Question 513 Marks
Let $\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$ and
$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that
$\big[\text{G}(\beta)\big]^{-1}=\text{G}(-\beta)$
Answer$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}\Rightarrow\ |\text{G}(\beta)|=\cos^2\beta+\sin^2$
$\text{C}_{11}=\cos\beta, \text{C}_{21}=0,\text{C}_{31}=\sin\beta$
$\text{C}_{12}=0,\text{C}_{22}=1,\text{C}_{32}=0$
$\text{C}_{13}=\sin\beta, \text{C}_{23}=0,\text{C}_{33}=\cos\beta$
$\big[\text{G}(\beta)\big]^{-1}=\frac{\text{adj}(\text{G}(\beta))}{|\text{G}(\beta)|}=\frac{1}{1}\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(i)}$
Now, $\text{G}(-\beta)=\begin{bmatrix} \cos(-\beta) & 0 & \sin(-\beta) \\ 0 & 1 & 0 \\ -\sin(-\beta) & 0 & \cos(-\beta) \end{bmatrix}$
$=\begin{bmatrix} \cos\beta & 0 & -\sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(ii)}$
From (i) & (ii)
$\big[\text{G}(\beta)\big]^{1}=\text{G}(-\beta)$
View full question & answer→Question 523 Marks
By using properties of determinants, show that:$\begin{vmatrix}x+y+2z&x&y\\z&y+z+2x&y\\z&x&z+x+2y\end{vmatrix}=2(x+y+z)^3$
Answer$ \text{L.H.S.}=\begin{vmatrix}x+y+2z&x&y\\z&y+z+2x&y\\z&x&z+x+2y\end{vmatrix}$
$=\begin{vmatrix}2(x+y+z)&x&y\\2(x+y+z)&y+z+2x&y\\2(x+y+z)&x&z+x+2y\end{vmatrix}\ \left[\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\right]$
$=2(x+y+z)\begin{vmatrix}1&x&y\\1&y+z+2x&y\\1&x&z+x+2y\end{vmatrix}$
$=2(x+y+z)\begin{vmatrix}1&x&y\\0&x+y+z&0\\0&0&x+y+z\end{vmatrix}\ \left[\text{R}_2\rightarrow\text{R}_2-\text{R}_1\ \text{and R}_3\rightarrow\text{R}_3-\text{R}_1\right]$
$=2(x+y+z)\begin{vmatrix}1&x&y\\0&x+y+z&0\\0&0&x+y+z\end{vmatrix}$
$=2\left(x + y + z\right).1.\begin{vmatrix}x + y + z&0\\0&x + y + z\end{vmatrix}$
$=2(x+y+z)\left[(x+y+z)^2-0\right]$
$=2(x+y+z)^3=\text{R.H.S.}$ Proved.
View full question & answer→Question 533 Marks
Find the integral value of x, if $\begin{vmatrix}\text{x}^2&\text{x}&1\\0&2&1\\3&1&4 \end{vmatrix}=28.$
AnswerGiven, $\begin{vmatrix}\text{x}^2&\text{x}&1\\0&2&1\\3&1&4 \end{vmatrix}=28$
$\Rightarrow\text{x}^2(8-1)-\text{x}(0-3)+1(0-6)$
$\Rightarrow8\text{x}^2-\text{x}^2+3\text{x}-6=28$
$\Rightarrow7\text{x}^2+3\text{x}-6=28$
$\Rightarrow7\text{x}^2+3\text{x}-34=0$
$\Rightarrow(7\text{x}+17)(\text{x}-2)=0$
$\Rightarrow\text{x}=2$
Integral value of x is 2. Thus, $\text{x}=\frac{-17}{7}$ is not an integer.
