top}$
Thus, $\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$.
Assume that $P(n)$ is true for $n=k \in N$.
$\text { i.e., } P(k):\left(A^{\top}\right)^k=\left(A^k\right)^{\top}$
To prove that $P(k+1)$ is true, we have
$\left(A^T\right)^{k+1}=\left(A^{\top}\right)^k \cdot\left(A^T\right)^1$
$=\left(A^k\right)^{\top} \cdot\left(A^1\right)^{\top}$
$=\left(A^{k+1}\right)^{\top}$
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 1045 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 1055 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 1065 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 1075 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 1085 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 1095 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$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) = R.H.S.
View full question & answer→Question 1105 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 1115 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 1125 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 1135 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 1145 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 1155 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 1165 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 1175 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 1185 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 1195 Marks
Let $\text{A}=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$ Use the principle of mathematical induction to show that
$\text{A}^\text{n}=\begin{bmatrix}1&\text{n}&\frac{\text{n}(\text{n}+1)}{2}\\0&1&\text{n}\\0&0&1\end{bmatrix}$ 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
$\text{A}^1=\begin{bmatrix}1&1&\frac{1(1+1)}{2}\\0&1&1\\0&0&1\end{bmatrix}=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}=\text{A}$
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}1&\text{m}&\frac{\text{m}(\text{m}+1)}{2}\\0&1&\text{m}\\0&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}1&\text{m}+1&\frac{\text{m}+1(\text{m}+1+1)}{2}\\0&1&\text{m}+1\\0&0&1\end{bmatrix}$
$ \text{A}^\text{m+1}=\begin{bmatrix}1&\text{m}+1&\frac{(\text{m}+1)(\text{m}+2)}{2}\\0&1&\text{m}+1\\0&0&1\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}1&\text{m}&\frac{\text{m}(\text{m}+1)}{2}\\0&1&\text{m}\\0&0&1\end{bmatrix}\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}$ [From eq. (1)]
$ \Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}1+0+0&1+\text{m}+0&1+\text{m}+\frac{\text{m}(\text{m}+1)}{2}\\0+0+0&0+1+0&0+1+\text{m}\\0+0+0&0+0+0&0+0+1\end{bmatrix}$
$ \Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}1&1+\text{m}&\frac{(2+2\text{m}+\text{m}^2+\text{m})}{2}\\0&1&1+\text{m}\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}1&1+\text{m}&\frac{(\text{m}^2+3\text{m}+2)}{2}\\0&1&1+\text{m}\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^\text{m+1}=\begin{bmatrix}1&1+\text{m}&\frac{(\text{m}+1)(\text{m}+2)}{2}\\0&1&1+\text{m}\\0&0&1\end{bmatrix}$
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 1205 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 1215 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 1225 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 1235 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}$$[\because\ \text{R}_2\rightarrow\ \frac{1}{22}\text{R}_2]$
$\Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}\frac{7}{22}&\frac{-3}{22}\\\frac{5}{22}&\frac{1}{22}\end{bmatrix}\text{A}$ $[\because\ \text{R}_2\rightarrow\ \text{R}_1+3\text{R}_2]$ $\Rightarrow\ \begin{bmatrix}1&0\\0&1\end{bmatrix}=\frac{1}{22}\begin{bmatrix}7&-3\\5&1\end{bmatrix}\text{A}$ $\Rightarrow\ \text{A}^{-1}=\frac{1}{22}\begin{bmatrix}7&-3\\5&1\end{bmatrix}$
View full question & answer→Question 1245 Marks
If $\text{A}=\begin{bmatrix}0&1\\1&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&-1\\1&0\end{bmatrix},$ then show that $(\text{A}+\text{B})(\text{A}-\text{B})\neq\text{A}^2-\text{B}^2.$
AnswerWe have, $\text{A}=\begin{bmatrix}0&1\\1&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$\therefore\ (\text{A}+\text{B})=\begin{bmatrix}0+0&1-1\\1+1&1+0\end{bmatrix}$
$=\begin{bmatrix}0&0\\2&1\end{bmatrix}$
and $(\text{A}-\text{B})=\begin{bmatrix}0-0&1+1\\1-1&1-0\end{bmatrix}$
