Question 13 Marks
Evaluate $\triangle=\begin{vmatrix}0&\sin\alpha&-\cos\alpha\\-\sin\alpha&0&\sin\beta\\\cos\alpha&-\sin\beta&0 \end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}0&\sin\alpha&-\cos\alpha\\-\sin\alpha&0&\sin\beta\\\cos\alpha&-\sin\beta&0 \end{vmatrix}$
$\triangle=(-1)^{1+1}0(0+\sin^2\beta)+(-1)^{1+2}\sin\alpha(0-\sin\beta\cos)\beta\\+(-1)^{1+3}(-\cos\alpha)(\sin\alpha\sin\beta-0)$ [Expanding along $R_1$]
$=0(0+\sin^2\beta)-\sin\alpha(0-\sin\beta\cos\alpha)-\cos\alpha(\sin\alpha\sin\beta-0)$
$=\sin\alpha\sin\beta\cos\alpha-\sin\alpha\sin\beta\cos\alpha$
$=0$
View full question & answer→Question 23 Marks
Show that $\begin{vmatrix}\sin10^\circ&-\cos10^\circ\\\sin80^\circ&\cos80^\circ \end{vmatrix}=1$
AnswerLet $\triangle=\begin{vmatrix}\sin10^\circ&-\cos10^\circ\\\sin80^\circ&\cos80^\circ \end{vmatrix}$
$\Rightarrow\triangle=\sin10^\circ\cos80^\circ+\cos10^\circ\sin80^\circ$
$=\sin10^\circ\cos(90^\circ-10^\circ)+\cos10^\circ\sin(90^\circ-10^\circ)$ $\big[\because\cos\theta=\sin(90-\theta)\big]$
$\Rightarrow\triangle=\sin10^\circ\sin10^\circ+\cos10^\circ\cos10^\circ$
$=\sin^210^\circ+\cos^210^\circ$
$\Rightarrow\triangle=1$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
View full question & answer→Question 33 Marks
Evaluate $\begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}^2$
AnswerLet $\text{A}=\begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}$
$\Rightarrow\text{A}=2(204-100)-3(156-75)+7(260-255)$
$\Rightarrow\text{A}=2(104)-3(81)+7(5)$
$\Rightarrow\text{A}=208-243+35$
$\Rightarrow\text{A}=243-243=0$
$\because\begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}^2=0^2=0$ $\big[\therefore\text{det } \text{A}^2 = (\text{det} \text{A})^2\big]$
View full question & answer→Question 43 Marks
Using determinants, find the equation of the line joining the points:
(1, 2) and (3, 6)
AnswerGiven: a = (1, 2) and B = (3, 6)
Let the point p be (x, y). So,
Area of triangle ABP = 0
$\Rightarrow\triangle=\frac{1}{2}\begin{vmatrix}1&2&1\\3&6&1\\\text{x}&\text{y}&1\end{vmatrix}=0$
⇒ 1(6 - y) - 2(3 - x) + 1(3y - 6x) = 0
⇒ 6 - y - 6 + 2x + 3y - 6x = 0
⇒ 2y - 4x = 0
⇒ y = 2x
View full question & answer→Question 53 Marks
Using determinants, find the value of k so that the points $(k, 2 - 2k), (-k + 1, 2k)$ and $(-4 - k, 6 - 2k)$ may be collinear.
AnswerIf the points (k, 2 - 2k), (-k + 1, 2k) and (-4 - k, 6 - 2k) are collinear, then
$\triangle=\begin{vmatrix}\text{k}&2-2\text{k}&1\\-\text{k}+1&2\text{k}&1\\-4-\text{k}&6-2\text{k}&1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{k}&2-2\text{k}&1\\-2\text{k}+1&4\text{k}-2&0\\-4-\text{k}&6-2\text{k}&1\end{vmatrix}=0$ [Applying $R_2 → R_2 - R_1$]
$\Rightarrow\begin{vmatrix}-2\text{k}+1&4\text{k}-2\\-4-\text{k}&6-2\text{k}\end{vmatrix}=0$
$\Rightarrow-8\text{k}+4+16\text{k}-8+8\text{k}^2-4\text{k}=0$
$\Rightarrow8\text{k}^2+4\text{k}-4=0$
$\Rightarrow(8\text{k}-4)(\text{k}+1)=0$
$\Rightarrow\text{k}=-1$ or $\text{k}=\frac{1}{2}$
View full question & answer→Question 63 Marks
Using determinants, find the equation of the line joining the points:
$(3, 1)$ and $(9, 3)$
AnswerLet A(x, y), B(3, 1) and C(9, 3) are 3 points in a line.
