Question 13 Marks
Using the properties of determinants, prove that$ \begin{vmatrix} \text{a + b} & \text{b + c} & \text{c + a} \\ \text{b + c} & \text{c + a} & \text{a + b} \\ \text{c + a} & \text{a + b} & \text{b + c} \end{vmatrix}=2 \begin{vmatrix} \text{a} & \text{b} & \text{c} \\ \text{b} & \text{c} & \text{a} \\ \text{c} & \text{a} & \text{b} \end{vmatrix}$.
AnswerUsing $C_1$
$\rightarrow C_1 + C_2 + C_{3}$ we get $\text{LHS} = 2 \begin{vmatrix} \text{a + b + c} & \text{b + c} & \text{c + a} \\ \text{a + b + c} & \text{c + a} & \text{a + b} \\ \text{a + b + c} & \text{a + b} & \text{b + c} \end{vmatrix}C_1$
$\rightarrow C_1 – C_2$ to get $= 2 \begin{vmatrix} \text{a } & \text{b + c} & \text{c + a} \\ \text{b } & \text{c + a} & \text{a + b} \\ \text{c } & \text{a + b} & \text{b + c} \end{vmatrix}$
Using $C_{3}$
$\rightarrow C_3 – C_1 = 2 \begin{vmatrix} \text{a } & \text{b + c} & \text{c} \\ \text{b } & \text{c + a} & \text{a} \\ \text{c } & \text{a + b} & \text{b } \end{vmatrix}$
Using $C_2$
$\rightarrow C_2 – C_3 = 2 \begin{vmatrix} \text{a } & \text{b} & \text{c} \\ \text{b } & \text{c } & \text{a} \\ \text{c } & \text{a } & \text{b } \end{vmatrix}=\text{RHS}$.
View full question & answer→Question 23 Marks
If $\text{A}= \begin{bmatrix} 3 & 1 \\ -1 & 2 \\ \end{bmatrix},$ show the $\text{A}^{2}-\text{5A}+\text{7I}=0$. Hence find $A^{-1}.$
AnswerFor $A^2 = \begin{pmatrix} 3 & 1 \\ -1 & 2 \\ \end{pmatrix} \begin{pmatrix} 3 & 1 \\ -1 & 2 \\ \end{pmatrix}= \begin{pmatrix} 8 & 5 \\ -5 & 3 \\ \end{pmatrix}$
For $A^2 – 5A + 7 I = \begin{pmatrix} 8 & 5 \\ -5 & 3 \\ \end{pmatrix}-\begin{pmatrix} 15 & 5 \\ -5 & 10 \\ \end{pmatrix}+\begin{pmatrix} 7 & 0 \\ 0 & 7 \\ \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$
$A^{-1} = \frac{1}{7}[\text{5I - A]}=\frac{1}{7}\Bigg[\begin{pmatrix} 5 & 0 \\ 0 & 5 \\ \end{pmatrix}-\begin{pmatrix} 3 & 1 \\ -1 & 2 \\ \end{pmatrix}\Bigg]=\frac{1}{7}\begin{pmatrix} 2 & -1 \\ 1 & 3 \\ \end{pmatrix}.$ View full question & answer→Question 33 Marks
Using properties of determinants, prove the following:
$\begin{vmatrix} 3a & -a + b & -a + c \\ a - b & 3b & c - b \\ a - c & b - c & 3c \end{vmatrix} = 3(a + b + c) (ab + bc + ca) $
Answer$\text{C}_{1} \rightarrow \text{C}_{1} + \text{C}_{2} + \text{C}_{3}, \triangle = \begin{bmatrix} a + b + c & -a + b & -a + c \\ a + b + c & 3b & c - b \\ a + b + c & b - c & 3c \end{bmatrix} = ( a + b + c) \begin{bmatrix} 1 & -a + b & -a + c \\ 1 & 3b & c - b \\ 1 & b - c & 3c \end{bmatrix} $
$\triangle = ( a + b + c) \begin{bmatrix} 1 & -a + b & -a + c \\ 0 & 2b + a & c - b \\ 0 & - c + a & 2c + a \end{bmatrix} $
$\begin{matrix} R_{2} \rightarrow & R_{2} - & R_{1} \\ R_{3} \rightarrow & R_{3}- & R_{1} \\ \end{matrix} $
$\text{Expanding to get} \triangle = \text{(a + b + c) (3ab + 3bc + 3ca) = 3(a + b + c) (ab + bc + ca)}$
View full question & answer→Question 43 Marks
If $A = \begin{bmatrix} 2 & -3 & \\ 3 & 4 & \\ \end{bmatrix} $ show that $\text{A^{2} - 6 A + 17 I = 0.}$ Hense find $A^{-1}.$
