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}$$\Delta = \begin{vmatrix} \text{a}^{2} - 1 & \text{a - 1} & 0 \\ 2\text{(a - 1)} & \text{a - 1} & 0 \\ 3 & 3 & 1 \end{vmatrix}$
$ = \text{(a - 1)}^{2}\begin{vmatrix} \text{a + 1} & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix}$ Expanding $\text{(a - 1)}^{2}. \text{(a - 1)} = \text{(a - 1)}^{3}.$
View full question & answer→Question 25 Marks
Using properties of determinants, prove that:
$\begin{vmatrix} \text{1 + a } & \text{1} & \text{1} 0.3em] \text{1} & \text{1 + b} & \text{1} 0.3em]\text{1} & 1 &\text{1 + c} \end{vmatrix}= \text{ abc + bc + ca + ab}$
Answer$R_1$_$\rightarrow$
$\frac{1}{\text{a}}R_{1,}R_2$
$\rightarrow \frac{1}{\text{b}}R_{2,}R_3$
$\rightarrow \frac{1}{\text{c}}R_3$
$\therefore\ \text{LHS} =\text{abc} \begin{bmatrix} \frac{1}{\text{a}}+1 & \frac{1}{\text{a}}& \frac{1}{\text{a}} 0.3em] \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} 0.3em] \frac{1}{\text{c}} & \frac{1}{\text{c}} &\frac{1}{\text{c}}+1 \end{bmatrix}
R_1$
$\rightarrow R_1 + R_{2 +}R_3$
$\Rightarrow$ $\text{ LHS} =\text{abc} \begin{vmatrix} 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}} &1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}& 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}} 0.3em] \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} 0.3em] \frac{1}{\text{c}} & \frac{1}{\text{c}} &\frac{1}{\text{c}}+1 \end{vmatrix}$
$=\text{abc}\Bigg( 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Bigg)$$\begin{vmatrix} 1 & 1& 1 0.3em] \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} 0.3em] \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
$\begin{vmatrix} \text{b + c } & \text{c + a} & \text{a + b} 0.3em] \text{q } + \text{r} & \text{r + p} & \text{p + q} 0.3em] \text{y + z} & \text{z + x} &\text{x + y} \end{vmatrix}= \text{2}$
$\begin{vmatrix} \text{a } & \text{b} & \text{c} 0.3em] \text{p} & \text{q} & \text{r} 0.3em] \text{x} & \text{y} &\text{z} \end{vmatrix}$
AnswerOperating $C → C_{1 —}(C_2 + C_3)$, we get
$\text{LHS}=\begin{vmatrix} \text{-2a } & \text{c + a} & \text{a + b} 0.3em] \text{-2p } & \text{r + p} & \text{p + q} 0.3em] \text{-2x} & \text{z + x} &\text{x + y} \end{vmatrix}=-2$
$\begin{vmatrix} \text{a } & \text{c + a} & \text{a + b} 0.3em] \text{p } & \text{r + p} & \text{p + q} \$0.3em] \text{x} & \text{z + x} &\text{x + y} \end{vmatrix}$
$ \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} \$0.3em] \text{p } & \text{r} & \text{q} \$0.3em] \text{x} & \text{z} &\text{ y} \end{vmatrix}$
$\text{C}_2\leftrightarrow\text{C}_3$
$=+2\begin{vmatrix} \text{a } & \text{c} & \text{b} \$0.3em] \text{p } & \text{r} & \text{q} \$0.3em] \text{x} & \text{z} &\text{ y} \end{vmatrix}=\text{RHS}$
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 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}). $
AnswerL.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}[\text{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}\text{[ 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}\text{[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
$= 3 (x + y) {1 (2y^2 + y^2 )}$
$= 9y^2 (x + y) = RHS.$
View full question & answer→Question 65 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 75 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] = 0
12x = 48 i.e. x = 4
View full question & answer→Question 85 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 95 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 105 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 115 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 125 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 135 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 $\begin{vmatrix} 1 & 1 & \text{1 + x} \\ 1 & \text{1 + y} & 1 \\ \text{1 + z} & 1 & 1 \end{vmatrix}$
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 145 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 155 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 195 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
$\triangle = \text{(x + y + z)}^{2}\text{(xy + zy + zx - x}^{2} - \text{y}^{2} - \text{z}^{2})^{2}$
$\text{Hence}\triangle \text{is divisible by (x + y + z) and}$