Since these points are collinear, hence the area of the triangle ABC must be zero.
$\frac{1}{2}\begin{vmatrix}\text{x}&\text{y}&1\\3&1&1\\9&3&1\end{vmatrix}=0$
Expanding along $R_1$
$\Rightarrow x(-2) - y(-6) + 1(0) = 0$
$\Rightarrow -2x + 6y = 0$
$\Rightarrow x - 3y = 0$
Hence the equation of the line is $x - 3y = 0$
View full question & answer→Question 73 Marks
If a, b, c are non-zero real numbers and if the system of equations
(a - 1)x = y + z
(b - 1)y = z + x
(c - 1)z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.
Answer$\text{D}=\begin{vmatrix}(\text{a}-1)&-1&-1\\-1&(\text{b}-1)&-1\\-1&-1&(\text{c}-1)\end{vmatrix} $
Now for non-trivial solution, D = 0
0 = (a + 1)[(b - 1)(c - 1) - 1] + 1[-c + 1 - 1] - [1 + b - 1]
0 = (a - 1)[bc - b - c + 1 - 1]-c - b
0 = abc - ab - ac + b + c - c - b
ab + bc + ac = abc
Hence proved.
View full question & answer→Question 83 Marks
Using determinants show that the following points are collinear:
$(1, -1), (2, 1)$ and $(4, 5)$
AnswerIf the points (1, -1), (2, 1) and (4, 5) are collinear, then
$\triangle=\begin{vmatrix}1&-1&1\\2&1&1\\4&5&1\end{vmatrix}=0$
$=\begin{vmatrix}1&-1&1\\1&2&0\\4&5&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\begin{vmatrix}1&-1&1\\1&2&0\\3&6&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=\begin{vmatrix}1&2\\3&6\end{vmatrix}$
$=6-6=0$
Thus, these points are collinear.
View full question & answer→Question 93 Marks
Find the value of the determinant $\begin{vmatrix}101&102&103\\104&105&106\\107&108&109\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}101&102&103\\104&105&106\\107&108&109\end{vmatrix}$
$=\begin{vmatrix}101&1&2\\104&1&2\\107&1&2\end{vmatrix}$ [Applying $C_2 → C_2 - C_1$ and $C_3 → C_3 - C_1$]
$=2\begin{vmatrix}101&1&1\\104&1&1\\107&1&1\end{vmatrix}$
$=0$
Since two columns are identitical, the value of the determinant is zero.
$\triangle=\begin{vmatrix}101&102&103\\104&105&106\\107&108&109\end{vmatrix}=0$
View full question & answer→Question 103 Marks
For what value of x the matrix A is singular?
$\text{A}=\begin{vmatrix}\text{x}-1&1&1\\1&\text{x}-1&1\\1&1&\text{x}-1 \end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}\text{x}-1&1&1\\1&\text{x}-1&1\\1&1&\text{x}-1 \end{vmatrix}$
Matrix A will be singular if
$\Rightarrow|\text{A}|=0$
$\Rightarrow(\text{x}-1)\big[(\text{x}-1)^2-1\big]-1(\text{x}-1-1)\\+1\big[1-(\text{x}-1)\big]=0$
$\Rightarrow(\text{x}-1)(\text{x}^2-2\text{x})-1(\text{x}-2)+1(2-\text{x})-0$
$\Rightarrow\text{x}^3-2\text{x}^2-\text{x}^2+2\text{x}-\text{x}+2-\text{x}+2=0$
$\Rightarrow\text{x}^3-3\text{x}^2+4=0$
$\Rightarrow(\text{x}-2)^2(\text{x}+1)=0$
$\Rightarrow\text{x}=2$ or $\text{x}=-1$
View full question & answer→Question 113 Marks
Find the area of the triangle with vertices at the points:
$(2, 7), (1, 1)$ and $(10, 8)$
AnswerThe area is given by:
$\triangle=\frac{1}{2}\begin{vmatrix}2&7&1\\1&1&1\\10&8&1\end{vmatrix}$
Expanding along $R_1$
$=\frac{1}{2}\big[2(-7)-7(-9)+1(-2)\big]$
$=\frac{1}{2}\big[-14+63-2]$
$=\frac{47}{2}\text{sq. units}$
The area of the $\triangle$ is $\frac{47}{2}\text{sq. units}$
View full question & answer→Question 123 Marks
Using determinants, find the area of the triangle whose vertices are $(1, 4), (2, 3)$ and $(-5, -3)$. Are the given points collinear?