View full question & answer→Question 543 Marks
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}-a^{2}&ab&ac\\ba&-b^{2}&bc\\ca&cb&-c^{2}\end{vmatrix}=4a^2b^2c^2$
Answer$\text{L.H.S.}=\begin{vmatrix}-a^2&ab&ac\\ba&-b^2&bc\\ca&cb&-c^2\end{vmatrix}$
Taking common a,b,c from $R_1,R_2,R_3$ respectively,
$=abc\begin{vmatrix}-a&b&c\\a&-b&c\\a&b&-c\end{vmatrix}$
$=abc\begin{vmatrix}-a&b&c\\0&0&2c\\0&2b&0\end{vmatrix}\left[\text{operating}\ \text{R}_3\rightarrow\text{R}_3+\text{R}_1\ \text{and R}_2\rightarrow\text{ R}_2+\text{ R}_1\right]$
$=abc[-a(0-4bc)]$
$=abc(4abc)$
$=4a^2b^2c^2\ \text{R.H.S}$
View full question & answer→Question 553 Marks
Evaluate the following determinant:
$\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
AnswerLet $\text{A}=\begin{vmatrix}\text{a}+\text{ib}&\text{c}+\text{id}\\-\text{c}+\text{id}&\text{a}-\text{ib}\end{vmatrix}$
$|\text{A}|=(\text{a}+\text{ib})(\text{a}-\text{ib})-(\text{c}+\text{id})(-\text{c}+\text{id})$ (Taking (-) sign common from -c + id)
$=(\text{a}^2+\text{b}^2)+(\text{c}+\text{id})(\text{c}-\text{id})$ (Also $(a + ib)(a - ib) = a^2 + b^2$)
$=\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2$
Hence, $|\text{A}|=\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2$
View full question & answer→Question 563 Marks
If $\text{A}=\begin{bmatrix}-4 & -3 & -3 \\1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix},$ show that adj A = A.
Answer$\text{A}=\begin{bmatrix}-4 & -3 & -3 \\1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$
Now,
$\text{C}_{11}=\begin{vmatrix}0 & 1 \\4 & 3 \end{vmatrix}=-4,\ \text{C}_{12}=\begin{vmatrix}1 & 1 \\4 & 3 \end{vmatrix}=1$
$\text{and C}_{13}=\begin{bmatrix}1 & 0 \\4 & 4 \end{bmatrix}=4$
$\text{C}_{21}=\begin{vmatrix}-3 & -3 \\4 & 3 \end{vmatrix}=-3,\ \text{C}_{22}=\begin{vmatrix}-4 & -3 \\4 & 3 \end{vmatrix}=0$
$\text{and C}_{23}=\begin{bmatrix}-4 & -3 \\4 & 4 \end{bmatrix}=4$
$\text{C}_{31}=\begin{vmatrix}-3 & -3 \\0 & 1 \end{vmatrix}=-3,\ \text{C}_{32}=\begin{vmatrix}-4 & -3 \\1 & 1 \end{vmatrix}=1$
$\text{and C}_{33}=\begin{bmatrix}-4 & -3 \\1 & 0 \end{bmatrix}=3$
$\therefore\ \text{adj A}=\begin{bmatrix}-4 & 1 & 4\\-3 & 0 & 4 \\ -3 & 1 & 3 \end{bmatrix}^\text{T}=\begin{bmatrix}-4 & -3 & -3 \\1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}=\text{A}$
Hence proved.
View full question & answer→Question 573 Marks
Find A (adjoint A) for the matrix $\text{A}=\begin{bmatrix}1 & -2 & 3 \\0 & 2 & -1 \\ -4 & 5 & 2 \end{bmatrix}$.