Answer $\text{For writing}A^{2} = \begin{pmatrix} -5& -18\\ 18 & 7 \end{pmatrix}$$A^{2} - \text{6A + 17 } I = A^{2} = \begin{pmatrix} -5& -18\\ 18 & 7 \end{pmatrix} + \begin{pmatrix} -12& 18\\ -18 & -24 \end{pmatrix} + \begin{pmatrix} 17 & 0\\ 0& 17 \end{pmatrix} = 0$
$\text{Pre-multiplying (i) by}\stackrel{-1}{\hbox{A}}, \stackrel{-1}{\hbox{A}} = \frac{1}{17} [ \text{6 I - A] and getting} \Rightarrow \stackrel{-1}{\hbox{A}} = \frac{1}{17} \begin{bmatrix} 4 & 3\\ -3 & 2\\ \end{bmatrix} $
View full question & answer→Question 53 Marks
Using properties of determinants, prove the following:$ \begin{vmatrix} a - b -c & 2a & 2a \\ 2b & b- c - a & 2b \\ 2c & 2c & c- a -b \end{vmatrix} = (a + b + c)^{3}$
Answer$R_{1} \rightarrow R_{1} + R_{2} + R_{3} \Rightarrow \triangle= (a+ b+ c) \begin{vmatrix} 1 & 1 & 1\\ 2b & b- c - a & 2b \\ 2c & 2c & c- a -b \end{vmatrix} $$C_{2} \rightarrow C_{2}-C_{1}, C_{3}\rightarrow C_{3}-C_{1} \Rightarrow $
$\triangle = (a + b + c) \begin{vmatrix} 1& 0 & 0 \\ 2b & -(b+ c + a) & 0 \\ 2c & 0 & -(c+ a +b) \end{vmatrix} $
$= (a + b + c)^{3}$
View full question & answer→Question 63 Marks
Using properties of determinants, prove the following:$ \begin{vmatrix} a & b & c \\ a - b & b - c & c - a \\ b + c & c + a & a + b \end{vmatrix} = a^{3} + b^{3} + c^{3} - 3abc$
Answer$\triangle = \begin{vmatrix} a & b & c \\ a - b & b - c & c - a \\ b + c & c + a & a + b \end{vmatrix}$$C_1\rightarrow C_1 +C_2 +C_3\Rightarrow \triangle =\begin{vmatrix} a+b+c & b & c \\ 0 & b - c & c - a \\ 2(a + b + c) & c + a & a + b \end{vmatrix}$
$\therefore \triangle = (a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & b - c& c - a \\ 2 & c + a & a + b \end{vmatrix}$
$R_3\rightarrow R_3 - 2R_1\Rightarrow \triangle = (a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & b - c & c - a \\ 0 & c + a - 2b & a + b - 2c \end{vmatrix}$
$= (a + b+ c) [(b -c) (a + b -2c) - (c + a - 2b)(c - a)]$
$= ( a + b + c) (a^{2} + b^{2} + c^{2 } - ab - bc - ca)$
$= a^{3} + b^{3} + c^{3} - 3abc$
View full question & answer→Question 73 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 83 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 93 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 103 Marks
Using Cofactors of elements of second row, evaluate $\triangle=\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}.$
AnswerThe given determinant is $\begin{vmatrix}5&3&8\\2&0&1\\1&2&3\end{vmatrix}.$
we have:
$M_{21} =\begin{vmatrix}3&8\\2&3\end{vmatrix}=9-16=-7$
$\therefore A_{21} =$ cofactor of $a_{21} = (-1)^{2+1 }M_{21} = 7$
$M_{22} =\begin{vmatrix}5&8\\1&3\end{vmatrix}=15-8=7$
$\therefore A_{22} =$ cofactor of $a_{22} = (-1)^{2+2} M_{22} = 7$
$M_{23} =\begin{vmatrix}5&3\\1&2\end{vmatrix}=10-3=7$
$\therefore A_{23} =$ cofactor of $a_{22} = (-1)^{2+3} M_{23 }= -7$
We know that $\triangle$ is equal to the sum of the product of the elements of the second row with their corresponding cofactor.