Answer$\triangle-\frac{1}{2}=\begin{vmatrix}1&4&1\\2&3&1\\-5&-3&1\end{vmatrix}$
$=\frac{1}{2}\begin{vmatrix}1&4&1\\1&-1&0\\-5&-3&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\frac{1}{2}\begin{vmatrix}1&4&1\\1&-1&0\\-6&-7&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=\frac{1}{2}\begin{vmatrix}1&-1\\-6&-7\end{vmatrix}$
$=\frac{1}{2}(-7-6)$
$=\frac{13}{2}\text{square units}$ $[\because$ Area cannot be negative$]$
Thus, (1, 4), (2, 3) and (-5, 3) are not collinear because $\begin{vmatrix}1&4&1\\2&3&1\\-5&-3&1\end{vmatrix}$ is not equal to zero.
View full question & answer→Question 133 Marks
Using determinants show that the following points are collinear:
$(3, -2), (8, 8)$ and $(5, 2)$
AnswerIf the points (3, -2), (8, 8) and (5, 2) collinear, then
$\triangle=\begin{vmatrix}3&-2&1\\8&8&1\\5&2&1\end{vmatrix}=0$
$=\begin{vmatrix}3&-2&1\\5&10&0\\5&2&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\begin{vmatrix}3&-2&1\\5&10&0\\2&4&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=\begin{vmatrix}5&10\\2&4\end{vmatrix}$
$=20-20$
$=0$
Thus, points are collinear.
View full question & answer→Question 143 Marks
Using determinants show that the following points are collinear:
$(5, 5), (-5, 1)$ and $(10, 7)$
AnswerIf 3 points are collinear, then the area of the triangle then form will be zero.
Hence, $\frac{1}{2}\begin{vmatrix}5&5&1\\-5&1&1\\10&7&1\end{vmatrix}=0$
Expanding along $R_1$
$=\frac{1}{2}\big[5(-6)-5(-15)+1(-35-10)]$
$=\frac{1}{2}[-35+75-45]$
$=\frac{1}{2}[0]$
$=0$
Since the area of the triangle is zero, hence the points are collinear.
View full question & answer→Question 153 Marks
If A is a 3 × 3 matrix, |A| ≠ 0 and |3A| = |A| then write the value of k.
AnswerLet $\text{A}=\begin{bmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{bmatrix}$
then, $3\text{A}=\begin{bmatrix}3\text{a}_1&3\text{a}_2&3\text{a}_3\\3\text{b}_1&3\text{b}_2&3\text{b}_3\\3\text{c}_1&3\text{c}_2&3\text{c}_3 \end{bmatrix}$
$|3\text{A}|=\begin{vmatrix}3\text{a}_1&3\text{a}_2&3\text{a}_3\\3\text{b}_1&3\text{b}_2&3\text{b}_3\\3\text{c}_1&3\text{c}_2&3\text{c}_3 \end{vmatrix}$
$=3^3\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}$ [Taking 3 common from each row]
$=27|\text{A}|$
Hence, the value of k is 27
View full question & answer→Question 163 Marks
Write the value of the determinant $\begin{vmatrix}\text{x}+\text{y}&\text{y}+\text{z}&\text{z}+\text{x}\\\text{z}&\text{x}&\text{y}\\-3&-3&-3 \end{vmatrix}$
Answer$\begin{vmatrix}\text{x}+\text{y}&\text{y}+\text{z}&\text{z}+\text{x}\\\text{z}&\text{x}&\text{y}\\-3&-3&-3 \end{vmatrix}$
$=\begin{vmatrix}\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}&\text{z}+\text{x}+\text{y}\\\text{z}&\text{x}&\text{y}\\-3&-3&-3 \end{vmatrix}$ [Applying $R_1 → R_1 + R_2$]
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&1&1\\\text{z}&\text{x}&\text{y}\\-3&-3&-3 \end{vmatrix}$ [Taking (x + y + z) common from $R_1$]
