Question 15 Marks
Using properties of determinants, prove that
$\begin{vmatrix} \text{a}^{2} + \text{2a} & \text{2a + 1} & 1 \\ \text{2a + 1} & \text{a + 2} & 1 \\ 3 & 3 & 1 \end{vmatrix} = \text{(a - 1)}^{3}$
Answer$\Delta = \begin{vmatrix} \text{a}^{2} + \text{2a} & \text{2a + 1} & 1 \\ \text{2a + 1} & \text{a + 2} & 1 \\ 3 & 3 & 1 \end{vmatrix}$ $\text{R}_{1} \rightarrow \text{R}_{1} - \text{R}_{2} \text{ and } \text{R}_{2} \rightarrow \text{R}_{2} - \text{R}_{3}$$\Delta = \begin{vmatrix} \text{a}^{2} - 1 & \text{a - 1} & 0 \\ 2\text{(a - 1)} & \text{a - 1} & 0 \\ 3 & 3 & 1 \end{vmatrix}$
$ = \text{(a - 1)}^{2}\begin{vmatrix} \text{a + 1} & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix}$ Expanding $\text{(a - 1)}^{2}. \text{(a - 1)} = \text{(a - 1)}^{3}.$
View full question & answer→Question 25 Marks
Using properties of determinants, prove that:
$\begin{vmatrix} \text{1 + a } & \text{1} & \text{1} \\ \text{1} & \text{1 + b} & \text{1} \\\text{1} & 1 &\text{1 + c} \end{vmatrix}= \text{ abc + bc + ca + ab}$
Answer$R_{1 }\rightarrow$ $\frac{1}{\text{a}} R_{1, }R_{2 }\rightarrow \frac{1}{\text{b}} R_{2, }R_{3 }\rightarrow \frac{1}{\text{c}}R_{3 }$
$\therefore\ \text{ LHS} =\text{abc} \begin{bmatrix} \frac{1}{\text{a}}+1 & \frac{1}{\text{a}}& \frac{1}{\text{a}} \\ \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} \\ \frac{1}{\text{c}} & \frac{1}{\text{c}} &\frac{1}{\text{c}}+1 \end{bmatrix}$
$R_{1 }\rightarrow R_1 + R_{2 +}R_3 \Rightarrow \text{ LHS} =\text{abc} \begin{vmatrix} 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}} &1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}& 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}} \\ \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} \\ \frac{1}{\text{c}} & \frac{1}{\text{c}} &\frac{1}{\text{c}}+1 \end{vmatrix}$
$=\text{abc}\Bigg( 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Bigg)$$\begin{vmatrix} 1 & 1& 1 \\ \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} \\ \frac{1}{\text{c}} & \frac{1}{\text{c}} &\frac{1}{\text{c}}+1 \end{vmatrix}$
$\begin{matrix} \text{c}_2\rightarrow\text{c}_2-\text{c}_1\\ \text{c}_3\rightarrow\text{c}_3-\text{c}_1 \end{matrix}=\text{abc}\Bigg( 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Bigg)\begin{vmatrix} 1&0&0\\ \frac{1}{\text{b}}&1&0\\ \frac{1}{\text{c}}&0&1\\\end{vmatrix}$
$=\text{abc}\Bigg( 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Bigg)$.
$1 = \text{ abc + bc + ca + ab}$ = RHS
View full question & answer→Question 35 Marks
Using properties of determinants, prove that
$\begin{vmatrix} \text{b + c } & \text{c + a} & \text{a + b} \\ \text{q } + \text{r} & \text{r + p} & \text{p + q} \\ \text{y + z} & \text{z + x} &\text{x + y} \end{vmatrix}= \text{2}\begin{vmatrix} \text{a } & \text{b} & \text{c} \\ \text{p} & \text{q} & \text{r} \\ \text{x} & \text{y} &\text{z} \end{vmatrix}$
AnswerOperating $C \rightarrow C_{1 — }(C_2 + C_3),$ we get
$\text{LHS}=\begin{vmatrix} \text{-2a } & \text{c + a} & \text{a + b} \\ \text{-2p } & \text{r + p} & \text{p + q} \\ \text{-2x} & \text{z + x} &\text{x + y} \end{vmatrix}$
$=-2\begin{vmatrix} \text{a } & \text{c + a} & \text{a + b} \\ \text{p } & \text{r + p} & \text{p + q} \\ \text{x} & \text{z + x} &\text{x + y} \end{vmatrix}$
$ \begin{matrix} \text{C}{_2} → C_{2} — C_{1} \ \ \ \\ \text{C}{_3} → C_{3} — C_{1}\ \ \ \end{matrix}$
$\Rightarrow\text{LHS}=-2\begin{vmatrix} \text{a } & \text{c} & \text{b} \\ \text{p } & \text{r} & \text{q} \\ \text{x} & \text{z} &\text{ y} \end{vmatrix}$
$\text{C}_2\leftrightarrow\text{C}_3$ $=+2\begin{vmatrix} \text{a } & \text{c} & \text{b} \\ \text{p } & \text{r} & \text{q} \\ \text{x} & \text{z} &\text{ y} \end{vmatrix}=\text{RHS}$
View full question & answer→Question 45 Marks
Using properties of determinants, show that $\triangle\text{ABC}$ is isosceles if:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos\text{A} & 1 + \cos\text{B} & 1 + \cos\text{C} \\ \cos^{2}\text{A} + \cos\text{A} & \cos^{2}\text{B}+\cos\text{B} & \cos^{2}\text{C} + \cos\text{C} \end{vmatrix} = 0 $
Answer$\begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos\text{A} & 1 + \cos\text{B} & 1 + \cos\text{C} \\ \cos^{2}\text{A} + \cos\text{A} & \cos^{2}\text{B}+\cos\text{B} & \cos^{2}\text{C} + \cos\text{C} \end{vmatrix} = 0 $
$\text{Apply }\text{C}_{2}\rightarrow\text{C}_{2} -\text{C}_{1}, \text{C}_{3}\rightarrow\text{C}_{3} - \text{C}_{1}$
$\Leftrightarrow \begin{vmatrix} 1 & 0 & 0 \\ 1 +\cos\text{A} & \cos\text{B} - \cos\text{A} & \cos\text{C} - \cos\text{A} \\ \cos^{2}\text{A}+\cos\text{A} & (\cos\text{B} - \cos\text{A})(\cos\text{B} + \cos\text{A} + 1) & (\cos\text{C} - \cos\text{A}) (\cos\text{C} + \cos\text{A} + 1) \end{vmatrix} = 0 $
$\text{Taking}(\cos\text{B} - \cos\text{A}), (\cos\text{C} - \cos\text{A}) \text{common from C}_{2} \& \text{ C}_{3}$
$\Leftrightarrow(\cos\text{B} - \cos\text{A})(\cos\text{C} - \cos\text{A}) \begin{vmatrix} 1 & 0 & 0 \\ 1 + \cos\text{A} & 1 & 1 \\ \cos^{2}\text{A} + \cos\text{A} & \cos\text{B} + \cos\text{A} + 1 & \cos\text{C} + \cos\text{A} + 1 \end{vmatrix} = 0 $
$\text{Expand along R}_{1}$
$\Leftrightarrow(\cos\text{B} - \cos\text{A})(\cos\text{C} - \cos\text{A}) (\cos\text{C} - \cos\text{B}) = 0$
$ \begin{matrix} \Leftrightarrow\cos\text{A} = \cos\text{B} & \Leftrightarrow\text{A = B} &\Leftrightarrow\triangle\text{ABC is an isosceles triangle} \\ \text{or} & \text{or} \\ \cos\text{B} = \cos\text{C} & \text{B = C} \\ \text{or} & \text{or} \\ \cos\text{C} = \cos\text{A} & \text{C = A} \end{matrix} $
View full question & answer→Question 55 Marks
Using properties of determinants, prove that
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AnswerOperating $R_2$ ; background:rgba(220,220,220,0.5))
$R_{2 }- 4R_1$_{ } and $R_{3 } R_{3 }-8R_1,$ we get
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Expanding along $C_2,$ we get
$– x (– 5x^2 + 4x^2) = x^3.$ View full question & answer→Question 65 Marks
Using properties of determinants, prove the following: $ \begin{bmatrix} \text{ x}&\text{x + y }&\text{x} + 2\text{y}\\ \text{x} + 2\text{y} & \text{x}& \text{x + y }\\\text{x + y}&\text{x} + 2\text{y}& \text{x} \end{bmatrix} = 9\text{y}^{2}(\text{x} + \text{y}). $
Answer$\text{L.H.S.} =\begin{bmatrix} \text{ x}&\text{x + y }&\text{x} + 2\text{y}\\ \text{x} + 2\text{y} & \text{x}& \text{x + y }\\\text{x + y}&\text{x} + 2\text{y}& \text{x} \end{bmatrix} $
$ =\begin{bmatrix} 3(\text{x + y})&3(\text{x + y})&3(\text{x + y})\\ \text{x} + 2\text{y}&\text{x}&\text{x + y}\\ \text{x + y}&\text{x} + 2\text{y}&\text{x} \end{bmatrix}[$ Applying $R_{1} = \text{R}_{1} + \text{R}_{2}+ \text{R}_{3}] $
$ =3(\text{x + y })\begin{bmatrix} 1 &1&1\\ \text{x} + 2\text{y}&\text{x}&\text{x + y}\\ \text{x + y}&\text{x} + 2\text{y}&\text{x} \end{bmatrix}$ Taking $[3 (x + y)$ common from $R_{1}]$
$ =3(\text{x + y })\begin{bmatrix} 0 &0&1\\ \text{y}&-\text{y}&\text{x + y} \\ \text{ y} & 2\text{y} &\text{x} \end{bmatrix}[$Applying $C_{1}\rightarrow\text{C}_{1} - \text{C}_{3} , \text{C}_{2}\rightarrow\text{C}_{2} - \text{C}_{3} ]$
Expanding along $R_1$ we get
$= 3 (x + y) {1 (2y^2 + y^2 )}$
$= 9y^2 (x + y) = \text{RHS}.$
View full question & answer→Question 75 Marks
Using properties of determinants, show that$\Delta=\begin{vmatrix} \text{b+c} & \text{a} & \text{a} \\ \text{b} & \text{c+a} & \text{b} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}=\text{4 abc}.$
Answer$\text{LHS }=\Delta=\begin{vmatrix} \text{b+c} & \text{a} & \text{a} \\ \text{b} & \text{c+a} & \text{b} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix};\DeclareMathOperator*{\median}{\text{ performing}} \median_{\text{R}_{1}\rightarrow\text{R}_{1}-\text{R}_{2}-\text{R}_{3}}\Delta=\begin{vmatrix} \text{0} & \text{-2c} & \text{-2b} \\ \text{b} & \text{c+a} & \text{b} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}$ $=\frac{1}{\text{c}}\begin{vmatrix} 0 & \text{-2c} & \text{-2b} \\ \text{bc} & \text{(c+a)c} & \text{bc} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}$
using $R_2\rightarrow R_2-bR_3$ gives $\Delta=\frac{1}{\text{c}}$ $\begin{vmatrix} 0 & \text{-2c} & \text{-2b} \\ \text{0} & \text{c(a+c-b)} & \text{b(c-a-b)} \\ \text{c} & \text{c} & \text{a+b} \end{vmatrix}$ $=\frac{\text{1}}{\text{c}}\cdot\text{c}\begin{vmatrix} \text{-2c} & \text{-2b} \\ \text{(c+a-b)c} & \text{(c-a-b)b} \\ \end{vmatrix}$
$= 2bc [(-c+a+b) + (c+a-b)] = 4 $
$abc= RHS.$
View full question & answer→Question 85 Marks
Using properties of determinants, solve the following for $x:$
$ \begin{vmatrix} \text{x - 2} & \text{2x - 3 } & \text{3x - 4 } \\ \text{x - 4} & \text{2x - 9} & \text{2x - 16} \\ \text{x -8} & \text{2x - 27} & \text{3x -64} \end{vmatrix}=0$
AnswerApplying $C_2 \rightarrow C_2 - 2C_1 $ and $C_3 \rightarrow C_3 - 3C_1,$ we get $\begin{vmatrix} \text{x - 2} & \text{1 } & \text{2 } \\ \text{x - 4} & \text{-1} & \text{-4}\\ \text{x -8} & \text{-11} & \text{40} \end{vmatrix}=0$
Applying $R_1 \rightarrow R_1 + R_2$ and $R_3 \rightarrow R_3 - 11R_2,$ we get $\begin{vmatrix} \text{2x - 6} & \text{0 } & \text{-2 } \\ \text{x - 4} & \text{-1} & \text{-4}\\ \text{-10x + 36} & \text{0} & \text{4} \end{vmatrix}=0$
Expanding along $C_2,$ we get $–1[8x – 24 – 20x + 72] = 012x = 48$ i.e. $x = 4$
View full question & answer→Question 95 Marks
Using properties of determinants, prove the following:$\begin{vmatrix} \text{x} &\text{x}^{2} & \text{1 + px}^{3} \\ \text{y} & \text{y}^{2} & \text{1 + py}^{3} \\ \text{z} & \text{z}^{2} & \text{1 + pz}^{3} \end{vmatrix}=\text{(1 + pxyz) (x - y)(y - z)(z - x)}$.