$\therefore\triangle = a_{21}A_{21} + a_{22}A_{22 }+ a_{23}A_{23 }= 2(7) + 0(7) + 1(-7) = 14 - 7 = 7$
View full question & answer→Question 113 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 123 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 133 Marks
Using properties of determinants, prove that: $\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}=0$
Answer$\triangle=\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$=\frac{1}{\sin\delta\cos\delta}\begin{vmatrix}\sin\alpha\sin\delta&\cos\alpha\cos\delta&\cos\alpha\cos\delta-\sin\alpha\sin\delta\\\sin\beta\sin\delta&\cos\beta\cos\delta&\cos\beta\cos\delta-\sin\beta\sin\delta\\\sin\gamma\sin\delta&\cos\gamma\cos\delta&\cos\gamma\cos\delta-\sin\gamma\sin\delta\end{vmatrix}$
Applying $C_1 \rightarrow C_1 + C_3,$ we have$:$
$\triangle=\frac{1}{\sin\delta\cos\delta}\begin{vmatrix}\cos\alpha\cos\delta&\cos\alpha\cos\delta&\cos\alpha\cos\delta-\sin\alpha\sin\delta\\\cos\beta\cos\delta&\cos\beta\cos\delta&\cos\beta\cos\delta-\sin\beta\sin\delta\\\cos\gamma\cos\delta&\cos\gamma\cos\delta&\cos\gamma\cos\delta-\sin\gamma\sin\delta\end{vmatrix}$
Here, two columns $C_1$ and $C_1 + C_3,$ are identical.
$\therefore\triangle=0.$
Hence, the given result is proved.
View full question & answer→Question 143 Marks
By using properties of determinants, show that:
$\begin{vmatrix}a-b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}=(a+b+c)^3$
Answer$\text{L.H.S.}=\begin{vmatrix}a-b-c&2a&2a\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$$=\begin{vmatrix}a+b+c&a+b+c&a+b+c\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}\ \left[\text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\right]$
$=(a+b+c)\begin{vmatrix}1&1&1\\2b&b-c-a&2b\\2c&2c&c-a-b\end{vmatrix}$
$=(a+b+c)\begin{vmatrix}1&0&0\\2b&-b-c-a&0\\2c&0&-c-a-b\end{vmatrix}\ \left[\text{C}_2\rightarrow\text{C}_2-\text{C}_1\ \text{and C}_3\rightarrow\text{C}_3-\text{C}_1\right]$
$=(a+b+c)1[(a+b+c)(a+b+c)-0]$
$=(a+b+c)(a+b+c)^2$
$=(a+b+c)^3=\text{R.H.S.}$
View full question & answer→Question 153 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 163 Marks
Show that the points (a + 5, a - 4), (a - 2, a + 3) and (a, a) do not lie on a straight line for any value if a.
AnswerGiven, the points are (a + 5, a - 4), (a - 2, a + 3) and (a, a).
We have to prove that these points do not lie on a straight line.
So, we have to prove that these points form a triangle.
Area $\triangle=\frac{1}{2}\begin{vmatrix}\text{a}+5&\text{a}-4&1\\\text{a}-2&\text{a}+3&1\\\text{a}&\text{a}&1 \end{vmatrix}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1-\text{R}_3\text{ and R}_2\rightarrow\text{R}_2-\text{R}_3\big]$
$=\frac{1}{2}\begin{vmatrix}5&-4&0\\-2&3&0\\\text{a}&\text{a}&1\end{vmatrix}=\frac{1}{2}\big[1.(15-8)\big]=\frac{7}{2}\neq0$
Hence, given points from. a triangle i.e., points do not lie on a straight line.