Answer$\text{A}=\begin{bmatrix}1 & -2 & 3 \\0 & 2 & -1 \\ 4 & 5 & 2 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{bmatrix}2 & -1 \\5 & 2 \end{bmatrix}=9,\text{C}_{12}=-\begin{bmatrix}0 & -1 \\-4 & 2 \end{bmatrix}=4$
$\text{and C}_{13}=\begin{bmatrix}0 & 2 \\-4 & 5 \end{bmatrix}=8$
$\text{C}_{21}=\begin{bmatrix}-2 & 3 \\5 & 2 \end{bmatrix}=19,\text{C}_{22}=\begin{bmatrix}1 & 3 \\-4 & 2 \end{bmatrix}=14$
$\text{and C}_{23}=\begin{bmatrix}1 & -2 \\-4 & 5 \end{bmatrix}=3$
$\text{C}_{31}=\begin{bmatrix}-2 & 3 \\ 2 & -1 \end{bmatrix}=4,\text{C}_{32}=-\begin{bmatrix}1 & 3 \\0 & -1 \end{bmatrix}=1$
$\text{and C}_{33}=\begin{bmatrix}1 & -2 \\0 & 2 \end{bmatrix}=2$
$\text{adj A}=\begin{bmatrix}9 & 4 & 8 \\19 & 14 & 3 \\ -4 & 1 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix}9 & 19 & -4 \\4 & 14 & 1 \\ 8 & 3 & 2 \end{bmatrix}$
$\therefore\ \text{A(adj A)}=\begin{bmatrix}25 & 0 & 0 \\0 & 25 & 0 \\ 0 & 0 & 25 \end{bmatrix}$
View full question & answer→Question 583 Marks
If $\text{A}=\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}$, then show that $\left|3\text{A}\right|=27\left|\text{A}\right|$
AnswerThe given matrix is $\text{A}=\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}$.
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column $(C_1)$ for easier calculation.
$|\text{A}|=1\begin{vmatrix}1&2\\0&4\end{vmatrix}-0\begin{vmatrix}0&1\\0&4\end{vmatrix}+0\begin{vmatrix}0&1\\1&2\end{vmatrix}=1\left(4-0\right)-0+0=4$
$ \therefore27|\text{A}|=27(4)=108 \dots\dots(1)$
$ \text{Now},3\text{A}=3\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}=\begin{bmatrix}3&0&3\\0&3&6\\0&0&12\end{bmatrix}$
$\therefore|3\text{A}|=3\begin{vmatrix}3&6\\0&12\end{vmatrix}-0\begin{vmatrix}0&3\\0&12\end{vmatrix}+0\begin{vmatrix}0&3\\3&6\end{vmatrix}$
$=3\left(36-0\right)=3\left(36\right)=108 \dots\dots(2)$
From equations (1) and (2), we have:
$\left|3A\right|=27\left|A\right|$
Hence, the given result is proved.
View full question & answer→Question 593 Marks
If A is a square matrix of order 3 such that |A| = 3, then write the value of adj (adj A).
AnswerIf A is a non-singular matrix of order n, then
$|\text{adj (adj A)}|=|\text{A}|^{(\text{n}-1)^{2}}$
Now, ATQ
$|\text{A}|=3$
$\text{n}=3$
$|\text{adj (adj A)}|=|3|^{(3-1)^{2}}$
$=(3)^{2^3}$
$=(3)^4$
$=81$
Hence,
$|\text{adj (adj A)}|=81$
View full question & answer→Question 603 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}5&20\\0&-1 \end{vmatrix}$
Answer$\text{M}_{11}=-1$
$\text{M}_{20}=20$
$\text{C}_{\text{ij}}=(-1)^{\text{i}+\text{j}}\text{M}_{\text{ij}}$
$\text{C}_{11}=(-1)^{1+1}(-1)=-1$
$\text{C}_{21}=(-1)^{1+2}(20)=-20$
$\text{D}=(-1\times5)-(20\times0)=-5$
View full question & answer→Question 613 Marks
If $a_1, a_2, a_3, ...,$ ar are in G.P., then prove that the determinant $\begin{bmatrix}\text{a}_{\text{r}+1}&\text{a}_{\text{r}+5}&\text{a}_{\text{r}+9}\\\text{a}_{\text{r}+7}&\text{a}_{\text{r}+11}&\text{a}_{\text{r}+15}\\\text{a}_{\text{r}+11}&\text{a}_{\text{r}+17}&\text{a} _{\text{r}+21}\end{bmatrix}$ is independent of r.