Answer$\text{LHS} = \begin{vmatrix} \text{x} &\text{x}^{2} & \text{1 + px}^{3} \\ \text{y} &\text{y}^{2} & \text{1 + py}^{3}\\ \text{z} &\text{z}^{2} & \text{1 + pz}^{3} \end{vmatrix}=\begin{vmatrix} \text{x} & \text{x}^{2} & 1 \\ \text{y} & \text{y}^{2} & 1 \\ \text{z} & \text{z}^{2} & 1 \end{vmatrix}+\text{p}\begin{vmatrix} \text{x} & \text{x}^{2} & \text{x}^{3} \\ \text{y} & \text{y}^{2} & \text{y}^{3} \\ \text{z} & \text{z}^{2} & \text{z}^{3} \end{vmatrix}$=$\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{1} & \text{y} & \text{y}^{2} \\ \text{1} & \text{z} & \text{z}^{2} \end{vmatrix}+\text{pxyz}\begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{1} & \text{y} & \text{y}^{2} \\ \text{1} & \text{z} & \text{z}^{2} \end{vmatrix}$
$= (1 + pxyz) \begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{1} & \text{y} & \text{y}^{2} \\ \text{1} & \text{z} & \text{z}^{2} \end{vmatrix}$
$= (1 + pxyz) \begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{y - x} & \text{y}^{2}-\text{x}^{2} \\ \text{0} & \text{z - x} & \text{z}^{2}-\text{x}^{2} \end{vmatrix} \begin{matrix} \text{R}_{2}\rightarrow & \text{R}_{2}\text{ }-& \text{R}_{1} \\ \text{R}_{3}\rightarrow & \text{R}_{3}\text{ }-& \text{R}_{1} \\ \end{matrix}$
$= (1 + pxyz) (x - y) (z - x) \begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{-1} &-\text{(x + y)} \\ \text{0} & \text{1} &\text{z + x} \end{vmatrix} $
$= (1 + pxyz) (x - y) (z - x) \begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{0} &\text{z - y} \\ \text{0} & \text{1} &\text{z + x} \end{vmatrix}\text{R}_{2}\rightarrow\text{R}_{2}+\text{R}_{3} $
$= (1 + pxyz) (x - y) (y - z) (z - x) \begin{vmatrix} \text{1} & \text{x} & \text{x}^{2} \\ \text{0} & \text{0} &-\text{1} \\ \text{0} & \text{1} &\text{z + x} \end{vmatrix} $
$= (1 + pxyz) (x - y) (y - z) (z - x) 1 = RHS.$
View full question & answer→Question 105 Marks
Using properties of determinants, prove the following:$\begin{vmatrix} 1 & \text{1 + P} & \text{1 + p + q} \\ 2 & \text{3 + 2p} & \text{1 + 3p + 2q} \\ 3 & \text{6 + 3p} & \text{1 + 6p + 3q} \end{vmatrix}=1.$
Answer$\Delta=\begin{vmatrix} 1 & \text{1 + P} & \text{1 + p + q} \\ 2 & \text{3 + 2p} & \text{1 + 3p + 2q} \\ 3 & \text{6 + 3p} & \text{1 + 6p + 3q} \end{vmatrix}$
$ \text{R}_{2}\rightarrow\text{R}_{2}-2\text{ R}_{1}$
$\Delta=\begin{vmatrix} 1 & \text{1 + P} & \text{1 + p + q} \\ 0 & \text{1} & \text{p - 1} \\ 3 & \text{6 + 3p} & \text{1 + 6p + 3q} \end{vmatrix}$
$\text{R}_{3}\rightarrow\text{R}_{3}-3\text{ R}_{1}$
$\Delta=\begin{vmatrix} 1 & \text{1 + P} & \text{1 + p + q} \\ 0 & \text{1} & \text{p - 1} \\ 0 & \text{3} & \text{3p - 2} \end{vmatrix}$
$=1[(\text{3p - 2) - 3 (p - 1)]}$
$=\text{3p - 2 - 3p + 3 = 1 = RHS}.$
View full question & answer→Question 115 Marks
Obtain the Inverse of the following matrix using elementary operations:$A= \begin{vmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{vmatrix}.$
Answer$\begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1}-\text{R}_{2}:\begin{bmatrix} 1 & -3 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2}-\text{2R}_{1}:\begin{bmatrix} 1 & -3 & -1 \\ 0 & 9 & 2 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1}-\text{R}_{3}:\begin{bmatrix} 1 & 1 & 0 \\ 0 & 9 & 2 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 & 1 \\ -2 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2}-\text{2R}_{3}:\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} 1 & -1 & 1 \\ -2 & 3 & -2 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1}-\text{R}_{2}:\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 4 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{R}_{3}\rightarrow\text{R}_{3}-\text{4R}_{2}:\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}\text{A}$
$\therefore\text{A}^{-1}=\begin{bmatrix} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix}.$
View full question & answer→Question 125 Marks
If x, y, z are different and $\begin{vmatrix} \text{x} & \text{x}^{2} & \text{1 + x}^{3} \\ \text{y} & \text{y}^{2} & \text{1 + y}^{3} \\ \text{z} & \text{z}^{2} & \text{1 + z}^{3} \end{vmatrix}$= 0, show that xyz = -1.
Answer$\begin{vmatrix} \text{x} & \text{x}^{2} & \text{1 + x}^{3} \\ \text{y} & \text{y}^{2} & \text{1 + y}^{3} \\ \text{z} & \text{z}^{2} & \text{1 + z}^{3} \end{vmatrix}=0$ $\Delta=\begin{vmatrix} \text{x} & \text{x}^{2} & \text{1 } \\ \text{y} & \text{y}^{2} & \text{1} \\ \text{z} & \text{z}^{2} & \text{1} \end{vmatrix}+\begin{vmatrix} \text{x} & \text{x}^{2} & \text{x}^{3} \\ \text{y} & \text{y}^{2} & \text{y}^{3} \\ \text{z} & \text{z}^{2} & \text{z}^{3} \end{vmatrix}$ = $\Delta=\begin{vmatrix} \text{x} & \text{x}^{2} & \text{1 } \\ \text{y} & \text{y}^{2} & \text{1} \\ \text{z} & \text{z}^{2} & \text{1} \end{vmatrix}+\text{xyz}\begin{vmatrix} \text{1} & \text{x}& \text{x}^{2} \\ \text{1} & \text{y}& \text{y}^{2} \\ \text{1} & \text{z}& \text{z}^{2} \end{vmatrix}$ = (1 + xyz) $\begin{vmatrix} \text{1} & \text{x}& \text{x}^{2} \\ \text{1} & \text{y}& \text{y}^{2} \\ \text{1} & \text{z}& \text{z}^{2} \end{vmatrix}$ $\text{R}_{2}\rightarrow\text{R}_{2}-\text{R}_{1},\text{ }\text{R}_{3}\rightarrow\text{R}_{3}-\text{R}_{1}$ $\begin{vmatrix} \text{1} & \text{x}& \text{x}^{2} \\ \text{1} & \text{y}& \text{y}^{2} \\ \text{1} & \text{z}& \text{z}^{2} \end{vmatrix}=\begin{vmatrix} \text{1} & \text{x}& \text{x}^{2} \\ \text{0} & \text{y - x}& \text{y}^{2}-\text{x}^{2} \\ \text{0} & \text{z - x}& \text{z}^{2}-\text{x}^{2} \ \end{vmatrix}=\text{(y - x)}\text{(z - x)}\begin{vmatrix} \text{1} & \text{x}& \text{x}^{2} \\ \text{0} & \text{1}& \text{y + x} \\ \text{0} & \text{1}& \text{z + x} \end{vmatrix}$= (y - x) (z - x) (z + x - y - x)
= (y - x) (z - x) (z - y)
= (x - y) (y - z) (z - x)
$\therefore\text{ }\Delta$ = (1 + xyz) (x - y) (y - z) (z - x) = 0 (given)As x - y $\neq$ 0, y - z $\neq$ 0, z - x $\neq$ 0 $\Rightarrow$ 1 + xyz = 0.
