Question 515 Marks
Show that $\text{AB}\neq\text{BA}$ in the following cases:
$\text{A}=\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix},\ \text{B}=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$$\text{AB}=\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix}\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}$
$=\begin{bmatrix}10-12-1&20-16-3&10-8-2\\-11+15+0&-22+20+0&-11+10+0\\9-15+1&18-20+3&9-10+2\end{bmatrix}$
$\text{AB}=\begin{bmatrix}-3&1&0\\4&-2&-1\\-5&1&1\end{bmatrix}\ \dots(\text{i})$
$\text{BA}=\begin{bmatrix}1&2&1\\3&4&2\\1&3&2\end{bmatrix}\begin{bmatrix}10&-4&-1\\-11&5&0\\9&-5&1 \end{bmatrix}$
$=\begin{bmatrix}10-22+9&-4+10-5&-9+0+1\\30-44+10&-12+20-10&-3+0+2\\10-33+18&-4+15-10&-1+0+2\end{bmatrix}$
$\text{BA}=\begin{bmatrix}-3&1&0\\4&-2&-1\\-5&1&1\end{bmatrix}\ \dots(\text{ii})$
From equation (i) and (ii),
$\text{AB}\neq\text{BA}$
View full question & answer→Question 525 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 535 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 545 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 555 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 565 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 575 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 585 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,
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) = $RHS
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 595 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 605 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 615 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 625 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 635 Marks
If $\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix},$ find $A^{16}$.
AnswerGiven,
$\text{A}=\begin{bmatrix}0&0\\4&0\end{bmatrix}$
$ \text{A}^2=\text{A}\times\text{A}$
$ =\begin{bmatrix}0&0\\4&0\end{bmatrix}\begin{bmatrix}0&0\\4&0\end{bmatrix}$
$ =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
$ \text{A}^4=\text{A}^2\times\text{A}^2$
$=0\times0$
$=0$
$ \text{A}^{16}=\text{A}^4\times\text{A}^4$
$=0\times0$
$=0$
So,
$A^{16}$ is a null matrix
View full question & answer→Question 645 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 655 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 665 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 675 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 685 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 695 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 705 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 715 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 725 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 735 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 745 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 755 Marks
Find the value of x if $\begin{bmatrix}1&\text{x}&1\end{bmatrix}\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}=0$
AnswerWe have, $\begin{bmatrix}1&\text{x}&1\end{bmatrix}_{1\times3}\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix}_{3\times3}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}_{3\times1}=0$
$\Rightarrow\ \begin{bmatrix}1+2\text{x}+15&3+5\text{x}+3&2+\text{x}+2\end{bmatrix}_{1\times3}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}_{3\times1}=0$
$\Rightarrow\ \begin{bmatrix}16+2\text{x}&5\text{x}+6&\text{x}+4\end{bmatrix}_{1\times3}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}_{3\times1}=0$
$\Rightarrow\ \begin{bmatrix}16+2\text{x}+10\text{x}+12+\text{x}^2+4\text{x}\end{bmatrix}=0$
$\Rightarrow\ [\text{x}^2+16\text{x}+28]=0$
$\Rightarrow\ \text{x}^2+16\text{x}+28=0$
$\Rightarrow\ (\text{x}+2)(\text{x}+4)=0$
$\Rightarrow\ \text{x}=-2,-14$
View full question & answer→Question 765 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 775 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 785 Marks
Show that the matrix $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ satisfies the equation $A^3 - 4A^2 + A = 0$.
AnswerGiven, $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^3=\text{A}^2.\text{A}$
$ =\begin{bmatrix}7&12\\4&7\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}14+12&21+24\\8+7&12+14\end{bmatrix}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}$
Hence, $\text{A}^3-4\text{A}^2+\text{A}$
$=\begin{bmatrix}26&45\\15&26\end{bmatrix}-4\begin{bmatrix}7&12\\4&7\end{bmatrix}+\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}26-28+2&45-48+3\\15-16+1&26-28+2\end{bmatrix}$
$ =\begin{bmatrix}0&0\\0&0\end{bmatrix}$
So, $ \text{A}^3-4\text{A}+\text{A}=0$
View full question & answer→Question 795 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 805 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→Question 815 Marks
If possible, find BA and AB, where $\text{A}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}.$
AnswerWe have, $\text{A}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}_{2\times3}$ and $\text{B}=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}_{3\times2}$
Since, in both AB and BA, the number of columns of first is equal to the number of rows of second.
So, AB and BA both are possible.