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}1&1&1\\\text{z}&\text{x}&\text{y}\\0&0&0 \end{vmatrix}$
$=0$ [Expanding along the last row]
Hence, the value of the determinant $\begin{vmatrix}\text{x}+\text{y}&\text{y}+\text{z}&\text{z}+\text{x}\\\text{z}&\text{x}&\text{y}\\-3&-3&-3 \end{vmatrix}$ is 0
View full question & answer→Question 173 Marks
Find the area of the triangle with vertices at the points:
$(3, 8), (-4, 2)$ and $(5, -1)$
Answer$\triangle=\frac{1}{2}\begin{vmatrix}3&8&1\\-4&2&1\\5&-1&1\end{vmatrix}$
$=\frac{1}{2}\begin{vmatrix}3&8&1\\-7&6&0\\5&-1&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\frac{1}{2}\begin{vmatrix}3&8&1\\-7&-6&0\\2&-9&0\end{vmatrix}$ [Applying $R_3 → R_2 - R_1$]
$=\frac{1}{2}\begin{vmatrix}-7&-6\\2&-9\end{vmatrix}$
$=\frac{1}{2}(63+12)$
$=\frac{1}{2}(75)$
$=\frac{75}{2}\text{ Square units.}$
View full question & answer→Question 183 Marks
Solve the following system of homogeneous linear equations:
2x + 3y + 4z = 0,
x + y + z = 0,
2x - y + 3z = 0
AnswerHere,
$\text{D}=\begin{vmatrix}2&3&4\\1&1&1\\2&-1&3\end{vmatrix}$
$=2(4)-3(1)+4(-3) $
$=8-3-7$
$=-2$
$\neq0$
So, the given system of equations has only the triveal solutuion i.e., x = 0 = y = z
Hence, x = 0, y = 0, z = 0
View full question & answer→Question 193 Marks
Using determinants show that the following points are collinear:
(2, 3), (-1, -2) and (5, 8)
AnswerIf given points are collinear, then area of the triangle must be zero.
Hence,
$=\frac{1}{2}\begin{vmatrix}2&3&1\\-1&-2&1\\5&8&1\end{vmatrix}$
$=\frac{1}{2}\big[2(-10)-3(-6)+1(2)\big]$
$=\frac{1}{2}[-20+18+2]$
$=\frac{1}{2}[0]$
$=0$
Hence the given points are collinear.
View full question & answer→Question 203 Marks
Find the value of x if the area of $\triangle$ is $35$ square cms with vertices $(x, 4), (2, -6)$ and $(5, 4)$.
Answer$\triangle=\frac{1}{2}\begin{vmatrix}\text{x}&4&1\\2&-6&1\\5&4&1\end{vmatrix}=\pm35$
$=\frac{1}{2}\begin{vmatrix}\text{x}&4&1\\2-\text{x}&-10&0\\5&4&1\end{vmatrix}=\pm35$ [Applying $R_2 → R_2 - R_1$]
$=\frac{1}{2}\begin{vmatrix}\text{x}&4&1\\2-\text{x}&-10&0\\5-\text{x}&0&0\end{vmatrix}=\pm35$ [Applying $R_3 → R_3 - R_1$]
$=\frac{1}{2}\begin{vmatrix}2-\text{x}&-10\\5-\text{x}&0\end{vmatrix}=\pm35$
$=0+10(5-\text{x})=\pm70$
$\Rightarrow50-10\text{x}=70$ or $50-10\text{x}=-70$
$\Rightarrow-10\text{x}=20$ or $-10\text{x}=-120$
$\Rightarrow\text{x}=-2$ or $\text{x}=12$
View full question & answer→Question 213 Marks
Evaluate the following determinant:
$\begin{vmatrix}\cos15^\circ&\sin15^\circ\\\sin75^\circ&\cos75^\circ \end{vmatrix}$
Answer$\triangle=\cos15^\circ\cos75^\circ-\sin15^\circ\sin75^\circ$
$=\cos15^\circ\cos75^\circ-\sin(90^\circ-75^\circ)\sin(90^\circ-15^\circ)$ $[\because\sin(90^\circ-\theta)=\cos\theta]$
$=\cos15^\circ\cos75^\circ-\cos75^\circ\cos15^\circ$
$=\cos15^\circ\cos75^\circ-\cos15^\circ\cos75^\circ$
$=0$
View full question & answer→Question 223 Marks
If A is a $3 \times 3$ invertible matrix, then what will be the value of k if det $(A^{-1}) = (det A)^k$.