$=\begin{bmatrix}0&2\\0&1\end{bmatrix}$
$(\text{A}+\text{B}).(\text{A}-\text{B})=\begin{bmatrix}0&0\\2&1\end{bmatrix}\begin{bmatrix}0&2\\0&1\end{bmatrix}$
$=\begin{bmatrix}0+0&0+0\\0+0&4+1\end{bmatrix}=\begin{bmatrix}0&0\\0&5\end{bmatrix}\ ....(\text{i})$
Also, $\text{A}^2=\text{A}.\text{A}$
$=\begin{bmatrix}0&1\\1&1\end{bmatrix}.\begin{bmatrix}0&1\\1&1\end{bmatrix}$
$=\begin{bmatrix}0+1&0+1\\0+1&1+1\end{bmatrix}=\begin{bmatrix}1&1\\1&2\end{bmatrix}$
and $\text{B}^2=\text{B}.\text{B}$
$=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}$
$=\begin{bmatrix}0-1&0+0\\0+0&-1+0\end{bmatrix}=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$
$\therefore\ \text{A}^2-\text{B}^2=\begin{bmatrix}1&1\\1&2\end{bmatrix}-\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$
$=\begin{bmatrix}2&1\\1&3\end{bmatrix}\ ....(\text{ii})$
From (i) and (ii), $(\text{A}+\text{B})(\text{A}-\text{B})\neq\text{A}^2-\text{B}^2$
View full question & answer→Question 1255 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 1265 Marks
If $A =$ diag$(a, b, c)$, show that $A^n =$ diag$(a^n, b^n, c^n) $for all positive integer n.
AnswerGiven, $A=\operatorname{diag}(a, b, c)$ Show that, $A^n=\operatorname{diag}\left(a^n, b^n, c^n\right)$
Step 1: Put $n=1 A^1=\operatorname{diag}\left(a^1, b^1, c^1\right) A=\operatorname{diag}(a, b, c)$
So, $A^n$ is true for $n=1$
Step 2: Let, $A n$ be true for $n=k$, so, $A^k=\operatorname{diag}\left(a^k, b^k, c^k\right) \ldots$ (i)
Step 3: Now, we have to show that, $A^{k+1}=\operatorname{diag}\left(a^{k+1}, b^{k+1}, c^{k+1}\right)$
Now, $A ^{ k +1}= A ^{ k } \times k ^3=\operatorname{diag}\left( a ^{ k }, b ^{ k }, c ^{ k }\right) \times \operatorname{diag}( a , b , c )\{ using$ equation (i) and given $\}$
$\textA}^{\text{k}+1}=\begin{bmatrix}\text{a}^\text{k}&0&0\\0&\text{b}^\text{k}&0\\0&0&\text{c}^\text{k}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
$ =\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 1275 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 1285 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 1295 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 1305 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 1315 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 1325 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 1335 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 1345 Marks
If AB = BA for any two sqaure matrices, prove by mathematical induction that $(AB)^n = A^nB^n.$
AnswerLet $P(n) : (AB)^n = A^nB^n$
$\therefore$ $P(1) : (AB)^1 = A^1B^1$
$\Rightarrow AB = AB$
So, 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 1355 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 1365 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 × 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 × 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$
Solution:
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
Solution:
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.
Solution:
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 1375 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 1385 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 $\text{x}^2+\text{y}^2+\text{z}^2=1$ $\Rightarrow\text{x}^2=1-\text{y}^2-\text{z}^2=1-\frac{1}{6}-\frac{1}{3}$
$=1-\frac{3}{6}=\frac{1}{2}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt{2}}$
v b$\therefore\text{x}=\pm,\frac{1}{\sqrt{2}},\text{y}=\pm\frac{1}{\sqrt{6}}$ and $\text{z}=\pm\frac{1}{\sqrt{3}}$
View full question & answer→Question 1395 Marks
For the matrices A and B, verify that (AB)' = B'A' where
-
$\text{A}=\begin{bmatrix}1\\-4\\3\end{bmatrix},\text{B}=\begin{bmatrix}-1&2&1\end{bmatrix}$
- $\text{A}=\begin{bmatrix}0\\1\\2\end{bmatrix},\text{B}=\begin{bmatrix}1&5&7\end{bmatrix}$
Answer
- $\text{AB}= \begin{bmatrix}1\\-4\\3\end{bmatrix}\begin{bmatrix}-1&2&1\end{bmatrix}=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}$
$\therefore\ \text{(AB)}'=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix}$
Now, $\text{A}'=\begin{bmatrix}1&-4&3\end{bmatrix},\text{B}'=\begin{bmatrix}-1\\2\\1\end{bmatrix}$
$\therefore\ \text{B}'\text{A}'=\begin{bmatrix}-1&\\2\\1\end{bmatrix}\begin{bmatrix}1&-4&3\end{bmatrix}=\begin{bmatrix}-1&4&-3\\2&-8&6\\1&-4&3\end{bmatrix} $
Hence, we have verified that (AB)' = B'A'.