$\therefore\ \text{AB}=\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}$
$=\begin{bmatrix}8+2+2&2+3+4\\4+4+4&1+6+8\end{bmatrix}=\begin{bmatrix}12&9\\12&15\end{bmatrix}$
Also, $\text{BA}=\begin{bmatrix}4&1\\2&3\\1&2\end{bmatrix}\begin{bmatrix}2&1&2\\1&2&4\end{bmatrix}$
$=\begin{bmatrix}8+1&4+2&8+4\\4+3&2+6&4+12\\2+2&1+4&2+8\end{bmatrix}=\begin{bmatrix}9&6&12\\7&8&16\\4&5&10\end{bmatrix}$
View full question & answer→Question 825 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 835 Marks
If $\text{A}=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix},$ then show that A is a root of the polynomial $f(x) = x^3 - 6x^2 + 7x + 2$.
AnswerGiven: $f(x) = x^3 - 6x^2 + 7x + 2$
$f(A) = A^3 - 6A^2 + 7A + 2I_3$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}$
$ \Rightarrow\text{A}^3=\begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}$
$\Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}-\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}+\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}+\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$
$ \Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2\end{bmatrix}$
$ \Rightarrow\text{A}^3-6\text{A}^2+7\text{A}+2\text{I}_3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Since f(A) = 0, A is the root of $f(x) = x^3 - 6x^2 + 7x + 2$.
View full question & answer→Question 845 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 855 Marks
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ show that $\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $\text{A}^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}.$
AnswerGiven, $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1 \end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0&1+1\\0+0&0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$
$ \text{A}^3=\text{A}^2\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1&2\\0&1 \end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1+0&1+2\\0+0&0+1 \end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}1&3\\0&1 \end{bmatrix}$
Hence proved.
View full question & answer→Question 865 Marks
If $\text{A}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix},$ find $A^T - B^T$.
AnswerGiven: $\text{A}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}$
Now,
$\text{A}^\text{T}-\text{B}^\text{T}=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}-\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}$
$=\begin{bmatrix}3+1&4-1\\-1-2&2-2\\0-1&1-3\end{bmatrix}$
$=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
Therefore,
$\text{A}^\text{T}-\text{B}^\text{T}=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}$
View full question & answer→Question 875 Marks
If $f(x) = x^2 - 2x$, find f(A), where $\text{A}=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
AnswerGiven: $f(x) = x^2 - 2x$
$f(A) = A^2 - 2A$
Now,
$\text{A}^2=\text{AA}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}0+4+0&0+5+4&0+0+6\\0+20+0&4+25+0&8+0+0\\0+8+0&0+10+6&0+0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}$
$\text{f(A)}=\text{A}^2-2\text{A}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-2\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-\begin{bmatrix}0&2&4\\8&10&0\\0&4&6\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4-0&9-2&6-4\\20-8&29-10&8-0\\8-0&16-4&9-6\end{bmatrix}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&7&2\\12&19&8\\8&12&3\end{bmatrix}$
View full question & answer→Question 885 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 895 Marks
For the matrices $A$ and $B$, verify that $(AB)^T = B^TA^T$, where $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix},\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}1+6&4+15\\2+8&8+20\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}7&19\\10&28\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(1)$
Also,
$\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1+6&2+8\\4+15&8+20\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(2)$
$\therefore\ (\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [From eqs. (1) and (2)]
View full question & answer→Question 905 Marks
Solve the matrix equations:
$\begin{bmatrix}\text{x}&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
AnswerHere,
$\begin{bmatrix}\text{x}&1\end{bmatrix}\begin{bmatrix}1&0\\-2&-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}\text{x}-2&0-3\end{bmatrix}\begin{bmatrix}\text{x}\\5\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}(\text{x}-2)\text{x}-15\end{bmatrix} =0$
$\Rightarrow\text{x}^2-2\text{x}-15=0$
$ \Rightarrow\text{x}^2-5\text{x}+3\text{x}-15=0$
$ \Rightarrow\text{x}(\text{x}-5)+3(\text{x}-5)=0$
$\Rightarrow(\text{x}-5)(\text{x}+3)=0$
$ \Rightarrow\text{x}-5=0\ \text{or}\ \text{x}+3=0$
$ \Rightarrow\text{x}=5\ \text{or}\ \text{x}=-3$
So,
$\text{x}=5\text{ or }-3$
View full question & answer→Question 915 Marks
If $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -l$, then show that $(A + B)^2 = A^2 + B^2$.