View full question & answer→Question 173 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 $R_2 \rightarrow R_2 - R_1]$
$=\frac{1}{2}\begin{vmatrix}\text{x}&4&1\\2-\text{x}&-10&0\\5-\text{x}&0&0\end{vmatrix}=\pm35 [$Applying $R_3 \rightarrow R_3 - R_1]$
$=\frac{1}{2}\begin{vmatrix}2-\text{x}&-10\\5-\text{x}&0\end{vmatrix}=\pm35$
$=0+10(5-\text{x})=\pm70$
$\Rightarrow50-10\text{x}=70$ 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 183 Marks
Find the inverse of the following matrices: $\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}$
$|\text{A}|=\cos^3\theta+\sin^2\theta=1\neq0$
$A$ is a singular matrix; therefore, it is invertible Let $C_{ij}$ be a cofactor of $a_{ij}$ in $A.$
Now,
$\text{C}_{11}=\cos\theta$
$\text{C}_{12}=\sin\theta$
$\text{C}_{21}=-\sin\theta$
$\text{C}_{22}=\cos\theta$
$\text{adj A}=\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}^\text{T}=\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
View full question & answer→Question 193 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 203 Marks
Using the properties of determinants: $\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
AnswerWe have, $\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&\text{x}&\text{x}\\\text{y}&0&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$ $\big[\text{Taking x}^2,\text{y}^2,\text{z}^2\text{ common from C}_1,\text{C}_2\text{ and C}_3\big]$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&0&\text{x}\\\text{y}&-\text{y}&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$ $\big[\text{C}_2\rightarrow\text{C}_2-\text{C}_3\big]$
Expanding along $R_1,$ we get
$=\text{x}^2\text{y}^2\text{z}^2[\text{x(yz}+\text{yz})]$
$=\text{x}^2\text{y}^2\text{z}^2.2\text{xyz}$
$=2\text{x}^3\text{y}^3\text{z}^3$
View full question & answer→Question 213 Marks
By using properties of determinants, show that:
$\begin{vmatrix}y+k&y&y\\y&y+k&y\\y&y&y+k\end{vmatrix}=k^2(3y+k)$
Answer$\text{L.H.S.}=\begin{vmatrix}y+k&y&y\\y&y+k&y\\y&y&y+k\end{vmatrix}$$=\begin{vmatrix}3y+k&y&y\\3y+k&y+k&y\\3y+k&y&y+k\end{vmatrix}\ \left[\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\right]$
$=(3y+k)\begin{vmatrix}1&y&y\\1&y+k&y\\1&y&y+k\end{vmatrix}$
$=(3y+k)\begin{vmatrix}1&y&y\\0&k&0\\0&0&k\end{vmatrix}\ \left[\text{operating}\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1\ \text{and R}_3\rightarrow\text{R}_3-\text{R}_1\right]$
$ =(3y+k).1\begin{vmatrix}k&0\\0&k\end{vmatrix}=(3y+k)k^2=k^2(3y+k)=\text{R.H.S.}$ Proved.
View full question & answer→Question 223 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 \times 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 233 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 243 Marks
If $[\cdot]$ and $\{\cdot\}$ denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_0\sin\{\text{x}\}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{\frac{\pi}{4}}_0\sin\{\text{x}\}\text{dx}$
We know that,
$\{\text{x}\}=\text{x},\text{ when }0<\text{x}<\frac{\pi}{4}$ $\big(\text{As }\pi=3.14\Rightarrow\frac{\pi}{4}=0.784<1\big)$
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{4}}_0\sin\{\text{x}\}\text{dx}$
$=\big[-\cos\text{x}\big]^{\frac{\pi}{4}}_{0}$
$=-\Big(\cos\frac{\pi}{4}-\cos\frac{\pi}{4}\Big)$
$=\cos0-\cos\frac{\pi}{4}$
$=1-\frac{1}{\sqrt{2}}$
$=\frac{\sqrt{2}-1}{\sqrt{2}}$
View full question & answer→Question 253 Marks
Find the inverse of the following matrices:
$\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$
Answer$\text{C}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$
$|\text{C}|=1+\text{bc}-\text{bc}=1\neq0$
$C$ is a singular matrix; therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $C_{ij}$ in $C$.