$\text{the quotient is (x + y + z) (xy + yz + zx - x}^{2} - \text{y}^{2} - \text{z}^{2})^{2}$
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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 225 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 $\begin{vmatrix} \text{-a} & \text{b} & \text{c} \\ \text{a} & \text{-b} & \text{c} \\ \text{a} & \text{b} & \text{-c} \end{vmatrix}$
TX
$LHS = a^2b^2c^2$ $\begin{vmatrix} \text{-1} & \text{1} & \text{1} \\ \text{1} & \text{-1} & \text{1} \\ \text{1} & \text{1} & \text{-1} \end{vmatrix}$
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 275 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 285 Marks
Using properties of determinants, prove the following:
$\begin{vmatrix}\text{a} & \text{b} & \text{c} \\\text{a}-\text{b} & \text{b}-\text{c} & \text{c}-\text{a}\\\text{b}+\text{c} & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}.$
AnswerLHS: 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 $C_1 → C_1 + C_2 + C_3$
$=\begin{vmatrix}\text{a}+\text{b}+\text{c} & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\2(\text{a}+\text{b}+\text{c}) & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}$
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: $R_3 → R_3 - 2R_1$
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1 & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\0 & \text{c}+\text{a} -\text{2b}& \text{a}+\text{b} -2\text{c}\end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})[(\text{b}-\text{c})(\text{a}+\text{b}-2\text{c})-(\text{c}-\text{a})(\text{c}+\text{a}-2\text{b})]$
$=\text{a}^3+\text{b}^3+\text{c}^{3}-3\text{abc}$
$=\text{RHS}$
View full question & answer→Question 295 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 305 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 315 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$\text{A}=\begin{bmatrix}2 & -3 & 5 \\3 & 2 & -4\\1 & 1 & -2 \end{bmatrix}$
⇒ |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 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
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 345 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 $R_2 → R_2 → R_1$]
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 2\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\0 & 1 & 0\\0 & -2 & 1 \end{bmatrix}$ [Applying $R_2 → R_2 → R_3$]
$\text{A}^{-1}=\begin{bmatrix}-1 & -2 & -4\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & -2\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$ [Applying $R_1 → R_1 + (-2)R_2, R_3 → R_3 + 2R_2$]
$\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$ [Applying $R_2 → R_2 → R_3$]
Hence, $\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}$
View full question & answer→Question 355 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 365 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 $C_1 → C_1 + C_2$]
$=\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=(-1)\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=0$
View full question & answer→Question 375 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 $\text{A}=\begin{bmatrix}5 & 4 \\1 & 1 \end{bmatrix}\text{ and B}=\begin{bmatrix}1 & -2 \\1 & 3 \end{bmatrix}$
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 385 Marks
Prove that: $\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
Answer$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
$\text{L.H.S}=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}$ Apply $R_3 → R_3 - R_2$
$=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)(\text{a}+3)&\text{a}+3&1\\(\text{a}+3)2&1&0\end{vmatrix}$ Apply $R_2 → R_2 - R_1$
$=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\\(\text{a}+2)2&1&0\\(\text{a}+3)2&1&0\end{vmatrix}$
$=[(2\text{a}+4)(1)-(1)(2\text{a}+6)]$
$=-2$
$=\text{R.H.S}$
View full question & answer→Question 395 Marks
Evaluate:
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
When a = b, the first two rows become identical. Hence, a - b is a factor. Similarly, when b = c and c = a, the second and third and third and first rows become indetical. Hence, b - c and c - a are also factors. The degree of product of the diagonal elements is 3. Hence, there are no other factors.