AnswerWe have, $\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ and $x^2 = -1$
$\therefore\ (\text{A}+\text{B})=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$
$\therefore\ (\text{A}+\text{B})^2=\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}\begin{bmatrix}0&-\text{x}+1\\\text{x}+1&0\end{bmatrix}$ $=\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ ....(\text{i})$ Also, $\text{A}^2=\text{A}.\text{A}=\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}\begin{bmatrix}0&-\text{x}\\\text{x}&0\end{bmatrix}$$=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}$
And $\text{B}^2=\text{B}.\text{B}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}$$=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\therefore\ \text{A}^2+\text{B}^2=\begin{bmatrix}-\text{x}^2&0\\0&-\text{x}^2\end{bmatrix}+\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$=\begin{bmatrix}1-\text{x}^2&0\\0&1-\text{x}^2\end{bmatrix}\ ....(\text{ii})$
From eq. (i) and (ii), we have $(\text{A}+\text{B})^2=\text{A}^2+\text{B}^2$
View full question & answer→Question 925 Marks
If $\text{A} = \begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix},$ prove that $\text{A}'' = \begin{bmatrix}3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\end{bmatrix}n \in \text{N}.$
AnswerGiven: $\text{A}=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
Let p(n): $\text A^{n}=\begin{bmatrix}3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\\3^{n-1}&3^{n-1}&3^{n-1}\end{bmatrix}\ $
p(1): $\Rightarrow \ \ \ \text A=\begin{bmatrix}3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0}\\3^{0}&3^{0}&3^{0}\end{bmatrix}\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
$\therefore$ p(1) is true for n = 1.
Now p(k): $\text A^{k}=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}...\text{(ii)}$
Multiplying eq. (ii) by eq. (i) $\text{A}^{k}\text{A}=\begin{bmatrix}3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\\3^{k-1}&3^{k-1}&3^{k-1}\end{bmatrix}\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$
$\Rightarrow \text{ A}^ \text{k+1}=\begin{bmatrix}3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k}\\3^{k}&3^{k}&3^{k}\end{bmatrix}=\text{p(k+1)}$
Therefore, p(n) is true for all natural numbers by P.M.I.
View full question & answer→Question 935 Marks
Let $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix},$ Find $A^T, B^T$ and verify that.$(\text{A}\text{B})^\text{T}=\text{B}^\text{T}+\text{A}^\text{T}$
AnswerGiven: $\text{A}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$ $\text{A}^\text{T}=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ and $\text{B}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}$ $\text{AB}=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}$ $\Rightarrow\ \text{AB}=\begin{bmatrix}1-2+0&2-1+0&3-3+0\\2+2+0&4+1+3&6+3+3\\1+4+0&2+2+1&3+6+1\end{bmatrix}$$\Rightarrow\text{AB}=\begin{bmatrix}-1&1&0\\4&8&12\\5&5&10\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(1)$
Now, $\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}$ $\Rightarrow\text{B}^\text{A}\text{A}^\text{T}=\begin{bmatrix}1-2+0&2+2+0&1+4+0\\2-1+0&4+1+3&2+2+1\\3-3+0&6+3+3&3+6+1\end{bmatrix}$ $\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}\ \dots(2)$ $\Rightarrow(\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [from eqs. (1) and (2)]
View full question & answer→Question 945 Marks
Without using the concept of inverse of a matrix, find the matrix $\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}$ such that $\begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}5&-7\\-2&3\end{bmatrix}\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}5\text{x}-7\text{z}&5\text{y}-7\text{u}\\-2\text{x}+3\text{z}&-2\text{y}+3\text{u}\end{bmatrix}=\begin{bmatrix}-16&-6\\7&2\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$ \therefore\ 5\text{x}-7\text{z}=-16\ \dots(1)$
$5\text{y}-7\text{u}=-6\ \dots(2)$
$-2\text{y}+3\text{u}=2$
$ \Rightarrow3\text{u}=2+2\text{y}$
$\Rightarrow\text{u}=\frac{2+2\text{y}}{3}\ \dots(3)$
$-2\text{x}+3\text{z}=7$
$\Rightarrow3\text{z}=7+2\text{x}$
$ \Rightarrow\text{z}=\frac{7+2\text{x}}{3}\ \dots(4)$
Putting the value of z in eq. (1), we get
$5\text{x}-7\Big(\frac{7+2\text{x}}{3}\Big)=-16$
$\Rightarrow5\text{x}-\frac{49+14\text{x}}{3}=-16$
$ \Rightarrow\frac{15\text{x}-49+14\text{x}}{3}=-16$
$\Rightarrow\text{x}-49=-48$
$\Rightarrow\text{x}=-48+49$
$\therefore\ \text{x}=1$
Putting the value of x in eq. (4), we get
$\text{z}=\frac{7+2(1)}{3}$
$\text{z}=\frac{9}{3}=3$
Putting the value of u in eq. (2), we get
$5\text{y}-7\Big(\frac{2+2\text{y}}{3}\Big)=-6$
$\Rightarrow5\text{y}-\frac{14+14\text{y}}{3}=-6$
$\Rightarrow\frac{15\text{y}-14+14\text{y}}{3}=-6$
$ \Rightarrow\text{y}-14=-18$
$\Rightarrow\text{y}=-18+14$
$ \Rightarrow\text{y}=-4$
Putting the value of y in eq. (3), we get
$ \text{u}=\frac{2+2(-4)}{3}$
$ \Rightarrow\text{u}=-2$
$\therefore\ \begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{u}\end{bmatrix}=\begin{bmatrix}1&-4\\3&-2\end{bmatrix}$
View full question & answer→Question 955 Marks
If the matrix $\begin{bmatrix}0&\text{a}&3\\2&\text{b}&-1\\\text{c}&1&0\end{bmatrix}$ is a skew-symmetric matrix, then find the values of a, b and c.