AnswerWe know that, $\text{a}_{\text{r}+1}=\text{AR}^{(\text{r}+1)}=\text{AR}^\text{r}$ where $r = r^{th}$ term of a GA, A = First term of a GP and R = Common ratio of GP We have $\begin{vmatrix}\text{a}_{\text{r}+1}&\text{a}_{\text{r}+5}&\text{a}_{\text{r}+9}\\\text{a}_{\text{r}+7}&\text{a}_{\text{r}+11}&\text{a}_{\text{r}+15}\\\text{a}_{\text{r}+11}&\text{a}_{\text{r}+17}&\text{a} _{\text{r}+21}\end{vmatrix}$ $=\text{AR}^{\text{r}}\text{AR}^{\text{r}+6}\text{AR}^{\text{r}+10}\begin{vmatrix}1&\text{AR}^4&\text{AR}^8\\1&\text{AR}^4&\text{AR}^8\\1&\text{AR}^6&\text{AR}^{10}\end{vmatrix}$[Taking $\text{AR}^\text{r}.\text{AR}^{\text{r}+6}.\text{AR}^{\text{r}+10}$ common from $\text{R}_1,\text{R}_2\text{ and R}_2$ respectively]
$=0\ [\text{Since, R}_1\text{ and R}_2\text{ are identicals}]$
View full question & answer→Question 623 Marks
By using properties of determinants, show that:
$\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}=(a-b)(b-c)(c-a)$
Answer$\text{L.H.S.}=\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}\ \text{operating}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1\ \text{and}\ \text{R}_3\rightarrow\text{R}_3-\text{R}_1,$
$=\begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{vmatrix}=1\begin{vmatrix}b-a&b^2-a^2\\c-a&c^2-a^2\end{vmatrix}$
$=(b-a)(c-a)\begin{vmatrix}1&b+a\\1&c+a\end{vmatrix}=(b-a)(c-a)(c+a-b-a)$
$=(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)=\text{R.H.S.}$ Proved.
View full question & answer→Question 633 Marks
Solve system of linear equations, using matrix method.
4x - 3y = 3
3x - 5y = 7
AnswerMatrix form of given equations AX = B $\Rightarrow\ \begin{bmatrix}4&-3\\3&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}$ $\text{Here}\ \text{A}=\begin{bmatrix}4&-3\\3&-5\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\end{bmatrix}\text{and B}=\begin{bmatrix}3\\7\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}4&-3\\3&-5\end{vmatrix}=-20-(-9)=-20+9=-11\neq0$ Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$ $\Rightarrow\ \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{-11}\begin{bmatrix}-5&3\\-3&4\end{bmatrix}\begin{bmatrix}3\\7\end{bmatrix}$ $=\frac{1}{-11}\begin{bmatrix}-15+21\\-9+28\end{bmatrix}=\frac{1}{-11}\begin{bmatrix}6\\19\end{bmatrix}=\begin{bmatrix}\frac{-6}{11}\\\frac{-19}{11}\end{bmatrix}$ Therefore, $x=\frac{-6}{11}\text{and}\ y=\frac{-19}{11}$
View full question & answer→Question 643 Marks
If $\text{A}=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\text{ and A (adj A =)}\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix},$ then find the value of k.
Answer$\text{A}=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
$\therefore|\text{A}|=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
$=\cos^2\theta+\sin^2\theta=1\neq0$
Thus, $A^{-1}$ exists.