AnswerAs we know that
$\text{A}^{-1}=\frac{\text{Adj A}}{|\text{A}|}$
$\therefore\big|\text{A}^{-1}\big|=\frac{|\text{Adj A}|}{|\text{A}|}$
$=\frac{|\text{A}|^{3-1}}{|\text{A}|}$ $\big[\because$ If A is a non singular matrix of order n, then $|\text{adj(A)} = |\text{A}|^\text{n-1}\big]$
$=\frac{|\text{A}|^2}{|\text{A}|}$
$=|\text{A}|$
As we are given that $|\text{A}^{-1}|=|\text{A}|^{\text{k}}$
$\therefore\text{ k}=1$
View full question & answer→Question 233 Marks
For what value of x is the matrix $\begin{bmatrix}6-\text{x}&4\\3-\text{x}&1\end{bmatrix}$ singular?
Answer$\begin{bmatrix}6-\text{x}&4\\3-\text{x}&1\end{bmatrix}$ is singular when its determinant is 0.
$\Rightarrow\begin{bmatrix}6-\text{x}&4\\3-\text{x}&1\end{bmatrix}=0$
$\Rightarrow(6-\text{x})-4(3-\text{x})=0$
$\Rightarrow6-\text{x}-12+4\text{x}=0$
$\Rightarrow3\text{x}-6=0$
$\Rightarrow3\text{x}=6$
$\Rightarrow\text{x}=\frac{6}{2}$
$\Rightarrow\text{x}=2$
View full question & answer→Question 243 Marks
If $\text{A}\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix},$ then show that |3A| = 27|A|.
Answer$\text{A}\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}$
$\Rightarrow3\text{A}=\begin{bmatrix}3&0&3\\0&3&6\\0&0&12\end{bmatrix}$ [Multiplying each element of A by 3]
$\Rightarrow|3\text{A}|=(-1)^{1+1}3(36-0)+(-1)^{1+2}0(0-0)+(-1)^{1+3}3(0-0)\\=3(36-0)-0(0-0)+3(0-0)\ ...(\text{i})\ \ \ [\text{Expanding along}{\text{ R}_1]}$
$|\text{A}|=(-1)^{1+1}1(4-0)+(-1)^{1+2}0(0-0)+(-1)^{1+3}1(0-0)\\=1(4-0)-0(0-0)+1(0-0)=4\ \ \ \ \ \ \ \ [\text{Expanding along}{\text{ R}_1]}$
$\Rightarrow27|\text{A}|=27\times4=108 ....(\text{ii})$
$\therefore|3\text{A}|=27|\text{A}|$ [From eqs. (i) and (ii)]
View full question & answer→Question 253 Marks
If $\begin{vmatrix}\text{x}&\sin\theta&\cos\theta\\-\sin\theta&-\text{x}&1\\\cos\theta&1&\text{x} \end{vmatrix}=8,$ write the value of $x$.
Answer$\begin{vmatrix}\text{x}&\sin\theta&\cos\theta\\-\sin\theta&-\text{x}&1\\\cos\theta&1&\text{x} \end{vmatrix}=8$
Expanding along $R_1$ we get
$\text{x}(-\text{x}^2-1)-\sin\theta(-\text{x}\sin\theta-\cos\theta)\\+\cos\theta(-\sin\theta+\text{x}\cos\theta)=8$
$\Rightarrow-\text{x}^3-\text{x}+\text{x}\sin^2\theta+\sin\theta\cos\theta\\-\sin\theta\cos\theta+\text{x}\cos^2\theta=8$
$\Rightarrow-\text{x}^3-\text{x}+\text{x}(\sin^2\theta+\cos^2\theta)=8$
$\Rightarrow-\text{x}^3-\text{x}+\text{x}=8$
$\Rightarrow\text{x}^3+8=0$
$\Rightarrow(\text{x}+2)(\text{x}^2-2\text{x}+4)=0$
$\Rightarrow\text{x}+2=0$ $[\because\text{x}^2-2\text{x}+4>0\ \forall\text{ x}]$
$\Rightarrow\text{x}=-2$
View full question & answer→Question 263 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}1&-3&2\\4&-1&3\\3&5&2\end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}-1&2\\5&2 \end{vmatrix}=-2-10=-12$
$\text{M}_{21}=\begin{vmatrix}-3&2\\5&2 \end{vmatrix}=-6-10=-6$
$\text{M}_{31}=\begin{vmatrix}-3&2\\-1&2 \end{vmatrix}=-6+2=-4$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=-12$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=-(-16)=16$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=-4$
$\text{D}=1(-12)+3(8-6)+2(20+3)$
$\text{D}=-12+6+46=40$
View full question & answer→Question 273 Marks
If $\text{A}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix},$ then for any natural number, find the value of Det $(A^n)$.