- $\text{AB}=\begin{bmatrix}0\\1\\2\end{bmatrix}\begin{bmatrix}1&5&7\end{bmatrix}=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}$
$\therefore\ \text{(AB)}'=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}$
Now, $\text{A}'=\begin{bmatrix}0&1&2\end{bmatrix},\text{B}'=\begin{bmatrix}1\\5\\7\end{bmatrix}$
$\therefore\ \text{B}'\text{A}'=\begin{bmatrix}1\\5\\7\end{bmatrix}\begin{bmatrix}0&1&2\end{bmatrix}=\begin{bmatrix}0&1&2\\0&5&10\\0&7&14\end{bmatrix}$
Hence, we have verified that (AB)' = B'A'. View full question & answer→Question 1405 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 1415 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 1425 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}$$\therefore\ \text{A}^2=\text{A}.\text{A}$
$=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}.\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}$
$=\begin{bmatrix}1+0+0&0+0-1&-1+0-1\\2+2+0&0+1+3&-2+3+3\\0+2+0&0+1+1&0+3+1\end{bmatrix}$
$=\begin{bmatrix}1&-1&-2\\4&4&4\\2&2&4\end{bmatrix}$
$\therefore\ \text{A}^2+\text{A}=\begin{bmatrix}1&-1&-2\\4&4&4\\2&2&4\end{bmatrix}+\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}$
$=\begin{bmatrix}2&-1&-3\\6&5&7\\2&3&5\end{bmatrix}\ ....(\text{i})$
Now, $\text{A}+\text{I}=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}+\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}2&0&-1\\2&2&3\\0&1&2\end{bmatrix}$
So, $\text{A}(\text{A}+\text{I})=\begin{bmatrix}1&0&-1\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}2&0&-1\\2&2&3\\0&1&2\end{bmatrix}$
$=\begin{bmatrix}2+0+0&0+0-1&-1+0-2\\4+2+0&0+2+3&-2+3+6\\0+2+0&0+2+1&0+3+2\end{bmatrix}$
$=\begin{bmatrix}2&-1&-3\\6&5&7\\2&3&5\end{bmatrix}\ ....(\text{ii})$
From (i) and (ii), we get $\text{A}^{2}+\text{A}=\text{A}(\text{A}+\text{I}).$
View full question & answer→Question 1435 Marks
If $\text A = \begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix},\ \text B = \begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix} \text{and}\ \text C = \begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix},$ then compute (A + B) and (B - C).
Also, verify that A + (B - C) = (A + B) - C.
Answer$\text{A + B}=\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix}+\begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix}$$=\begin{bmatrix}1+3&2-1&-3+2\\5+4&0+2&2+5\\1+2&-1+0&1+3\end{bmatrix}$$=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix} $
$\text{B - C}=\begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix}-\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix}$
$=\begin{bmatrix}3-4&-1-1&2-2\\4-0&2-3&5-2\\2-1&0+2&3-3\end{bmatrix}$$=\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}$
Now, A + (B - C) = (A + B) - C
$\Rightarrow\begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix}+\begin{bmatrix}-1&-2&0\\4&-1&3\\1&2&0\end{bmatrix}$
$=\begin{bmatrix}4&1&-1\\9&2&7\\3&-1&4\end{bmatrix}-\begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix} $
$\Rightarrow \begin{bmatrix}1-1&2-2&-3+0\\5+4&0-1&2+3\\1+1&-1+2&1+0\end{bmatrix}$
$=\begin{bmatrix}4-4&1-1&-1-2&\\9-0&2-3&7-2\\3-1&-1+2&4-3\end{bmatrix} $
$\Rightarrow \begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix}=\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix} $
$\Rightarrow$ L.H.S = R.H.S
View full question & answer→Question 1445 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 1455 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 1465 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 1475 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 1485 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→