Now,
$\text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\text{adj A}$
$\Rightarrow\text{AA}^{-1}=\text{A adj A}$
$\Rightarrow\text{AA}^{-1}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix} \ \big[\because\ \text{AA}^{-1}=\text{I}\big]$
$\Rightarrow\text{k}=1$
View full question & answer→Question 653 Marks
By using properties of determinants, show that:
$\begin{vmatrix}x+4&2x&2x\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}=(5x+4)(4-x)^2$
Answer$\text{L.H.S.}=\begin{vmatrix}x+4&2x&2x\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}$
$=\begin{vmatrix}5x+4&5x+4&5x+4\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}\ \left[\text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\right]$
$=(5x+4)\begin{vmatrix}1&1&1\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}$
$=(5x+4)\begin{vmatrix}1&0&0\\2x&4-x&0\\2x&0&4-x\end{vmatrix}\ \left[\text{operating C}_2\rightarrow\text{C}_2-\text{C}_1\ \text{and C}_3\rightarrow\text{C}_3-\text{C}_1\right]$
$=(5x+4)\begin{vmatrix}4-x&0\\0&4-x\end{vmatrix}=(5x-4)(4-x)^2=\text{R.H.S.}$
View full question & answer→Question 663 Marks
Given $\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix},$ compute $A^{-1}$ and show that $2A^{-1} = 9I - A$.
AnswerWe have, $\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$
Now, $\text{adj (A)}=\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
and $|A| = 2$
$\therefore\ \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
Now, $2A^{-1} = 9I - A$
$\text{L.H.S}=2\text{A}^{-1}=\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
$\text{R.H.S}=9\text{I}-\text{A}=9\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$
$=\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}=\text{L.H.S}$
Hence proved.
View full question & answer→Question 673 Marks
If $\text{A}=\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}$ be sech that $A^{-1} = kA$, then find the value of k.
Answer$\text{A}=\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}$
$\therefore|\text{A}|=\begin{vmatrix}2 & 3 \\ 5 & -2 \end{vmatrix}=-14-15=-19$
The value is non-zero, so $A^{-1}$ exists.
By definition, we have
$A^{-1} A = I$ [I is the identity matrix]
$\Rightarrow kAA = I$ [Substituting $A^{-1} = kA$]
$\Rightarrow\text{k}\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow\text{k}\begin{bmatrix} 4+15 & 6-6 \\ 10-10 & 15+4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow\text{k}\begin{bmatrix} 19 & 0 \\ 0 & 19 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow\text{k}=\frac{1}{19}$
View full question & answer→Question 683 Marks
Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}5&2\\3&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}3\\5\end{bmatrix}$
$\text{Here}\ \text{A}= \begin{bmatrix}5&2\\3&2\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\end{bmatrix}\text{and B}=\begin{bmatrix}3\\5\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}5&2\\3&2\end{vmatrix}=10-6=4\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{4}\begin{bmatrix}2&-2\\-3&5\end{bmatrix}\begin{bmatrix}3\\5\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}6-10\\-9+25\end{bmatrix}=\begin{bmatrix}1\\4\end{bmatrix}\begin{bmatrix}-4\\16\end{bmatrix}=\begin{bmatrix}-1\\4\end{bmatrix}$
Therefore, x = -1 and y = 4
View full question & answer→Question 693 Marks
Find the inverse of the following matrices:
$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Now $|\text{A}|=-1\neq0$
Hence $A^{-1}$ exists.
Cofactors of a are:
$C_{11} = 0 C_{12} = -1$
$C_{21} = -1 C_{22} = 0$
$\therefore\ \text{adj A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} \\ \text{C}_{21} & \text{C}_{22} \end{bmatrix}$
$=\begin{bmatrix}0 & -1 \\ -1 & 0 \end{bmatrix}$
Also, $\text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj A})$
$\text{A}^{-1}=\frac{1}{(-1)}\begin{bmatrix}0 & -1 \\ -1 & 0 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$
View full question & answer→Question 703 Marks
If $\text{A}=\begin{bmatrix}4 & 5 \\2 & 1 \end{bmatrix},$ then show that $A - 3I = 2 (I + 3A^{-1})$.