View full question & answer→Question 135 Marks
Using elementary transformations, find the inverse of the following matrix:$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix}$
Answer$\text{A}=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{pmatrix}=\begin{pmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_2 \rightarrow R_2 - 2R_1, R_3 \rightarrow R_3 + 2R_1,$ we get
$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 &0 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_1 \rightarrow R_1 - 2R_2,$ we get
$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 5 & -2 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_1 \rightarrow R_1 - R_3, $ we get
$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 3 & -2 & -1 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
Applying $R_2 \rightarrow R_2 - R_3, $ we get
$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix}\text{A}$
$\therefore \text{ A}^{-1}=\begin{pmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{pmatrix}.$
View full question & answer→Question 145 Marks
Using properties of determinants show that $\begin{vmatrix} 1 & 1 & \text{1 + x} \\ 1 & \text{1 + y} & 1 \\ \text{1 + z} & 1 & 1 \end{vmatrix} = \text{xyz + yz + zx + xy}.$
Answerto prove $\begin{vmatrix} 1 & 1 & \text{1 + x} \\ 1 & \text{1 + y} & 1 \\ \text{1 + z} & 1 & 1 \end{vmatrix}$
$\text{LHS}:$ Let $\triangle = \begin{vmatrix} 1 & 1 & \text{1 + x} \\ 1 & \text{1 + y} & 1 \\ \text{1 + z} & 1 & 1 \end{vmatrix}$
Take $x, y$ and $z$ common from $C_3, C_2$ and $C_1$ respectively.
$\therefore \triangle = xyz \begin{vmatrix} \frac{1}{z} & \frac{1}{y} & \frac{1}{x} + 1\\ \frac{1}{z} & \frac{1}{y} + 1 & \frac{1}{x} \\ \frac{1}{z} + 1 & \frac{1}{y} & \frac{1}{x} \end{vmatrix} $
$\text{C}_{3} \rightarrow \text{C}_{3} + \text{C}_{2} + \text{C}_{1}$
$\triangle = xyz \begin{vmatrix} \frac{1}{z} & \frac{1}{y} & 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{1}{z} & \frac{1}{y} + 1 & 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \\ \frac{1}{z} + 1 & \frac{1}{y} & 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \end{vmatrix}$
Taking $1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ Common
$\triangle = xyz \bigg(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\bigg) \begin{vmatrix} \frac{1}{z} & \frac{1}{y} & 1 \\ \frac{1}{z} & \frac{1}{y} + 1 & 1 \\ \frac{1}{z} + 1 & \frac{1}{y} & 1 \end{vmatrix}$
Applying $\text{R}_{2} \rightarrow \text{R}_{2} - \text{R}_{1}, \text{R}_{3} \rightarrow \text{R}_{3} - \text{R}_{1}$
$\triangle = xyz \bigg(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\bigg) \begin{vmatrix} \frac{1}{z} & \frac{1}{y} & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{vmatrix}$
On expanding we get
$ \triangle = xyz \bigg(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\bigg) = xyz + yz + xz + xy$
View full question & answer→Question 155 Marks
If $\text{A} = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix},$ find adj.A and verify that $\text{A(adj. A) = (adj.A)A} | \text{A} = | \text{I}_{3}.$
Answer$|\text{A}| = 1$
$\text{adj A} = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\text{A (adj A}) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \text{I}$
$|\text{A}| \text{I}_{3} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \text{I} $
View full question & answer→Question 165 Marks
$\text{A(adj. A) = (adj.A)A} | \text{A} = | \text{I}_{3}.$
If a, b and c are all non-zero and $ \begin{vmatrix} \text{1 + a} & 1 & 1 \\ 1 & \text{1 + b} & 1 \\ 1 & 1 & \text{1 + c} \end{vmatrix} = 0,$ then prove that $\frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} 1 = 0.$
Answer$\text{abc} \begin{vmatrix} \frac{1}{\text{a}} + 1 & \frac{1}{\text{b}} & \frac{1}{\text{c}} \\ \frac{1}{\text{a}} & \frac{1}{\text{b}} + 1 & \frac{1}{\text{c}} \\ \frac{1}{\text{a}} & \frac{1}{\text{b}} & \frac{1}{\text{c}} + 1 \end{vmatrix} = 0$
$ \text{C}_{1} \rightarrow \text{C}_{1} + \text{C}_{2} + \text{C}_{3}$
$\Rightarrow \text{abc} \begin{vmatrix} 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} & \frac{1}{\text{b}}& \frac{1}{\text{c}} \\ 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} & \frac{1}{\text{b}} + 1& \frac{1}{\text{c}} \\ 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} & \frac{1}{\text{b}}& \frac{1}{\text{c}} + 1 \end{vmatrix} = 0$
$\Rightarrow \text{abc} \bigg(1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}}\bigg) \begin{vmatrix} 1 & \frac{1}{\text{b}} & \frac{1}{\text{c}} \\ 1 & \frac{1}{\text{b}} + 1 & \frac{1}{\text{c}} \\ 1 & \frac{1}{\text{b}} & \frac{1}{\text{c}} + 1 \end{vmatrix} = 0$
$\text{R}_{2} \rightarrow - \text{R}_{2} - \text{R}_{1}. \text{R}_{3} \rightarrow \text{R}_{3}- \text{R}_{1}$
$\Rightarrow \text{abc} \bigg(1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}}\bigg) = 0$
$\because \text{a, b, c,}\neq 0$
$\therefore 1 + \frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} = 0$
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Using properties of determinants, prove that $ \begin{vmatrix} x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix} = 9\text{y}^{2} \text{(x + y)}.$
Answer$\begin{vmatrix} \text{x} & \text{x} + \text{y}& \text{x + 2y} \\ \text{x + 2y} & \text{x} & \text{x + y} \\ \text{x + y} & \text{x + 2y} & \text{x} \end{vmatrix} $
$\text{C}_{1} \rightarrow \text{C}_{1} + \text{C}_{2} + \text{C}_{3}$
$= 3\text{(x + y)} \begin{vmatrix} 1 & \text{x} + \text{y}& \text{x + 2y} \\ 1 & \text{x} & \text{x + y} \\ 1 & \text{x + 2y} & \text{x} \end{vmatrix}$
$\text{R}_{1} \rightarrow \text{R}_{1} - \text{R}_{2}, \text{R}_{3} \rightarrow \text{R}_{3} - \text{R}_{2}$
$= 3\text{(x + y)} \begin{vmatrix} 0 & \text{y} & \text{y} \\ 1 & \text{x} & \text{x + y} \\ 0 & \text{2y} & \text{-y} \end{vmatrix}$
$ = \text{-3(x + y)} \text{(-y}^{2} - \text{2y)}^{2} = \text{9y}^{2} \text{(x + y)}$
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$\text{If f(x)} = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^{2} & ax & a \end{vmatrix} , $ using properties of determinants find the value of $\text{f}(2x) - \text{f}(x).$
Answer$\text{ f(x)} = \begin{vmatrix} \text{a} & -1 & 0 \\ \text{ax} & \text{a} & -1 \\ \text{ax}^{2} & \text{ax} & \text{a} \end{vmatrix} , $
$\text{R}_{2} \rightarrow\text{R}_{2} - \text{x R}_{1}$ and $\text{R}_{3} \rightarrow\text{R}_{3} - \text{x}^{2} \text{ R}_{1}$
$\text{f (x)} = \begin{bmatrix} \text{a} & -1 & 0 \\ 0 & \text{a + x} & -1 \\ 0 & \text{ax} + \text{x}^{2} & \text{a} \end{bmatrix} \text{(For bringing 2 zeroes in any row/column}) $
$\therefore \text{f(x)} = \text{a (a}^{2} + \text{2ax + x}^{2}) = \text{a(x + a)}^{2}$
$\therefore\text{f (2x) - f(x)} = \text{a[2x + a]}^{2} - \text{a(x + a})^{2} $
$= \text{a x (3x + 2a)} $
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Using properties of determinants, prove the following:
$\begin{vmatrix} \text{x}^2+1 & \text{ xy } & \text{xz} \\ \text{xy} & \text{y}^2+1 & \text{yz} \\ \text{xz} & \text{yz} & \text{z}^2+1 \end{vmatrix} = 1+\text{x}^2+\text{y}^2+\text{z}^2.$
Answer$\text{LHS}=\frac{1}{\text{x.y.z}}\begin{vmatrix} \text{x}^3+\text{x} &\text{x}^2 \text{ y } & \text{x}^2\text{z} \\ \text{x}\text{y}^2 & \text{y}^3+\text{y} & \text{y}^2\text{z} \\ \text{x} \text{z}^2 & \text{y}\text{z}^2 & \text{z}^3+\text{z} \end{vmatrix} \begin{matrix} \text{R}_1\rightarrow\text{x} \text{R}_1,\ \ \text{R}_2\rightarrow \text{y} \text{R}_2\\ \text{R}_3\rightarrow \text{z} \text{R}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} $
$=\frac{\text{xyz}}{\text{xyz}}\begin{vmatrix} \text{x}^2+1 & \text{ x }^2 & \text{x}^2 \\ \text{y}^2 & \text{y}^2+1 & \text{y}^2 \\ \text{z}^2 & \text{z}^2 & \text{z}^2+1 \end{vmatrix}$
$=\begin{vmatrix}1+\text{x}^2+\text{y}^2+\text{z}^2 & 1+\text{x}^2+\text{y}^2+\text{z}^2 & 1+\text{x}^2+\text{y}^2+\text{z}^2 \\ \text{y}^2 & \text{y}^2+1 & \text{y}^2 \\ \text{z}^2 & \text{z}^2 & \text{z}^2+1 \end{vmatrix}\begin{matrix}\text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3 \end{matrix}$
$=\begin{vmatrix}1+\text{x}^2+\text{y}^2+\text{z}^2 & 0 & 0 \\ \text{y}^2 & 1 & 0 \\ \text{z}^2 & 0 &1 \end{vmatrix};\begin{matrix}\text{C}_2\rightarrow\text{C}_2-\text{C}_1\\ \text{C}_3 \rightarrow\text{C}_3-\text{C}_1\end{matrix}$
$= 1 + x^2 + y^2 + z^2 = RHS ($Expand along $C_1)$
View full question & answer→Question 205 Marks
Prove that $\begin{vmatrix} \text{yz -x}^{2} & \text{zx - y}^{2} & \text{xy - z}^{2} \\ \text{zx - y}^{2} & \text{xy - z}^{2} & \text{yz - x}^{2} \\ \text{xy - z}^{2} & \text{yz - x}^{2} & \text{zx - y}^{2} \end{vmatrix}$ is divisible by (x + y + z), and hence find the quotient.