AnswerLet $\text{A}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix}$
Then, $\text{A}^2=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix}$
$=\begin{bmatrix}\cos^2\theta-\sin^2\theta&\cos\theta\sin\theta+\sin\theta\cos\theta\\-\sin\theta\cos\theta-\cos\theta\sin\theta&-\sin^2\theta+\cos^2\theta \end{bmatrix}$
$=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta \end{bmatrix}$
Similarly, $\text{A}^{\text{n}}=\begin{bmatrix}\cos(\text{n}\theta)&\sin(\text{n}\theta)\\-\sin(\text{n}\theta)&\cos(\text{n}\theta) \end{bmatrix}$
Therefore,
$|\text{A}^{\text{n}}|=\begin{vmatrix}\cos(\text{n}\theta)&\sin(\text{n}\theta)\\-\sin(\text{n}\theta)&\cos(\text{n}\theta) \end{vmatrix}$
$=\cos^2(\text{n}\theta)+\sin^2(\text{n}\theta)$
$=1$
Hence, det $(A^n) = 1$
View full question & answer→Question 283 Marks
If the points (3, -2), (x, 2), (8, 8) are collinear, find x using determinant.
AnswerSince the points are collinear, hance the area of the traingle must be zero.
$\Rightarrow\frac{1}{2}\begin{vmatrix}3&-2&1\\\text{x}&2&1\\8&8&1\end{vmatrix}=0$
$\Rightarrow3(-6)+2(\text{x}-8)+1(8\text{x}-16)=0$
$\Rightarrow-18+2\text{x}-16+8\text{x}-16=0$
$\Rightarrow10\text{x}=50$
$\Rightarrow\text{x}=5$
Hence x = 5
View full question & answer→Question 293 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}-1&4\\2&3 \end{vmatrix}$
AnswerLet $M_{ij}$ and $C_{ij}$ represents the minor and co-factor respectively of element which present at the $i^{th}$ row and $j^{th}$ column.
In a 2 × 2 matrix, the minor of an element is obtained by deleting that row that column in which it is present.
Now, $\text{M}_{11}=3$
$\text{M}_{21}=4$
$\text{C}_{11}=(-1)^{1+1}\times\text{M}_{11}$ $[\text{C}_{\text{ij}}=(-1)^{\text{i}+\text{j}}\times\text{M}_{\text{ij}}]$
$\text{C}_{21}=(-1)^{2+1}\times\text{M}_{21}$
$\text{C}_{21}=(-1)^3\times4$
$\text{C}_{21}=-4$
Also, $|\text{A}|=(-1)\times(3)-(2)\times(4)$
$|\text{A}|=3+8$
$|\text{A}|=11$
View full question & answer→Question 303 Marks
Find the area of the triangle with vertices at the points:
$(-1, -8), (-2, -3)$ and $(3, 2)$
Answer$\triangle=\frac{1}{2}\begin{vmatrix}-1&-8&1\\-2&-3&1\\3&2&1\end{vmatrix}$
$\triangle=\frac{1}{2}\begin{vmatrix}-1&-8&1\\-1&5&0\\3&2&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$\triangle=\frac{1}{2}\begin{vmatrix}-1&-8&1\\-1&5&0\\4&10&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$\triangle=\frac{1}{2}\begin{vmatrix}-1&5\\4&10\end{vmatrix}$
$\triangle=\frac{1}{2}|-10-20|$
$\triangle=\frac{1}{2}(30)$
$\triangle=15\text{ square units}$
View full question & answer→Question 313 Marks
Find the integral value of x, if $\begin{vmatrix}\text{x}^2&\text{x}&1\\0&2&1\\3&1&4 \end{vmatrix}=28.$
View full question & answer→Question 323 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 333 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 343 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 353 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 $R_3 → R_3- R_1$]
$=\frac{1}{2}\begin{vmatrix}6&0\\4&3\end{vmatrix}$
$=\frac{1}{2}(18-0)$
$=\frac{1}{2}(18)$
$=9\text{ square units}$
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