Now,
$\text{C}_{11}=\frac{1+\text{bc}}{\text{a}}$
$\text{C}_{12}=-\text{C}$
$\text{C}_{21}=-\text{b}$
$\text{C}_{22}=\text{a}$
$\text{adj C}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{c} \\ -\text{b} & \text{a} \end{bmatrix}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
$\therefore\ \text{C}^{-1}=\frac{1}{|\text{C}|}\text{adj C}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
View full question & answer→Question 263 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 273 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 283 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 293 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 303 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. $|\text{A}|=1\begin{vmatrix}1&2\\0&4\end{vmatrix}-0\begin{vmatrix}0&1\\0&4\end{vmatrix}+0\begin{vmatrix}0&1\\1&2\end{vmatrix}=1\left(4-0\right)-0+0=4$
$\therefore27|\text{A}|=27(4)=108 \dots\dots(1)$
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 313 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 323 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 333 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$]$
$=0\ [$Since$, R_1$ and $R_2$ are identicals$]$
View full question & answer→Question 343 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,$$=\begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{vmatrix}=1\begin{vmatrix}b-a&b^2-a^2\\c-a&c^2-a^2\end{vmatrix}$
$=(b-a)(c-a)\begin{vmatrix}1&b+a\\1&c+a\end{vmatrix}=(b-a)(c-a)(c+a-b-a)$
$=(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)=\text{R.H.S.}$ Proved.
View full question & answer→Question 353 Marks
If $\text{A}=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\text{ and A (adj A =)}\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix},$ then find the value of $k.$
Answer$\text{A}=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
$\therefore|\text{A}|=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
$=\cos^2\theta+\sin^2\theta=1\neq0$
Thus, $A^{-1}$ exists.
Now,
$\text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\text{adj A}$
$\Rightarrow\text{AA}^{-1}=\text{A adj A}$
$\Rightarrow\text{AA}^{-1}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix} \ \big[\because\ \text{AA}^{-1}=\text{I}\big]$
$\Rightarrow\text{k}=1$
View full question & answer→Question 363 Marks
By using properties of determinants, show that:
$\begin{vmatrix}x+4&2x&2x\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}=(5x+4)(4-x)^2$
Answer$\text{L.H.S.}=\begin{vmatrix}x+4&2x&2x\\2x&x+4&2x\\2x&2x&x+4\end{vmatrix}$$=\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 373 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 383 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 393 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 403 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 413 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 $R_2 \rightarrow R_2 - R_1]$
$=\frac{1}{2}\begin{vmatrix}0&0&1\\6&0&0\\4&3&0\end{vmatrix} [$Applying $R_3 \rightarrow R_3- R_1]$
$=\frac{1}{2}\begin{vmatrix}6&0\\4&3\end{vmatrix}$
$=\frac{1}{2}(18-0)$
$=\frac{1}{2}(18)$
$=9$ square units
View full question & answer→Question 423 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 433 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 443 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 453 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 463 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 473 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 483 Marks
Examine the consistency of the system of equations:
3x - y - 2z = 2
2y - z = -1
3x - 5y = 3
AnswerMatrix from of given equations is AX = B $\Rightarrow\ \begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\3\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}3&-1&-2\\0&2&-1\\3&-5&0\end{vmatrix}$
$\Rightarrow\ \text{|A|}=3(0-5)-(-1)(0+3)+(-2)(-6)=3(-5)+3+12=15+15=0$
$\text{Now}\ \text{(adj. A)}=\begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}$
$\text{And}\ \text{(adj. A)B}=\begin{bmatrix}-5&10&5\\-3&6&3\\-6&12&6\end{bmatrix}\begin{bmatrix}2\\-1\\3\end{bmatrix}$
$=\begin{bmatrix}-10-10+15\\-6-6+9\\-12-12+18\end{bmatrix}=\begin{bmatrix}-5\\-3\\-6\end{bmatrix}\neq0$
Therefore, given equations are inconsistent.