Answer$\text{Using C}_{1}\rightarrow \text{C}_{1} - \text{C}_{3}$ and $\text{C}_{2} \rightarrow\text{C}_{2} - \text{C}_{3} \text{we get}$
$\triangle = \begin{vmatrix} \text{y(z - x)+z}^{2} - \text{x}^{2} & \text{x(z - y)+ z}^{2} - \text{y}^{2} & \text{xy - z}^{2} \\ \text{z(x - y)+ x}^{2} - \text{y}^{2} & \text{y(x - z)+x}^{2} - \text{z}^{2} & \text{yz - x}^{2} \\ \text{x(y - z)y}^{2} - \text{z}^{2} & \text{z(y - x)+y}^{2} - \text{x}^{2} & \text{zx - y}^{2} \end{vmatrix} $
$\text{Talking (x + y + z) common from C}_{1} \& \text{C}_{2}$
$\Rightarrow\triangle = \text{(x + y + z)}^{2} \begin{vmatrix} \text{z - x} & \text{z - y} & \text{xy - z}^{2} \\ \text{x - y} & \text{x - z} & \text{yz - x}^{2} \\ \text{y - z} & \text{y - x} & \text{zx - y}^{2} \end{vmatrix} $
$\text{R}_{1}\rightarrow\text{R}_{1} + \text{R}_{2} +\text{R}_{3}$
$\Rightarrow\triangle = \text{(x + y + z)}^{2} \begin{vmatrix} \text{0} & \text{0} & \text{xy + yz + zx - x}^{2} - \text{y}^{2} -\text{z}^{2} \\ \text{x - y} & \text{x - z} & \text{yz - x}^{2} \\ \text{y - z} & \text{y - x} & \text{zx - y}^{2} \end{vmatrix} $
Expanding to get
$\triangle = \text{(x + y + z)}^{2}\text{(xy + zy + zx - x}^{2} - \text{y}^{2} - \text{z}^{2})^{2}$
$\text{Hence}\triangle \text{is divisible by (x + y + z) and}$
$\text{the quotient is (x + y + z) (xy + yz + zx - x}^{2} - \text{y}^{2} - \text{z}^{2})^{2}$
View full question & answer→Question 215 Marks
Prove the following using properties of determinants:
$ \begin{vmatrix} \text{a + b + 2c} & \text{a} & \text{b} \\ \text{c} & \text{b + c + 2a} & \text{b} \\ \text{c} & \text{a} & \text{c + a + 2b} \end{vmatrix}= 2(\text{a + b + c})^3 $
Answer$ \text{LHS}=\begin{vmatrix} 2(\text{a + b + c}) & \text{a} & \text{b} \\ 2(\text{a + b + c}) & \text{b + c + 2a} & \text{b} \\ 2(\text{a + b + c}) & \text{a} & \text{c + a + 2b} \end{vmatrix}\ \begin{matrix} \text{Using}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3 \\ \end{matrix} $ $ \text{LHS}=\begin{vmatrix} 2(\text{a + b + c}) & \text{a} & \text{b} \\ 0 & \text{a + b + c} & 0 \\ 0 & 0 & \text{a + b + c} \end{vmatrix}\ \begin{matrix} \text{Using}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \text{R}_2\rightarrow\text{R}_2-\text{R}_1;\ \text{R}_3\rightarrow\text{R}_3-\text{R}_1 \\ \end{matrix} $
$= 2 (a + b + c) \{(a + b + c)^2 – 0\}$ Expanding along $C_1$
$= 2 (a + b + c)^3 = RHS$
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Using properties of determinants, prove the following:
$\begin{vmatrix} 1 & \text{x} & \text{x}^{2} \\ \text{x}^{2} & 1 & \text{x} \\ \text{x} & \text{x}^{2} & 1 \end{vmatrix} = (1 - \text{x}^{3})^{2}.$
AnswerLet $ | A| = \begin{vmatrix} 1 & \text{x} & \text{x}^{2} \\ \text{x}^{2} & 1 & \text{x} \\ \text{x} & \text{x}^{2} & 1 \end{vmatrix} $
Apply $C_1 \rightarrow C_1 + C_2 + C_3$
$ |\text{A}|= \begin{vmatrix} 1 +& \text{x} + \text{x}^{2} &\text{x} &\text{x}^{2}\\ 1 +& \text{x } + \text{x}^{2} & 1 & \text{x} \\ 1 +& \text{x} + \text{x}^{2} & \text{x}^{2}& 1 \end{vmatrix} $
$\Rightarrow |\text{A}|=(1 + \text{x} + \text{x}^{2}) \begin{vmatrix} 1 & \text{x} & \text{x}^{2} \\ 1 & 1 & \text{x} \\ 1 & \text{x}^{2} &1 \end{vmatrix} $
$\text{Apply R}_{2}\rightarrow\text{R}_{2}-\text{ R}_{1},\text{ R}_{3}\rightarrow\text{R}_{3}- \text{R}_{1}$
$\Rightarrow |\text{A}|=(1 + \text{x} + \text{x}^{2}) \begin{vmatrix} 1 & \text{x} & \text{x}^{2} \\ 0 & 1 - \text{x} & \text{x} - \text{x}^{2} \\ 0 & \text{x}^{2} - \text{x} &1 - \text{x}^{2} \end{vmatrix} $
Take $(1 - x)$ common from $R_2$ and $R_3$
$ |\text{A}|=(1 + \text{x} + \text{x}^{2})(1 - \text{x})^{2} \begin{vmatrix} 1 & \text{x} & \text{x}^{2} \\ 0 & 1 & \text{x} \\ 0 & - \text{x} &1 + \text{x} \end{vmatrix} $
Expanding along $C_1$
$|A|= (1 + x + x^2 )(1 - x)^2 (1 + x + x^2)$
$= (1 - x^3)^2$
$[\because 1 - x^3 = (1 - x)(1 + x + x^2)].$
View full question & answer→Question 235 Marks
Using properties of determinants, prove that $\begin{vmatrix} -\text{a}^{2} & \text{ab} & \text{ac} \\ \text{ba} & -\text{b}^{2} & \text{bc} \\ \text{ca} & \text{cb} & -\text{c}^{2} \end{vmatrix}=\text{4a}^{2}\text{b}^{2}\text{c}^{2} $.
AnswerTaking $a, b, c$ respectively common from $R_1, R_2, R_3$ to get
$LHS =$ Determinant $= abc \begin{vmatrix} \text{-a} & \text{b} & \text{c} \\ \text{a} & \text{-b} & \text{c} \\ \text{a} & \text{b} & \text{-c} \end{vmatrix}$
Taking $a, b, c,$ respectively common from $c_1, c_2, c_3$ to get
$LHS = a^2b^2c^2 \begin{vmatrix} \text{-1} & \text{1} & \text{1} \\ \text{1} & \text{-1} & \text{1} \\ \text{1} & \text{1} & \text{-1} \end{vmatrix}$
Applying $R_{2 }\rightarrow R_2+R_1, R_3 \rightarrow R_3+ R_1,$ to get
$LHS = a^2b^2c^2 \begin{vmatrix} \text{-1} & \text{1} & \text{1} \\ \text{0} & \text{0} & \text{2} \\ \text{0} & \text{2} & \text{0} \end{vmatrix}$
$= a^2b^2c^{2}(-1) (-4) = 4 a^2b^2c^2= RHS.$
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Using properties of determinants, show the following: $\begin{vmatrix} (\text{b}+\text{c})^2& \text{ab} & \text{ca} \\ \text{ab} & (\text{b}+\text{c})^2 & \text{bc} \\ \text{ac} & \text{bc} & \text{(a+b)}^2 \end{vmatrix} =2\text{abc}\ (\text{a+b+c})^3\dot{}$
Answer$\triangle=\begin{vmatrix} (\text{b}+\text{c})^2& \text{ab} & \text{ca} \\ \text{ab} & (\text{b}+\text{c})^2 & \text{bc} \\ \text{ac} & \text{bc} & \text{(a+b)}^2 \end{vmatrix} ; \text{Applying R}_1\rightarrow\text{a R}_1,\text{ R}_2\rightarrow\text{b R}_2,\text{ R}_3\rightarrow\text{c R}_3 \text{ we get}$$=\frac{1}{\text{a b c}}\ \begin{vmatrix} \text{a}(\text{b}+\text{c})^2& \text{b a}^2 & \text{c a}^2 \\ \text{a b}^2 & \text{b}(\text{c}+\text{a})^2 & \text{c b}^2 \\ \text{a c}^2 & \text{b c}^2 &\text{c} \text{(a+b)}^2 \end{vmatrix} $
$=\begin{vmatrix} (\text{b}+\text{c})^2& \text{a}^2 & \text{a}^2 \\ \text{b}^2 & (\text{c}+\text{a})^2 & \text{b}^2 \\ \text{c}^2 & \text{c}^2 & \text{(a+b)}^2 \end{vmatrix} \\ \text{Applying}\text{ C}_1\rightarrow\text{ C}_1-\text{ C}_3,\ \text{C}_2\rightarrow\text{ C}_2-\text{ C}_3, \text{we get}$
$=\triangle=\begin{vmatrix} (\text{b}+\text{c})^2-\text{a}^2& \text{0} & \text{a}^2 \\ \text{0} & (\text{c}+\text{a})^2-\text{b}^2 & \text{b}^2 \\ \text{c}^2-(\text{a+b})^2 & \text{c}^2-\text{(a+b)}^2 & \text{(a+b)}^2 \end{vmatrix}$
$=\text{(a+b+c)}^2\begin{vmatrix} \text{b}+\text{c}-\text{a}& \text{0} & \text{a}^2 \\ \text{0} & \text{c}+\text{a}-\text{b} & \text{b}^2 \\ \text{c }-\text{a} - \text{b} & \text{c}-\text{a}- \text{b}& \text{(a+b)}^2 \end{vmatrix}$
$\text{Applying}\text{ R}_3\rightarrow\text{ R}_3-\text{ (R}_1+\text{R}_2), \text{ we get}$
$ \triangle=\text{(a+b+c)}^2\begin{vmatrix} \text{b}+\text{c}-\text{a}& \text{0} & \text{a}^2 \\ \text{0} & \text{c}+\text{a}-\text{b} & \text{b}^2 \\ -\text{ 2 b}&- \text{ 2 a}& 2\text{ ab} \end{vmatrix}$
$\text{Applying}\text{ C}_1\rightarrow\text{ aC}_1\ \text{and}\text{ C}_2\rightarrow\text{ bC}_2 \text{we get}$
$\triangle=\frac{\text{(a+b+c)}}{\text{a b}}\begin{vmatrix} \text{a b}+\text{a c}-\text{a}^2& \text{0}&\text{a}^2\\ \text{0} & \text{b }(\text{c + a}-\text{b})&\text{b}^2\\ -\text{ 2 b a}&- \text{ 2 a b}&2\text{a b} \end{vmatrix}$
$\text{Applying}\text{ C}_1\rightarrow\text{ C}_1+\text{ C}_3,\text{ C}_2\rightarrow\text{C}_2\ +\ \text{C}_3\ \text{we get}$
$\triangle=\frac{\text{(a+b+c)}^3}{\text{a b}}\begin{vmatrix} \text{a }\text{(b+c)}&\text{a}^2&\text{a}^2\\ \text{b}^2 & \text{b }(\text{a + c})&\text{b}^2\\ \text{0}& \text{0}&\text{2 a b} \end{vmatrix}$
$=\frac{\text{(a+b+c)}}{\text{a b}}\times\text {a b}\times\text {2a b}\begin{vmatrix} \text{b+c}&\text{a}&\text{a}\\ \text{b} & \text{c + a}&\text{b}\\ \text{0}& \text{0}&\text{1} \end{vmatrix}$
$= 2 \text{ab } \text{(a + b + c)}^2 \ [\text{(b + c) (c + a) – a b}]$
$= 2 \text{ ab (a + b + c)}^2 \ \text{[bc + c}^2\text{ + a b + a c – a b]}$
$\text{= 2 abc (a + b + c)}^3$$\dot{}$
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By using properties of determinants, prove the following:$\begin{vmatrix} \text {x + 4} & \text{2x} & \text{2x} \\ \text{ 2x} & \text{x + 4} & \text{2x} \\ \text{2x} & \text{2x} & \text{x + 4} \end{vmatrix} = ( 5\text{x} + 4) (4 -\text{x}) = 1. $