AnswerLet $\text{A}=\begin{bmatrix}0&\text{a}&3\\2&\text{b}&-1\\\text{c}&1&0\end{bmatrix}$ Since, A is skew-symmetric matrix. $\therefore\ \text{A}'=-\text{A}$$\Rightarrow\ \begin{bmatrix}0&2&\text{c}\\\text{a}&\text{b}&1\\3&-1&0\end{bmatrix}=\begin{bmatrix}0&\text{a}&3\\2&\text{b}&-1\\\text{c}&1&0\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}0&2&\text{c}\\\text{a}&\text{b}&1\\3&-1&0\end{bmatrix}=\begin{bmatrix}0&-\text{a}&-3\\-2&-\text{b}&1\\-\text{c}&-1&0\end{bmatrix}$ By equality of matrices, we get a = -2, c = -3 and b = -b ⇒ b = 0 $\therefore$ a = -2, b = 0 and c = -3
View full question & answer→Question 965 Marks
If $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then find $\lambda,\mu$ so that $\text{A}^2=\lambda\text{A}+\mu\text{I}$
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\1&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&3\\1&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4+3&6+6\\2+2&3+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}7&12\\4&7\end{bmatrix}$
$ \text{A}^2=\lambda\text{A}=\mu\text{I}$
$\Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\lambda\begin{bmatrix}2&3\\1&2\end{bmatrix}+\mu\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda&3\lambda\\1\lambda&2\lambda\end{bmatrix}+\begin{bmatrix}\mu&0\\0&\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda+0\\\lambda+0&2\lambda+\mu\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}7&12\\4&7\end{bmatrix}=\begin{bmatrix}2\lambda+\mu&3\lambda\\\lambda&2\lambda+\mu\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$ \therefore\ 7=2\lambda+\mu\ \dots(1)$
$ 12=3\lambda$
$\Rightarrow\lambda=\frac{12}{3}=4$
Putting the value of $\lambda$ in eq. (1), we get
$7=2(4)+\mu$
$\Rightarrow7-8=\mu$
$\therefore\ \mu=-1$
View full question & answer→Question 975 Marks
Find a 2 × 2 matrix A such that.
$\text{A}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
AnswerLet $\text{A}=\begin{bmatrix}\text{w}&\text{x}\\\text{y}&\text{z}\end{bmatrix}$
Now,
$\begin{bmatrix}\text{w}&\text{x}\\\text{y}&\text{z}\end{bmatrix}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
$ \Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=6\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=\begin{bmatrix}6&0\\0&6\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ w + x = 6
⇒ w = 6 - x ...(1)
-2w + 4x = 0 ...(2)
Putting the value of w in eq. (2), we get
-2(6 - x) + 4x = 0
⇒ -12 + 2x + 4x = 0
⇒ -12 + 6x = 0
⇒ 6x = 12
⇒ x = 2
Putting the value of x in eq. (1), we get
w = 6 - 2
⇒ w = 4
Now,
y + z = 0
⇒ y = -z ...(3)
-2y + 4z = 6 ...(4)
Putting the value of y in eq. (4), we get
-2(-z) + 4z = 6
⇒ 2z + 4z = 6
⇒ 6z = 6
⇒ z = 1
Putting the value of z in eq. (3), we get
y = -1
$ \therefore\ \text{A}=\begin{bmatrix}4&2\\-1&1\end{bmatrix}$
View full question & answer→Question 985 Marks
$ \text{Find}\ \frac{1}{2}(\text{A}+\text{A}')\ \text{and}\frac{1}{2}(\text{A}-\text{A}'),\ \text{when}\ \text{A}=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}$
AnswerThe given matrix is $\text{A}=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix} $
$\therefore \text{A}'=\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}$
$\text{A + A}'=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}+\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\therefore\ \frac{1}{2}\text{(A + A}')=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
Now, $\text{A} - \text{A}'=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}-\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}$$=\begin{bmatrix}0&2\text{a}&2\text{b}\\-2\text{a}&0&2\text{c}\\-2\text{b}&-2\text{c}&0\end{bmatrix}$
$\therefore\ \frac{1}{2}\text{(A} -\text{A}')=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}$
View full question & answer→Question 995 Marks
If $\text{A}=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix},$ then verify $A^2 + A = A(A + I)$, where I is the identity matrix.