View full question & answer→Question 493 Marks
Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x - 2y + z = -4
3x - y - 2z = 3
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\-4\\3\end{bmatrix}$
$\text{Here}\ \text{A}= \begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix},\text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}5\\-4\\3\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{vmatrix}=2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{40}\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3\end{bmatrix}$
$=\frac{1}{40}\begin{bmatrix}25-12+27\\25+52+3\\25-44-21\end{bmatrix}=\frac{1}{40}\begin{bmatrix}40\\80\\-40\end{bmatrix}=\begin{bmatrix}1\\2\\-1\end{bmatrix}$
Therefore, x = 1, y = 2 and z = -1
View full question & answer→Question 503 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 $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=2\begin{vmatrix}101&1&1\\104&1&1\\107&1&1\end{vmatrix}$
$=0$
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 513 Marks
Without expanding the determinant, prove that $\begin{vmatrix}a&a^2&bc\\b&b^2&ca\\c&c^2&ab\end{vmatrix}=\begin{vmatrix}1&a^2&a^3\\1&b^2&b^3\\1&b^2&b^3\end{vmatrix}$.
Answer$\text{L.H.S.}=\begin{vmatrix}a&a^2&bc\\b&b^2&ca\\c&c^2&ab\end{vmatrix}$
$=\frac{1}{abc}\begin{vmatrix}a^2&a^3&abc\\b^2&b^3&abc\\c^2&c^3&abc\end{vmatrix}\ \ \ \ \ \ \bigg[\text{R}_1\rightarrow a\text{R}_1,\text{R}_2\rightarrow b\text{R}_2,\text{and}\ \text{R}_3\rightarrow c\text{R}_3\bigg]$
$=\frac{1}{abc}.abc\begin{vmatrix}a^2&a^3&1\\b^2&b^3&1\\c^2&c^3&1\end{vmatrix} [$Taking out factor abc from $C_3]$
$=\begin{vmatrix}a^2&a^3&1\\b^2&b^3&1\\c^2&c^3&1\end{vmatrix}$
$=\begin{vmatrix}1&a^2&a^3\\1&b^2&b^3\\1&c^2&c^3\end{vmatrix} [$Applying $C_1 \leftrightarrow C_3$ and $C_2 \leftrightarrow C_3 ]$
$=\text{R.H.S.}$
Hence, the given result is proved.
View full question & answer→Question 523 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 533 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 $R_2 \rightarrow R_2 - R_1]$
$=\frac{1}{2}\begin{vmatrix}1&4&1\\1&-1&0\\-6&-7&0\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - R_1]$
$=\frac{1}{2}\begin{vmatrix}1&-1\\-6&-7\end{vmatrix}$
$=\frac{1}{2}(-7-6)$
$=\frac{13}{2}\text{square units}$ $[\because$ 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 543 Marks
Solve system of linear equations, using matrix method.
5x + 2y = 4
7x + 3y = 5
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}5&2\\7&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}4\\5\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}5&2\\7&3\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\end{bmatrix}\text{and B}=\begin{bmatrix}4\\5\end{bmatrix}$
$\therefore \ \text{|A|}=\begin{vmatrix}5&2\\7&5\end{vmatrix}=15-14=1\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{1}\begin{bmatrix}3&-2\\-7&5\end{bmatrix}\begin{bmatrix}4\\5\end{bmatrix}=\begin{bmatrix}12-10\\-28+25\end{bmatrix}=\begin{bmatrix}2\\-3\end{bmatrix}$
Therefore, x = 2 and y = -3
View full question & answer→Question 553 Marks
If $\text{adj A}=\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \text{and B}=\begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix},$ find adj AB.
AnswerIf A and B are non-singular square matrices of the same order, then
adj (AB) = (adj B)(adj A)
Now, ATQ
$\text{adj A}=\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}$
$\text{adj B}=\begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix}$
So, adj(AB) = (adj B)(adj A)
$=\begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}$
$=\begin{bmatrix} -6 & 5 \\ -2 & -10 \end{bmatrix}$
Hence, $\text{adj (AB)}=\begin{bmatrix} -6 & 5 \\ -2 & -10 \end{bmatrix}$
View full question & answer→Question 563 Marks
Using properties of determinants, prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+\text{2q}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
Answer$\triangle = \begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+\text{2q}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - 2R_1$ and $R_3 \rightarrow R_3 - 3R_1,$ we have:
$\triangle=\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\0&1&2+\text{p}\\0&3&7+3\text{p}\end{vmatrix}$
Applying $R_3 \rightarrow R_3 - 3R_2,$ we have:
$\triangle=\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\0&1&2+\text{p}\\0&0&1\end{vmatrix}$
Expanding along $C_1,$ we have:
$\triangle=1\begin{vmatrix}1&2+\text{p}\\0&1\end{vmatrix}=1(1-0)=1$
Hence, the given result is proved.