Answer$C_{1} \rightarrow C_{1} + C_{2} + C_{3} \text{gives Det} = (5\text{x} + 4) \begin{vmatrix} 1 & 2\text{x} & 2\text{x} \\ 1 & \text{x} + 4 & 2\text{x} \\ 1 & 2\text{x} & \text{x} + 4 \end{vmatrix} $$\begin{matrix} R_{2} \rightarrow & R_{2} - & R_{1} \\ R_{3} \rightarrow & R_{3} & R_{1} \\ \end{matrix} \Rightarrow \text{Det} = (5\text{x} + 4) \begin{vmatrix} 1 & 2\text{x} & 2\text{x} \\ 0 & 4 -\text{x} & 0 \\ 0 & 0 & 4 - \text{x} \end{vmatrix} $
$\text{ Expanding by C}_{1} \text{to get Det.} = ( 5\text{x} + 4) (4- \text{x})^{2}$
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$\text{If A} = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}, $ verify that $\text{A}^{2} - \text{4A - 5I = 0}$
Answer$A^{2} = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} $$\text{4 A} = \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} , 5 I = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} $
$\therefore \text{A}^{2} - \text{4 A - 5I} = \begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0& 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix} =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0$
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Using properties of determinants, prove the following:$\begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{vmatrix} = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)( \alpha + \beta + \gamma) $
Answer$\text{LHS:} \triangle = \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{vmatrix}$$\text{R}_{3} \rightarrow \text{R}_{3} + \text{R}_{1} = \triangle = \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \alpha +\beta+ \gamma &\alpha +\beta+ \gamma & \alpha +\beta+ \gamma \end{vmatrix}$
$= (\alpha + \beta + r) \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ 1 & 1 & 1 \end{vmatrix} $
$\text{Applying C}_{1} \rightarrow \text{C}_{1} -\text{C}_{2}$ $\text{and C}_{2} \rightarrow \text{C}_{2} - \text{C}_{3}$
$\triangle = ( \alpha + \beta + \gamma) \begin{vmatrix} \alpha- \beta & \beta - \gamma & \gamma \\ \alpha^{2} - \beta^{2} & \beta^{2} -\gamma^{2} & \gamma^{2} \\ 0 & 0 & 1 \end{vmatrix} $
Expanding by last row to get
$\triangle = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)( \alpha + \beta + \gamma) = \text{RHS} $
View full question & answer→Question 285 Marks
Using properties of determinants, prove that,
$\begin{vmatrix}1& 1&1+3\text{x} \\1+3\text{y} & 1&1\\1&1+3\text{z}&1 \end{vmatrix}=9(3\text{xyz}+\text{xy}+\text{yz}+\text{zx})$
Answer$\begin{vmatrix}1& 1&1+3\text{x} \\1+3\text{y} & 1&1\\1&1+3\text{z}&1 \end{vmatrix}$
$\text{xyz}\begin{vmatrix}\frac{1}{\text{x}}& \frac{1}{\text{x}}&\frac{1}{\text{x}}+3 \\\frac{1}{\text{y}}+3 & \frac{1}{\text{y}}&\frac{1}{\text{y}}\\\frac{1}{\text{z}}&\frac{1}{\text{z}}+3&\frac{1}{\text{z}} \end{vmatrix}$
$\text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3$
$=\text{xyz}\begin{vmatrix}\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3&\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3&\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3 \\\frac{1}{\text{y}}+3 & \frac{1}{\text{y}}&\frac{1}{\text{y}}\\\frac{1}{\text{z}}&\frac{1}{\text{z}}+3&\frac{1}{\text{z}} \end{vmatrix}$
$=(\text{xyz})\bigg(\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3\bigg)\begin{vmatrix}1 & 1&1 \\\frac{1}{\text{y}}+3 & \frac{1}{\text{y}}&\frac{1}{\text{y}}\\\frac{1}{\text{z}}&\frac{1}{\text{z}}+3&\frac{1}{\text{z}} \end{vmatrix}$
$\text{C}_2\rightarrow\text{C}_2-\text{C}_1\ \&\ \text{C}_3\rightarrow\text{C}_3-\text{C}_1$
$=\text{xyz}\bigg(\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}+3\bigg)\begin{vmatrix}1 & 0&0 \\\frac{1}{\text{y}}+3 & -3&-3\\\frac{1}{\text{z}}&3&0 \end{vmatrix}$
$=(\text{yz}+\text{zx}+\text{xy}+3\text{xyz})(0+9)$
$=9(3\text{xyz}+\text{xy}+\text{yz}+\text{zx})=\text{R.H.S.}$
View full question & answer→Question 295 Marks
Using properties of determinants, prove the following:
$\begin{vmatrix}\text{a} & \text{b} & \text{c} \\\text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a}\\\text{b}+\text{c} & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}.$
Answer$\text{LHS}:$ Let $\triangle=\begin{vmatrix}\text{a} & \text{b} & \text{c} \\\text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a}\\\text{b}+\text{c} & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}$
Apply $C_1 \rightarrow C_1 + C_2 + C_3$
$=\begin{vmatrix}\text{a}+\text{b}+\text{c} & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\2(\text{a}+\text{b}+\text{c}) & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}$
Taking $(a + b + c)$ common from $C_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1 & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\2 & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}$
Apply: $R_3 \rightarrow R_3 - 2R_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1 & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\0 & \text{c}+\text{a} -\text{2b}& \text{a}+\text{b} -2\text{c}\end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})[(\text{b}-\text{c})(\text{a}+\text{b}-2\text{c})-(\text{c}-\text{a})(\text{c}+\text{a}-2\text{b})]$
$=\text{a}^3+\text{b}^3+\text{c}^{3}-3\text{abc}$
$=\text{RHS}$
View full question & answer→Question 305 Marks
If $\text{A}=\begin{bmatrix}1 & 1 & 1 \\1 & 0 & 2\\3 & 1 & 1 \end{bmatrix},$ find $A^{–1}.$ Hence, solve the system of equations $x + y + z = 6, x + 2z = 7, 3x + y + z = 12.$
AnswerGiven Matrix $\text{A}=\begin{bmatrix}1 & 1 & 1 \\1 & 0 & 2\\3 & 1 & 1 \end{bmatrix}$
To find $A^{-1},$ we need cofactors of each element of matrix $A.$
Cofactor of $\text{a}_{11}=(-1)^{1+1}\begin{vmatrix}0 & 2 \\1 & 1 \end{vmatrix}=-2$
Cofactor of $\text{a}_{12}=(-1)^{1+2}\begin{vmatrix}1 & 2 \\3 & 1 \end{vmatrix}=-(1-6)=5$
Cofactor of $\text{a}_{13}=(-1)^{1+3}\begin{vmatrix}1 & 0 \\3 & 1 \end{vmatrix}=1$
Cofactor of $\text{a}_{21}=(-1)^{2+1}\begin{vmatrix}1 & 1 \\1 & 1 \end{vmatrix}=0$
Cofactor of $\text{a}_{22}=(-1)^{2+2}\begin{vmatrix}1 & 1 \\3 & 1 \end{vmatrix}=(1-3)=-2$
Cofactor of $\text{a}_{23}=(-1)^{2+3}\begin{vmatrix}1 & 1 \\3 & 1 \end{vmatrix}=-(1-3)=2$
Cofactor of $\text{a}_{31}=(-1)^{3+1}\begin{vmatrix}1 & 1 \\0 & 2 \end{vmatrix}=2$
Cofactor of $\text{a}_{32}=(-1)^{3+2}\begin{vmatrix}1 & 1 \\1 & 2 \end{vmatrix}=-(2-1)=-1$
Cofactor of $\text{a}_{33}=(-1)^{3+3}\begin{vmatrix}1 & 1 \\1 & 0 \end{vmatrix}=-1$
So, Cofactor of matrix of $\text{A}=\begin{bmatrix}-2 & 5 & 1 \\0 & -2 & 2\\2 & -1 & -1 \end{bmatrix}$
$\therefore\ $the trnspose of cofactor matrix $A$ is $adj\ (A)$
So adj $\text{A}=\begin{bmatrix}-2 & 0 & 2 \\5 & -2 & -1\\1 & 2 & -1 \end{bmatrix}$
Now, the given system of equation is
$x + y + z = 6$
$x + 2z = 7$
$3x + y + z = 12$
Writing the above equation in matrix form
$\begin{bmatrix}1 & 1 & 1 \\1 & 0 & 2\\3 & 1 & 1 \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}6\\7\\12\end{bmatrix}$
$\text{A}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}6\\7\\12\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\text{A}^{-1}\begin{bmatrix}6\\7\\12\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{4}\begin{bmatrix}-2 & 0 & 2 \\5 & -2 & -1\\1 & 2 & -1 \end{bmatrix}\begin{bmatrix}6\\7\\12\end{bmatrix}$
or, $\frac{1}{4}\begin{bmatrix}-12+0+24\\30-14-12\\6+14-12\end{bmatrix}$
or, $\frac{1}{4}\begin{bmatrix}12\\4\\8\end{bmatrix}$
or, $\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3\\1\\2\end{bmatrix}$
i.e $x = 3, y = 1, z = 2$
View full question & answer→Question 315 Marks
Find: $\int\frac{3\text{x}+5}{\text{x}^2+3\text{x}-18}\text{ dx}.$
Answer$\text{l}\int\frac{3\text{x}+5}{\text{x}^2+3\text{x}-18}\text{ dx}$
$=\int\frac{\frac{3}{2}(2\text{x}+3)-\frac{9}{2}+5}{\text{x}^2+3\text{x}-18}\text{ dx}$
$=\frac{3}{2}\int\frac{2\text{x}+3}{\text{x}^2+3\text{x}-18}\text{dx}+\int\frac{\frac{1}{2}}{\text{x}^2+3\text{x}-18}$
$=\frac{3}{2}\int\frac{\frac{\text{d}}{\text{dx}}(\text{x}^2+3\text{x}-18)}{\text{x}^2+3\text{x}-18}\text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\Big(\text{x}+\frac{3}{2}\Big)^2-18-\frac{9}{4}}$
$=\frac{3}{2}\text{ln}|\text{x}^2+3\text{x}-18|+\frac{1}{2}\int\frac{\text{dx}}{\Big(\text{x}+\frac{3}{2}\Big)^2-\big(\frac{9}{2}\big)^2}$
$=\frac{3}{2}\text{ln}|\text{x}^2+3\text{x}-18|+\frac{1}{18}\log\Big|\frac{\text{x}-3}{\text{x}+6}\Big|+\text{C}$
View full question & answer→Question 325 Marks
Obtain the inverse of the following matrix using elementary operations: $\text{A}=\begin{bmatrix}-1 & 1 & 2\\1 & 2 & 3\\3 & 1 & 1 \end{bmatrix}$