AnswerTo verify: $A^2 + A = A(A + I)$.
Given: $\text{A}=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$ =\begin{bmatrix}1+0+0&0+0-3&-3+0-3\\2+2+0&0+1+3&-6+3+3\\0+2+0&0+1+1&0+3+1\end{bmatrix}$
$ =\begin{bmatrix}1&-3&-6\\4&4&0\\2&21&4\end{bmatrix}$
LHS:
$ \text{A}^2+\text{A}=\begin{bmatrix}1&-3&-6\\4&4&0\\2&21&4\end{bmatrix}+\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}$
$=\begin{bmatrix}1+1&-3+0&-6-3\\4+2&4+1&0+3\\2+0&2+1&4+1\end{bmatrix}$
$=\begin{bmatrix}2&-3&-9\\6&5&3\\2&3&5\end{bmatrix}$
RHS:
$ \text{A}+\text{I}=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}+\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}1+1&0+0&-3+0\\2+0&1+0&3+0\\0+0&1+0&1+1\end{bmatrix}$
$ =\begin{bmatrix}2&0&-3\\2&2&3\\0&1&2\end{bmatrix}$
$ \text{A}(\text{A}+\text{I})=\begin{bmatrix}1&0&-3\\2&1&3\\0&1&1\end{bmatrix}\begin{bmatrix}2&0&-3\\2&2&3\\0&1&2\end{bmatrix}$
$ =\begin{bmatrix}2+0+0&0+0-3&-3+0-6\\4+2+0&0+2+3&-6+3+6\\0+2+0&0+2+1&0+3+2\end{bmatrix}$
$=\begin{bmatrix}2&-3&-9\\6&5&3\\2&3&5\end{bmatrix}$
Therefore, LHS = RHS
Hence, $A^2 + A = A(A + I)$ is verified.
View full question & answer→Question 1005 Marks
If $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix},$ prove that $\text{A}^\text{n}=\begin{bmatrix}\text{a}^\text{n}&\text{b}\Big(\frac{\text{a}^\text{n}-1}{\text{a}-1}\Big)\\0&1\end{bmatrix}$ for every positive integer n.
AnswerWe shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
$\text{A}^1=\begin{bmatrix}\text{a}^1&\text{b}\frac{(\text{a}^1-1)}{\text{a}-1}\\0&1\end{bmatrix}=\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix}=\text{A}$
So, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
$\text{A}^{\text{m}}=\begin{bmatrix}\text{a}^{\text{m}}&\text{b}\frac{(\text{a}^{\text{m}}-1)}{\text{a}-1}\\0&1\end{bmatrix}\ \dots(1)$
Now, we shall show that the result is true for n = m + 1.
Here,
$\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\text{b}\frac{(\text{a}^{\text{m}+1}-1)}{\text{a}-1}\\0&1\end{bmatrix}$
By definition of integral power of a matrix, we have
$\text{A}^{\text{m}+1}=\text{A}^\text{m}\text{A}$
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^\text{m}&\text{b}\frac{(\text{a}^\text{m}-1)}{\text{a}-1}\\0&1&\end{bmatrix}\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix}$ [From eq. (1)]
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^\text{m}\text{a}+0&\frac{\big\{\text{a}^\text{m}\text{b}+\text{b}(\text{a}^\text{m}-1)\big\}}{\text{a}-1}\\0+0&0+1\end{bmatrix}$
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\frac{(\text{a}^{\text{m}+1}\text{b}-\text{a}^\text{m}\text{b}+\text{a}^\text{m}\text{b}-\text{b})}{\text{a}-1}\\0&1\end{bmatrix}$
$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\text{b}\frac{(\text{a}^{\text{m}+1}-1)}{\text{a}-1}\\0&1\end{bmatrix}$
This shows that when the result is true for n = m, it is also true for n = m + 1.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.
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