View full question & answer→Question 573 Marks
Solve system of linear equations, using matrix method.
x - y + 2z=7
3x + 4y - 5z = -5
2x - y + 3z = 12
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}7\\-5\\12\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}7\\-5\\12\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&-1&2\\3&4&-5\\2&-1&3\end{vmatrix}=1(12-5)-(-1)(9+10)+2(-3-8)=7+9-22=4\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{4}\begin{bmatrix}7&1&-3\\-19&-1&11\\-11&-1&7\end{bmatrix}\begin{bmatrix}7\\-5\\12\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}49-5-36\\-133+5+132\\-77+5+84\end{bmatrix}=\frac{1}{4}\begin{bmatrix}8\\4\\12\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$
Therefore, x = 2, y = 1 and z = 3
View full question & answer→Question 583 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 593 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 603 Marks
Using the proprties of determinants in Exercise:
$\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}=(\text{a}-1)^3$
AnswerWe have to prove,
$\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}=(\text{a}-1)^3$
$\therefore\ \text{L.H.S}=\begin{vmatrix}\text{a}^2+2\text{a}&2\text{a}+1&1\\2\text{a}+1& \text{a}+2&1\\3&3&1\end{vmatrix}$
$=\begin{vmatrix}\text{a}^2+2\text{a}-2\text{a}-1&2\text{a}+1-\text{a}-2&0\\2\text{a}+1-3&\text{a}+2-3&0\\3&3&1\end{vmatrix}$
$[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_2\text{ and R}_2\rightarrow\text{R}_2-\text{R}_3]$
$=\begin{vmatrix}(\text{a}-1)(\text{a}+1)&(\text{a}-1)&0\\2(\text{a}-1)&(\text{a}-1)&0\\3&3&1\end{vmatrix}$
$=(\text{a}-1)^2\begin{vmatrix}(\text{a}+1)&1&0\\2&1&0\\3&3&1\end{vmatrix}$
$[\text{Taking (a}-1)\text{ common from R}_1\text{ and R}_2\text{ each}]$
$=(\text{a}-1)^2\big[1(\text{a}+1)-2\big]=(\text{a}-1)^3$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 613 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 $R_2 \rightarrow R_2 - R_1]$
$\Rightarrow\begin{vmatrix}-2\text{k}+1&4\text{k}-2\\-4-\text{k}&6-2\text{k}\end{vmatrix}=0$
$\Rightarrow-8\text{k}+4+16\text{k}-8+8\text{k}^2-4\text{k}=0$
$\Rightarrow8\text{k}^2+4\text{k}-4=0$
$\Rightarrow(8\text{k}-4)(\text{k}+1)=0$
$\Rightarrow\text{k}=-1$ or $\text{k}=\frac{1}{2}$
View full question & answer→Question 623 Marks
Solve system of linear equations, using matrix method.