Answer$\therefore\text{A}^{-1}=\text{IA}$
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}-1 & 1 & 2\\1 & 2 & 3\\3 & 1 & 1 \end{bmatrix}$
$\text{R}_1\rightarrow-\text{R}_1$
$\text{A}^{-1}=\begin{bmatrix}-1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & -1 & -2\\1 & 2 & 3\\3 & 1 & 1 \end{bmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\text{R}_1$
$\text{R}_3\rightarrow\text{R}_3-3\text{R}_1$
$\text{A}^{-1}=\begin{bmatrix}-1 & 0 & 0\\1 & 1 & 0\\3 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & -1 & -2\\0 & 3 & 5\\0 & 4 & 7 \end{bmatrix}$
$\text{R}_2\rightarrow-\text{R}_2+\text{R}_3$
$\text{A}^{-1}=\begin{bmatrix}-1 & 0 & 0\\2 & -1 & 1\\3 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & -1 & -2\\0 & 1 & 2\\0 & 4 & 7 \end{bmatrix}$
$\text{R}_1\rightarrow\text{R}_1+\text{R}_2$
$\text{R}_3\rightarrow\text{R}_3-4\text{R}_2$
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 2\\0 & 0 & -1 \end{bmatrix}\begin{bmatrix}1 & -1 & 1\\2 & -1 & 1\\-5 & 4 & -3 \end{bmatrix}$
$\text{R}_3\rightarrow-\text{R}_3$
$\text{A}^{-1}=\begin{bmatrix}1 & -1 & 1\\2 & -1 & 1\\5 & -4 & 3 \end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 2\\0 & 0 & 1 \end{bmatrix}$
$\text{R}_2\rightarrow\text{R}_2-2\text{R}_3$
$\text{A}^{-1}=\begin{bmatrix}1 & -1 & 1\\-8 & 7 & -5\\5 & -4 & 3 \end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$
Here $\text{I}=\begin{bmatrix}1 & -1 & 1\\-8 & 7 & -5\\5 & -4 & 3 \end{bmatrix}$
So, $\text{A}^{-1}==\begin{bmatrix}1 & -1 & 1\\-8 & 7 & -5\\5 & -4 & 3 \end{bmatrix}$
View full question & answer→Question 335 Marks
If $\text{A}=\begin{bmatrix}2 & -3 & 5 \\3 & 2 & -4\\1 & 1 & -2 \end{bmatrix},$ then find $A^{–1}.$ Hence solve the following system of equations: $2x - 3y + 5z = 11, 3x + 2y - 4z = -5, x + y - 2z = -3.$
Answer$\text{A}=\begin{bmatrix}2 & -3 & 5 \\3 & 2 & -4\\1 & 1 & -2 \end{bmatrix}$
$\Rightarrow |A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = -1$
$\text{Adj}(\text{A})=\begin{bmatrix}0 & 2 & 1 \\-1 & -9 & -5\\2 & 23 & 13 \end{bmatrix}^{\text{T}}=\begin{bmatrix}0 & -1 & 2 \\2 & -9 & 23\\1 & 5 & 13 \end{bmatrix}$
$\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}(\text{Adj }\text{A})=\begin{bmatrix}0 & 1 & -2 \\-2 & 9 & -23\\-1 & 5 & -13 \end{bmatrix}$
Given system of equations is
$2x - 3y + 5z = 11$
$3x + 2y - 4z = -5$
$x + y - 2z = -3$
$\Rightarrow\begin{bmatrix}2 & -3 & 5 \\3 & 2 & -4\\1 & 1 & -2 \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$\Rightarrow\text{A}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\text{A}^{-1}\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$=\begin{bmatrix}0 & 1 & -2 \\-2 & 9 & -23\\-1 & 5 & -13 \end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$
$\therefore x = 1, y = 2$ and $z = 3$ is the solution the given system of equations.
View full question & answer→Question 345 Marks
Using properties of determinants, prove that $\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\2\text{a}+1 & \text{a}+2 & 1\\3 & 3 & 1 \end{vmatrix}=(\text{a}-1)^3.$
Answer$\text{LHS}=\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\2\text{a}+1 & \text{a}+2 & 1\\3 & 3 & 1 \end{vmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\text{R}_1,\text{R}_3\rightarrow\text{R}_3-\text{R}_1$
$=\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\1-\text{a}^2 & -\text{a}+1 & 0\\3-\text{a}^2-2\text{a} & 3-2\text{a}-1 & 0 \end{vmatrix}$
$=\begin{vmatrix}\text{a}^2+2\text{a} & 2\text{a}+1 & 1 \\1-\text{a}^2 & 1-\text{a} & 0\\3-\text{a}^2-2\text{a} & 2-2\text{a} & 0 \end{vmatrix}$
Expanding along $C_3$
$=1\big[(1-\text{a}^2)(2-2\text{a})-(1-\text{a})(3-\text{a}^2-2\text{a})\big]$
$=2(1-\text{a})(1-\text{a})(1+\text{a})-(1-\text{a})(3-\text{a}^2-2\text{a})$
$=(1-\text{a})\big[2(1-\text{a}^2)-3+\text{a}^2+2\text{a}\big]$
$=(1-\text{a})(2\text{a}-\text{a}^2-1)$
$=(\text{a}-1)^3$
$=\text{RHS}$
View full question & answer→Question 355 Marks
Find the inverse of the following matrix using elementary operations. $\text{A}=\begin{bmatrix}1 & 2 & -2\\-1 & 3 & 0\\0 & -2 & 1 \end{bmatrix}$
AnswerWe know that
$\text{A}^{-1} = \text{I}^{}\text{A}$
Or, $\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\-1 & 3 & 0\\0 & -2 & 1 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\0 & 5 & -2\\0 & -2 & 1 \end{bmatrix}\ [$Applying $R_2 \rightarrow R_2 \rightarrow R_1]$
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 2\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\0 & 1 & 0\\0 & -2 & 1 \end{bmatrix}\ [$Applying $R_2 \rightarrow R_2 \rightarrow R_3]$
$\text{A}^{-1}=\begin{bmatrix}-1 & -2 & -4\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & -2\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\ [$Applying $R_1 \rightarrow R_1 + (-2)R_2, R_3 \rightarrow R_3 + 2R_2]$
$\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\ [$Applying $R_2 \rightarrow R_2 \rightarrow R_3]$
Hence, $\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}$
View full question & answer→Question 365 Marks
Use product $\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$ to solve the system of equations $x + 3z = 9, -x + 2y - 2z = 4, 2x - 3y + 4z = -3.$
AnswerSuppose, $\text{A}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\text{B}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$\text{A}\times\text{B}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
$=\begin{bmatrix}-2-9+12&0-2+2&1+3-4\\0+18-18&0+4-3&0-6+6\\-6-18+24&0-4+4&3+6-8\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Since, $A \times B = I,$
$\therefore B = A^{-1 }.....(1)$
Now, the given system of equations is
$x + 3z = 9$
$-x + 2y - 2z = 4$
$2x - 3y + 4z = -3$
This can also be presented as:
$\begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9\\4\\-3\end{bmatrix}$
Here, we can observe that $\begin{bmatrix}1&0&3\\-1&2&-2\\2&-3&4\end{bmatrix}=\text{A}^\text{T}$
So, $\text{A}^\text{T}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9\\4\\-3\end{bmatrix}$
Multiply the above expression by $(A^T)^{-1}.$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^\text{T})^{-1}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^{-1})^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\text{B}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$ $[$ Using $(1)]$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-2&9&6\\0&2&1\\1&-3&-2\end{bmatrix}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-18+36-18\\0+8-3\\9-12+6\end{bmatrix}$
$=\begin{bmatrix}0\\5\\3\end{bmatrix}$
Hence, $x = 0, y = 5$ and $z = 3.$
View full question & answer→Question 375 Marks
Solve the following system of equations by matrix method:$\frac{2}{\text{x}}+\frac{3}{\text{y}}+\frac{10}{\text{z}}=4,\frac{4}{\text{x}}-\frac{6}{\text{y}}+\frac{5}{\text{z}}=1,\frac{6}{\text{x}}+\frac{9}{\text{y}}-\frac{20}{\text{z}}=2:\text{x},\text{y},\text{z}\neq0$
AnswerLet
$\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
The above system can be written as:
$\begin{bmatrix}2&3&10\\ 4&-6&5\\ 6&9&-20\end{bmatrix}\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\begin{bmatrix}4\\ 1\\ 2\end{bmatrix}$
Or $AX = B$
$\text{|A|}=2{(75)}-3{(-110)}+10{(72)}=1200\neq0$
So, the above system has a unique solution, given by
$\text{X}=\text{A}^{-1}\text{B}$
Let $C_{ij}$ be the co-factors of $a_{ij}$ in $A$
$\text{C}_{11}=75\\ \text{C}_{21}=150\\ \text{C}_{31}=75$
$\text{C}_{12}=110\\ \text{C}_{22}=-100\\ \text{C}_{32}=30$
$\text{C}_{13}=72\\ \text{C}_{23}=0\\ \text{C}_{33}=-24$
$\text{adj A}=\begin{bmatrix}75&110&72\\ 150&-100&0\\ 75&30&-24\end{bmatrix}^\text{T}=\begin{bmatrix}75&150&75\\ 110&-100&30\\ 72&0&-24\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{B}$
$=\frac{1}{1200}\begin{bmatrix}75&150&75\\ 110&-100&30\\ 72&0&-24\end{bmatrix}\begin{bmatrix}4\\ 1\\ 2\end{bmatrix}$
$\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\frac{1}{1200}\begin{bmatrix}600\\ 400\\ 240\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{bmatrix}$
Hence, $x = 2, y = 3, z = 5$
View full question & answer→Question 385 Marks
An amount of $Rs. 10,000$ is put into three investments at the rate of $10, 12$ and $15\%$ per annum. The combined income is $Rs. 1310$ and the combined income of first and second investment is $Rs. 190$ short of the income from the third. Find the investment in each using matrix method.