AnswerWe have, $\text{A}=\begin{bmatrix}4 & 5 \\2 & 1 \end{bmatrix}$
Now,
$\text{adj (A)}\begin{bmatrix}1 & -5 \\-2 & 4 \end{bmatrix}$
and |A| = -6
$\therefore\ \text{A}^{-1}=-\frac{1}{6}\begin{bmatrix}1 & -5 \\-2 & 4 \end{bmatrix}$
Now, $A - 3I = I + 3A^{-1}$
$\text{L.H.S}=\text{A}-3\text{I}=\begin{bmatrix}4 & 5 \\2 & 1 \end{bmatrix}-3\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 5 \\2 & -2 \end{bmatrix}$
$\text{R.H.S}=2(\text{I}+3\text{A}^{-1})=2\left\{\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-3\times\frac{1}{3}\begin{bmatrix}1 & -5 \\-2 & 4 \end{bmatrix}\right\}$
$=2\begin{bmatrix}0.5 & 2.5 \\1 & -1 \end{bmatrix}=\begin{bmatrix}1 & 5 \\2 & -2 \end{bmatrix}\text{L.H.S}$
Hence proved.
View full question & answer→Question 713 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x - y + 3z = 6,
x + 3y - 3z = -4,
5x + 3y + 3z = 10
AnswerUsing the equations, we get
$\text{D}=\begin{vmatrix}1&-1&3\\1&3&-3\\5&3&3\end{vmatrix}$
$=1(9+9)+1(3+15)+3(3-15)$
$=18+18-36=0$
$\text{D}_1=\begin{vmatrix}6&-1&3\\-4&3&-3\\10&3&3\end{vmatrix}$
$=6(9+9)+1(-12+30)+3(-12-30)$
$=108+18-126=0$
$\text{D}_2=\begin{vmatrix}1&6&3\\1&-4&-3\\5&10&3\end{vmatrix}$
$=1(-12+30)-6(3+15)+3(10+20)$
$=18-108+90=0$
$\text{D}_3=\begin{vmatrix}1&-1&6\\1&3&-4\\5&3&10\end{vmatrix}$
$=42+30-72=0$
$\therefore\text{D}=\text{D}_1=\text{D}_2=\text{D}_3=0$
Hence, the system of equations has infinitely many solutions.
View full question & answer→Question 723 Marks
Find the area of the triangle with vertices at the points:
$(0, 0), (6, 0)$ and $(4, 3)$
Answer$\triangle=\frac{1}{2}\begin{vmatrix}0&0&1\\6&0&1\\4&3&1\end{vmatrix}$
$=\frac{1}{2}\begin{vmatrix}0&0&1\\6&0&0\\4&3&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\frac{1}{2}\begin{vmatrix}0&0&1\\6&0&0\\4&3&0\end{vmatrix}$ [Applying $R_3 → R_3- R_1$]
$=\frac{1}{2}\begin{vmatrix}6&0\\4&3\end{vmatrix}$
$=\frac{1}{2}(18-0)$
$=\frac{1}{2}(18)$
$=9\text{ square units}$
View full question & answer→Question 733 Marks
If $\text{A}=\begin{bmatrix}-1 & -2 & -2 \\2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix},$ show that $A = 3A^T$.
AnswerHere $\text{A}=\begin{bmatrix}-1 & -2 & -2 \\2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix},$
Cofactor of a are:
$C_{11} = -3, C_{21} = 6, C_{31} = 6$
$C_{12}= -6, C_{22} = 3, C_{32} = -6$
$C_{13} = -6, C_{23} = -6, C_{33} = 3$
$\therefore\ \text{adj A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} & \text{C}_{13} \\ \text{C}_{21} & \text{C}_{22} & \text{C}_{23} \\ \text{C}_{31} & \text{C}_{32} & \text{C}_{33} \end{bmatrix}$
$=\begin{bmatrix}-3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}$
$\therefore\ \text{adj A}=\begin{bmatrix}-3 & -6 & -6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}\ .....\text{(i)}$
Now, $3\text{A}^\text{T}=3\begin{bmatrix}-1& 2 & 2 \\-2 & 1 & -2 \\ -2 & -2 & 1 \end{bmatrix}=\begin{bmatrix}-3 & -6 & -6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}\ .....\text{(ii)}$
$\therefore\ \text{adj A}=3\text{A}^\text{T}$
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