2x - y = -2
3x + 4y = 3
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}2&-1\\3&4\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-2\\3\end{bmatrix}$ $\text{Here}\ \text{A}=\begin{bmatrix}2&-1\\3&4\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\end{bmatrix}\text{and B}=\begin{bmatrix}-2\\3\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}2&-1\\3&4\end{vmatrix}=8-(-3)=8+3=11\neq0$ Therefore, solution is Unique and X = $\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}(\text{adj. A)B}$ $\Rightarrow\ \begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{11}\begin{bmatrix}4&1\\-3&2\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-8+3\\6+6\end{bmatrix}=\frac{1}{11}\begin{bmatrix}-5\\12\end{bmatrix}=\begin{bmatrix}\frac{-5}{11}\\\frac{12}{11}\end{bmatrix}$ Therefore, $x=\frac{-5}{11}\text{and}\ y=\frac{12}{11}$
View full question & answer→Question 633 Marks
Find the adjoint of the following matrices: $\begin{bmatrix}-3 & 5 \\ 2 & 4 \end{bmatrix}$Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer$\text{A}=\begin{bmatrix}-3 & 5 \\ 2 & 4 \end{bmatrix}$
$\text{adjoint A}=\begin{bmatrix} 4 & -5 \\ -2 & -3 \end{bmatrix}$
$\text{adjoint A}=\begin{bmatrix} -22 & 0 \\ 0 & -22 \end{bmatrix}$
$|\text{A}|=-22$
$|\text{A}|\text{I}=\begin{bmatrix} -22 & 0 \\ 0 & -22 \end{bmatrix}$
$\text{A(adjoint A)}=\begin{bmatrix} -22& 0 \\ 0 & -22 \end{bmatrix}$
$\therefore\ \text{(adjoint A)A}=|\text{A}|\text{I}=\text{A(adjoint A)}$
Hence verified.
View full question & answer→Question 643 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 653 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 663 Marks
Solve system of linear equations, using matrix method.
x - y + z = 4
2x + y - 3z = 0
x + y + z = 2
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\0\\2\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}1&-1&1\\2&1&-3\\1&1&1\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}4\\0\\2\end{bmatrix}$
$\therefore\ \text{|A|}=\begin{vmatrix}1&-1&1\\2&1&-3\\1&1&1\end{vmatrix}=1(1+3)-(-1)(2+3)+1(2-1)=4+5+1=10\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{10}\begin{bmatrix}4&2&2\\-5&0&5\\1&-2&3\end{bmatrix}\begin{bmatrix}4\\0\\2\end{bmatrix}$
$=\frac{1}{10}\begin{bmatrix}16+0+4\\-20+0+10\\4-0+6\end{bmatrix}=\frac{1}{10}\begin{bmatrix}20\\-10\\10\end{bmatrix}=\begin{bmatrix}2\\-1\\1\end{bmatrix}$
Therefore, x = 2, y = -1 and z = 1
View full question & answer→Question 673 Marks
Examine the consistency of the system of equations:
5x - y + 4z = 5
2x - 3y + 5z = 2
5x - 2y + 6z = -1
AnswerMatrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\2\\-1\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}5&-1&4\\2&3&5\\5&-2&6\end{vmatrix}$
$\Rightarrow\ \text{|A|}=5(18+10)-(-1)(12-25)+4(-4-15)=140-13-76=140-89=51\neq0$
Therefore, Unique solution and hence equations are consistent.
View full question & answer→Question 683 Marks
Solve system of linear equations, using matrix method.
2x + y + z = 1
x - 2y - z = $\frac{3}{2}$
3y - 5z = 9
AnswerMatrix from of given equations is AX = B $\Rightarrow\ \begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix}$
$\therefore\ \text{|A|}= \begin{bmatrix}2&1&1\\1&-2&-1\\0&3&-5\end{bmatrix}=2(10+3)-1(-5-0)+1(3-0)=26+5+3=34\neq0$
Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B}$
$\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{34}\begin{bmatrix}13&8&1\\5&-10&3\\3&-6&-5\end{bmatrix}\begin{bmatrix}1\\\frac{3}{2}\\9\end{bmatrix}$
$=\frac{1}{34}\begin{bmatrix}13+12+9\\5-15+27\\3-9-45\end{bmatrix}=\frac{1}{34}\begin{bmatrix}34\\17\\-51\end{bmatrix}=\begin{bmatrix}1\\\frac{1}{2}\\\frac{-3}{2}\end{bmatrix}$
Therefore, $x=1,y=\frac{1}{2}\text{and}\ z=\frac{3}{2}$
View full question & answer→Question 693 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, $3\text{A}^\text{T}=3\begin{bmatrix}-1& 2 & 2 \\-2 & 1 & -2 \\ -2 & -2 & 1 \end{bmatrix}=\begin{bmatrix}-3 & -6 & -6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}\ .....\text{(ii)}$
$\therefore\ \text{adj A}=3\text{A}^\text{T}$
View full question & answer→Question 703 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 713 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 723 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 733 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.
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