AnswerLet the three investments are $x, y, z$
$\text{x}+\text{y}+\text{z}=10,000\ \dots(1)$
Also
$\frac{10}{100}\text{x}+\frac{12}{100}\text{y}+\frac{15}{100}\text{z}=1310$
$0.1\text{x}+0.12\text{y}+0.15\text{z}=1310\ \dots(2)$
Also
$\frac{10}{100}\text{x}+\frac{12}{100}\text{y}=\frac{15}{100}\text{z}-190$
$0.1\text{x}+0.12\text{y}+0.15\text{z}=-190\ \dots(3)$
The above system can be written as:
$\begin{bmatrix}1&1&1\\0.1&0.12&0.15\\0.1&0.12&-0.15\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}10000\\1310\\-190\end{bmatrix}$
Or $\text{AX}=\text{B}$
$|\text{A}|=1(-0.036)-1(-0.03)+1(0)=-0.006\neq0$
So, the above system has a unique solution, given by
$X = A^{-1}B$
Let $C_{ij}$ be the co-factor of $a_{ij}$ in $A$
$\text{C}_{11}=-0.036,\text{C}_{12}=0.03,\text{C}_{13}=0$
$\text{C}_{21}=0.27,\text{C}_{22}=-0.25,\text{C}_{23}=-0.02$
$\text{C}_{31}=0.03,\text{C}_{32}=-0.05,\text{C}_{33}=0.02$
$\text{adj}\text{A}=\begin{bmatrix}-0.036&0.03&0\\0.27&-0.25&-0.02\\0.03&-0.05&0.02\end{bmatrix}^\text{T}=\begin{bmatrix}-0.036&0.27&0.03\\0.03&-0.25&-0.05\\0&-0.02&0.02\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{|\text{A}|}(\text{Adj}\ \text{A})\times\text{B}$
$=\frac{1}{-0.006}\begin{bmatrix}-0.036&0.27&0.03\\0.03&-0.25&-0.05\\0&-0.02&0.02\end{bmatrix}\begin{bmatrix}10000\\1310\\-190\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{-0.006}\begin{bmatrix}-12\\-18\\-30\end{bmatrix}=\begin{bmatrix}2000\\3000\\5000\end{bmatrix}$
Hence, $x = Rs. 2000, y = Rs. 3000, z = Rs. 5000$
View full question & answer→Question 395 Marks
For the matrix $\text{A}=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}$Show that $\ce{A^3 - 6A^2 + 5A + 11I = 0.}$ Hence, find $A^{-1}.$
Answer$\text{Given:}\ \text{A}=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}\ \therefore\ \text{A}^2=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}$
$\Rightarrow\ \text{A}^2=\begin{bmatrix}1+1+2&1+2-1&1-3+3\\1+2-6&1+4+3&1-6-9\\2-1+6&2-2-3&2+3+9\end{bmatrix}=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}$
$\text{Now}\ \text{A}^3=\text{A}^2\text{A}=\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}$
$=\begin{bmatrix}4+2+2&4+4-1&4-6+3\\-3+8-28&-3+16+14&-3-24-42\\7-3+28&7-6-14&7+9+42\end{bmatrix}=\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}$
$\text{L.H.S.}=\text{A}^3-6\text{A}^2+5\text{A}+11\text{I}$
$=\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}-6\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}+5\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}+11\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}8&7&1\\-23&27&-69\\32&-13&58\end{bmatrix}-\begin{bmatrix}24&12&6\\-18&48&-84\\42&-18&84\end{bmatrix}+\begin{bmatrix}5&5&5\\5&10&-15\\10&-5&15\end{bmatrix}+\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}$
$=\begin{bmatrix}8-24+5&7-12+5&1-6+5\\-23+18+5&27-48+10&-69+84-15\\32-42+10&-13+18-5&58-84+15\end{bmatrix}+\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}$
$=\begin{bmatrix}-11&0&0\\0&-11&0\\0&0&-11\end{bmatrix}+\begin{bmatrix}11&0&0\\0&11&0\\0&0&11\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0=\text{R.H.S.}$
Now, to find $A^{-1}$, multiplying $\ce{A^3 - 6A^2 + 5A + 11I = 0}$ by $A^{-1}$
$\Rightarrow \ce{A^3A^{-1} - 6A^2A^{-1} + 5AA^{-1} + 11IA^{-1} = 0.A^{-1}}$
$\Rightarrow \ce{A^2 - 6A + 5I + 11A^{-1} = 0 }$
$\Rightarrow \ce{11A^{-1} = 6A - 5I - A^2}$
$\Rightarrow\ \ 11\text{A}^{-1}=6\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}-\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}$
$\Rightarrow\ \ 11\text{A}^{-1}=\begin{bmatrix}6&6&6\\6&12&-18\\12&-6&18\end{bmatrix}-\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}-\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}$
$\Rightarrow \ 11\text{A}^{-1}\begin{bmatrix}6-5-4&6-2&6-1\\6+3&12-5-8&-18+14\\12-7&-6+3&18-5-14\end{bmatrix}=\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}$
View full question & answer→Question 405 Marks
The sum of three numbers is $2.$ If twice the second number is added to the sum of first and third, the sum is $1.$ By adding second and third number to five times the first number, we get $6.$ Find the three numbers by using matrices.
AnswerLet the three number be $x, y$ and $a$
According to the question,
$x + y + 2$
$x + 2y + z = 1$
$5x + y + z = 6$
The given system of equations can be written in matrix form as follows:
$\begin{bmatrix}1&1&1\\1&2&1\\5&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2\\1\\6\end{bmatrix}$
$AX = B$
Here,
$\text{A}=\begin{bmatrix}1&1&1\\1&2&1\\5&1&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\1\\6\end{bmatrix}$
$|\text{A}|=1(2-1)-1(1-5)+1(1-10)$
$=1+4-9$
$=-4$
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}].$ Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}2&1\\1&1\end{vmatrix}=1,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}1&1\\5&1\end{vmatrix}=4,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}1&2\\5&1\end{vmatrix}=-9$
$\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\5&1\end{vmatrix}=-4,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\5&1\end{vmatrix}=4$
$\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\1&1\end{vmatrix}=0,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\1&2\end{vmatrix}=1$
$\text{adj }\text{A}=\begin{bmatrix}1&4&-9\\0&-4&4\\-1&0&1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$=\frac{1}{-4}\begin{bmatrix}1&0&-1\\4&-4&0\\-9&4&1\end{bmatrix}$
View full question & answer→Question 415 Marks
Solve the following system of equations by matrix method : $x - y + 2z = 7$
$3x + 4y - 5z = -5$
$2x - y + 3z = 12$
AnswerThe above system can be written as:
$\begin{bmatrix}1&-1&2\\ 3&4&-5\\ 2&-1&3\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}7\\ -5\\ 12\end{bmatrix}$
Or $AX = B$
$|\text{A}| = 1 (7) + 1 (19) + 2 (-11) = 4\neq0$
So, the above system has a unique solution, given by
$\text{X}=\text{A}^{-1}\text{B}$
Let $C_{ij}$ be the co$-$factors of $a_{ij}$ in $A$
$\text{C}_{11}=7\\ \text{C}_{21}=1\\ \text{C}_{31}=-3$
$\text{C}_{12}=-19\\ \text{C}_{22}=-1\\ \text{C}_{32}=11$
$\text{C}_{13}=-11\\ \text{C}_{23}=-1\\ \text{C}_{33}=7$
$\text{adj A}=\begin{bmatrix}7&-19&-11\\ 1&-1&-1\\ -3&11&7\end{bmatrix}^\text{T}=\begin{bmatrix}7&1&-3\\ -19&-1&11\\ -11&-1&7\end{bmatrix}$
Now,
$\text{X}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{B}$
$=\frac{1}{4}\begin{bmatrix}7&1&-3\\ -19&-1&11\\ -11&-1&7\end{bmatrix}\begin{bmatrix}7\\ -5\\ 12\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{4}\begin{bmatrix}8\\ 4\\ 12\end{bmatrix}=\begin{bmatrix}2\\ 1\\ 3\end{bmatrix}$
Hence$, x = 2, y = 1, z = 3$
View full question & answer→Question 425 Marks
If A + B + C = 0, then prove that $\begin{vmatrix}1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&1\end{vmatrix}=0.$
Answer$\text{L.H.S}=\begin{vmatrix}1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&1\end{vmatrix}$
$=(1-\cos^2\text{A})-\cos\text{C}(\cos\text{C}-\cos\text{A}.\cos\text{B})+\cos\text{B}(\cos\text{C}.\cos\text{A}-\cos\text{B})$
$=\sin^2\text{A}-\cos^2\text{C}+\cos\text{A}.\cos\text{B}.\cos\text{C}+\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{B}$
$=\sin^2\text{A}-\cos^2\text{B}+2\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{C}$
$=-\cos(\text{A}+\text{B}).\cos(\text{A}-\text{B})+2\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{C}$ $\big[\because\ \cos^2\text{B}-\sin^2\text{A}=\cos(\text{A}+\text{B}).\cos(\text{A}-\text{B}).\big]$
$=-\cos(-\text{C}).\cos(\text{A}-\text{B})+\cos\text{C}(2\cos\text{A}.\cos\text{B}-\cos\text{C})$ $[\because\ \cos(-\theta)=\cos\theta]$
$=-\cos\text{C}.(\cos\text{A}.\cos\text{B}+\sin\text{A}.\sin\text{B}-2\cos\text{A}.\cos\text{B}-\cos\text{C})$
$=\cos\text{C}(\cos\text{A}.\cos\text{B}-\sin\text{A}.\sin\text{B}-\cos\text{C})$
$=\cos\text{C}[\cos(\text{A}+\text{B})-\cos\text{C}]$
$=\cos\text{C}(\cos\text{C}-\cos\text{C})=0$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 435 Marks
Solve the following system of equations by matrix method:$\frac{2}{\text{x}}-\frac{3}{\text{y}}+\frac{3}{\text{z}}=10$
$\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=10$
$\frac{3}{\text{x}}-\frac{1}{\text{y}}+\frac{2}{\text{z}}=13$
AnswerLet $\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
2u - 3v + 3w = 10
u + v + w = 10
3u - v + 2w = 13
Which can be written as:
$\begin{bmatrix}2&-3&3\\ 1&1&1\\ 3&-2&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ 10\\ 13\end{bmatrix}$
$\text{|A|}=2{(3)}+3{(-1)}+3{(-4)}$
$=6-3-12=-9\neq0$
Hence, the system has a unique solution, given by
$\text{X}=\text{A}^{-1}\times\text{B}$
$\text{C}_{11}=3\ \text{C}_{21}=3\ \text{C}_{31}=-6$
$\text{C}_{12}=1\ \text{C}_{22}=-5\ \text{C}_{32}=1$
$\text{C}_{13}=-4\ \text{C}_{23}=-7\ \text{C}_{33}=5$
$\text{X}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{(B)}$
$=\frac{1}{-9}\begin{bmatrix}3&3&-6\\ 1&-5&1\\ -4&-7&5\end{bmatrix}\begin{bmatrix}10\\ 10\\ 13 \end{bmatrix}$
$\frac{1}{-9}\begin{bmatrix}30+30-78\\ 10-50+13\\ -40-70+65\end{bmatrix}$
$\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\frac{-1}{9}\begin{bmatrix}-18\\ -27\\ -45\end{bmatrix}=\begin{bmatrix}2\\ 3\\ 5\end{bmatrix}$
Hence, $\text{x}=\frac{1}{2},\text{y}=\frac{1}{3},\text{z}=\frac{1}{5}$
View full question & answer→Question 445 Marks
Given $\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$ , find $BA$ and use this to solve the system of equations $y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17$
AnswerHere,
$\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$ and $\text{B} =\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$
$\text{BA}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}$
$=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$
$\Rightarrow\text{BA}=6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{BA}=6\text{I}_3$
$\Rightarrow\text{B}\Big(\frac{1}{6}\text{A}\Big)=\text{I}_3$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\text{A}$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
Now$, BX = C$
where, $\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{C}=\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\therefore X = B^{-1}C$
$\Rightarrow\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}$
$\therefore x = 2, y = -1$ and $z = 4.$
View full question & answer→Question 455 Marks
Solve the following for x and y.$\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
AnswerHere,
$\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3\text{x}-4\text{y}\\9\text{x}-2\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow3 \text{x}-4\text{y}=10\ \dots(1)$
$9\text{x}+2 \text{y}=2\ \dots(2)$
Solving both the equation, we get
$\text{x}=\frac{14}{21}$
$=\frac{2}{3}$
Substituting the value of x in eq. (1), we get
$3\times\frac{2}{3}-4\text{y}=10$
$\Rightarrow2-4\text{y}=10$
$\Rightarrow4 \text{y}=-8$
$\Rightarrow\text{y}=-2$
$\therefore\ \text{x}=\frac{2}{3}\text{ and }\text{y}=-2$
View full question & answer→Question 465 Marks
Prove that :
$\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&\text{a}+\text{c}\\\text{a}+\text{b}&\text{b}&\text{a}\\\text{b}&\text{b}+\text{c}&\text{c}\end{vmatrix}$
$[$Taking out $a, b$ and $c$ common from $C_1, C_2$ and $C_3]$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&-2\text{b}\\\text{b}&\text{b}+\text{c}&-2\text{b}\end{vmatrix}$
$[$Applying $C_{3 }\rightarrow C_3 - C_2 - C_1]$
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&1\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
$[$Taking $(-2b)$ common from $C_3]$
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}&-\text{c}&0\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
$[$Applying $R_2 \rightarrow R_2 - R_1]$
$=(\text{abc})(-2\text{b})\times1\begin{vmatrix}\text{a}&\text{c}\\\text{a}&-\text{c}\end{vmatrix}$
$[$expanding along $C_3]$
$=(\text{abc})(-2\text{b})(-2\text{ac})$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 475 Marks
Using properties of determinants, prove that:
$\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\\text{y}&\text{y}^2&1+\text{py}^3\\\text{z}&\text{z}^2&1+\text{pz}^3\end{vmatrix}=(1+\text{pxyz})(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x}),$ where $p$ is any scalar.
Answer$\triangle=\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\\text{y}&\text{y}^2&1+\text{py}^3\\\text{z}&\text{z}^2&1+\text{pz}^3\end{vmatrix}$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1,$ we have:
$\triangle=\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\\text{y}-\text{x}&\text{y}^2-\text{x}^2&\text{p}(\text{y}^3-\text{x}^3)\\\text{z}-\text{x}&\text{z}^2-\text{x}^2&\text{p}(\text{z}^3-\text{x}^3)\end{vmatrix}$
$=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\1&\text{y}+\text{x}&\text{p}(\text{y}^2+\text{x}^2+\text{xy})\\1&\text{z+x}&\text{p}(\text{z}^2+\text{x}^2+\text{xz})\end{vmatrix}$
Applying $R_3 \rightarrow R_3 - R_2,$ we have:
$\triangle=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\1&\text{y}+\text{x}&\text{p}(\text{y}^2+\text{x}^2+\text{xy})\\0&\text{z}-\text{y}&\text{p}(\text{z}-\text{y})(\text{x+y+z})\end{vmatrix}$
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}-\text{y})\begin{vmatrix}\text{x}&\text{x}^2&1+\text{px}^3\\1&\text{y}+\text{x}&\text{p}(\text{y}^2+\text{x}^2+\text{xy})\\0&1&\text{p}(\text{x+y+z})\end{vmatrix}$
Expanding along $R_3,$ we have:
$\triangle = (x - y) (y - z) (z - x) [(-1) (p) (xy^2 + x^3 + x^2y) + 1 + px^3 + p(x + y + z) (xy)]$
$= (x - y) (y - z) (z - x) [-pxy^2 - px^3 - px^2y + 1 + px^3 + px^2y + pxy^2 + pxyz]$
$= (x - y) (y - z) (z - x) (1 + pxyz)$
Hence, the given result is proved.
View full question & answer→Question 485 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
Answer$\begin{vmatrix}\sin^223^{\circ}&\sin^267^{\circ}&\cos180^{\circ}\\-\sin^267^{\circ}&-\sin^223^{\circ}&\cos^2180^{\circ}\\\cos180^{\circ}&\sin^223^{\circ}&\sin^267^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\sin^2(90-23)^{\circ}&-1\\-\sin^2(90-23)^{\circ}&-\sin^223^{\circ}&1\\1&\sin^223^{\circ}&\sin^2(90-23)^{\circ}\end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}&-\sin^223^{\circ}&1\\-1&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=\begin{vmatrix}\sin^223^{\circ}+\cos^223^{\circ}&\cos^223^{\circ}&-1\\-\cos^223^{\circ}-\sin^223^{\circ}&-\sin^223^{\circ}&1\\-1+\sin^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix} \ [$Applying $C_1 \rightarrow C_1 + C_2]$
$=\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=(-1)\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=0$
View full question & answer→Question 495 Marks
Solve the matrix equation $\begin{bmatrix}5 & 4 \\1 & 1 \end{bmatrix}\text{X}=\begin{bmatrix}1 & -2 \\1 & 3 \end{bmatrix},$ where $X$ is a $2 \times 2$ matrix.
Answer$\begin{bmatrix}5 & 4 \\1 & 1 \end{bmatrix}\text{X}=\begin{bmatrix}1 & -2 \\1 & 3 \end{bmatrix}$
Let $\text{A}=\begin{bmatrix}5 & 4 \\1 & 1 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}1 & -2 \\1 & 3 \end{bmatrix}$
So$, AX = B$
or $X = A^{-1}B .....(i)$
$|\text{A}|=1\neq0$
Cofactors of $A$ are:
$C_{11} = 1, C_{12} = -1$
$C_{21} = -4, C_{22} = 5$
$\therefore\ \text{adj A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} \\ \text{C}_{21} & \text{C}_{22} \end{bmatrix}$
$=\begin{bmatrix}1 & -1 \\-4 & 5 \end{bmatrix}^\text{T}$
$=\begin{bmatrix}1 & -4 \\-1 & 5 \end{bmatrix}$
Now, $\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}$
$\text{A}^{1}=\frac{1}{2}=\begin{bmatrix}1 & -4 \\-1 & 5 \end{bmatrix}$
So from $(i)$
$\text{X}=\begin{bmatrix}1 & -4 \\-1 & 5 \end{bmatrix}=\begin{bmatrix}1 & 2 \\ 1 & 3 \end{bmatrix}$
$\text{X}=\begin{bmatrix}-3 & -14 \\ 4 & 17 \end{bmatrix}$
View full question & answer→Question 505 Marks
Prove that:
$\begin{vmatrix}1&1+\text{p}&1+\text{p}+\text{q}\\2&3+2\text{p}&4+3\text{p}+2\text{p}\\3&6+3\text{p}&10+6\text{p}+3\text{q}\end{vmatrix}=1$
Answer$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+\begin{vmatrix}1&1&\text{p}\\2&3&3\text{p}\\3&6&6\text{p}\end{vmatrix}+(\text{pq})\begin{vmatrix}1&1&1\\2&2&2\\3&3&3\end{vmatrix}$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+(\text{p})\begin{vmatrix}1&1&\text{p}\\2&3&3\\3&6&6\end{vmatrix}+0$
$=\begin{vmatrix}1&1&1\\2&3&4\\3&2&10\end{vmatrix}+0$
$[\because$ Value of determinant with two identical columns is zero$]$
$=\begin{vmatrix}1&0&0\\2&1&2\\3&3&7\end{vmatrix}\ [$Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1]$
$=\left\{1\times\begin{vmatrix}1&2\\3&7\end{vmatrix}\right\}\ [$Expanding along $R_1]$
$=7-6$
$=1$
$=\text{R.H.S}$
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