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=\lambda(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$ [Where $\lambda$ is a constant]
$\begin{vmatrix}1&0&2\\1&1&0\\1&2&0\end{vmatrix}=2\lambda$ $[$Putting a = 0, b = 1 and c = 2 to find $\lambda]$
$\Rightarrow2=2\lambda$
$\Rightarrow\lambda=1$
Hence,
$\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})$
View full question & answer→Question 405 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 $C_2 → C_2 - C_1$ and $C_3 → C_3 - C_1$]
$=\left\{1\times\begin{vmatrix}1&2\\3&7\end{vmatrix}\right\}$ [Expanding along $R_1$]
$=7-6$
$=1$
$=\text{R.H.S}$
View full question & answer→Question 415 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
Answer$\begin{vmatrix}\sin\alpha&\cos\alpha&\cos(\alpha+\delta)\\\sin\beta&\cos\beta&\cos(\beta+\delta)\\\sin\gamma&\cos\gamma&\cos(\gamma+\delta)\end{vmatrix}$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos\alpha\cos\delta&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos\beta\cos\delta&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos\gamma\cos\delta&\cos(\gamma+\delta)\end{vmatrix}$$[\text{Applying} \text{ C}_1\rightarrow\sin\delta\text{ C}_1\text{ and}\text{ C}_2\rightarrow\cos\delta\text{ C}_2]$
$=\begin{vmatrix}\sin\alpha\sin\delta&\cos(\alpha+\delta)&\cos(\alpha+\delta)\\\sin\beta\sin\delta&\cos(\beta+\delta)&\cos(\beta+\delta)\\\sin\gamma\sin\delta&\cos(\gamma+\delta)&\cos(\gamma+\delta)\end{vmatrix}$ [Applying $C_2 → C_2 - C_1$]
$=0$
View full question & answer→Question 425 Marks
Prove the following identities:
$\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(5\text{x}+\lambda)(\lambda-\text{x})^2$
Answer$\text{L.H.S}=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}-\text{x}-\lambda&\text{x}+\lambda-2\text{x}&0\\2\text{x}-\text{x}-\lambda&0&\text{x}+\lambda-2\text{x}\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$ and $R_3 → R_3 - R_1$]
$=\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-(\lambda-\text{x})&\lambda-\text{x}&0\\-(\lambda-\text{x})&0&\lambda-\text{x}\end{vmatrix}$
$=(\lambda-\text{x})^2\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\-1&1&0\\-1&0&1\end{vmatrix}$ [Taking $(\lambda-\text{x})$ common from $R_2$ and $(\lambda-\text{x})$ common from $R_3$]
$=(\lambda-\text{x})^2[-1(-2\text{x})+1(\text{x}+\lambda+2\text{x})]$ [Expanding along last row]
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
$=\text{R.H.S}$
$\because\begin{vmatrix}\text{x}+\lambda&2\text{x}&2\text{x}\\2\text{x}&\text{x}+\lambda&2\text{x}\\2\text{x}&2\text{x}&\text{x}+\lambda\end{vmatrix}$
$=(\lambda-\text{x})^2(\lambda+5\text{x})$
View full question & answer→Question 435 Marks
Prove the following identities:
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 → R_1 - R_2$
$=\begin{vmatrix}\text{y}&-\text{x}&\text{y}-\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
Applying $R_1 → R_1 - R_3$
$=\begin{vmatrix}0&-2\text{x}&-2\text{x}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=2\text{x}[\text{z}(\text{x}+\text{y})-\text{xy}]-2\text{x}[\text{zx}-\text{y}(\text{z}+\text{x})]$
$=2\text{x}[\text{zx}+\text{zy}-\text{xy}-\text{zx}+\text{yz}+\text{yx}]$
$=4\text{xyz}$
View full question & answer→Question 445 Marks
Prove that:
$\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}=2\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
Using the property of determinants that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants, we get
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)\begin{vmatrix}\text{a}&\text{c}&\text{b}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{b}&\text{a} \end{vmatrix}$
$[$Applying $\text{C}_1\leftrightarrow\text{C}_3$ in second determinant to get negative value of the deteminant$]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}+(-1)(-1)\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}$ $[$Applying $\text{C}_2\leftrightarrow\text{C}_3]$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b} \end{vmatrix}=\text{R.H.S}$
View full question & answer→Question 455 Marks
Solve the following systems of linear equations by cramer's rule:
3x + ay = 4,
2x + ay = 2, $\text{a}\neq0$
AnswerGiven, 3x + ay = 4
2x + ay = 2
Using Cramer's rule, we get
$\text{D}=\begin{vmatrix}3&\text{a}\\2&\text{a}\end{vmatrix}=3\text{a}-2\text{a}=\text{a}$
$\text{D}_1=\begin{vmatrix}4&\text{a}\\2&\text{a}\end{vmatrix}=4\text{a}-2\text{a}=2\text{a}$
$\text{D}_2=\begin{vmatrix}3&4\\2&2\end{vmatrix}=6-8=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{2\text{a}}{\text{a}}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{\text{a}}=-\frac{2}{\text{a}}$
$\therefore\text{x}=2$ and $\text{y}=-\frac{2}{\text{a}}$
View full question & answer→Question 465 Marks
If the points $(x, -2), (5, 2), (8, 8)$ are collinear, find x using determinants.
AnswerThe points $(k, -2), (5, 2), (8, 8)$ are collinear.
$\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}=0$
$\triangle=\begin{vmatrix}\text{x}&-2&1\\5&2&1\\8&8&1\end{vmatrix}$
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8&8&1\end{vmatrix}$ [Applying $R_2 → R_2 - R_1$]
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8-\text{x}&10&0\end{vmatrix}$ [Applying $R_3 → R_3 - R_1$]
$=\begin{vmatrix}5-\text{x}&4\\8-\text{x}&10\end{vmatrix}$
$=50-10\text{x}-32+4\text{x}$
$=18-6\text{x}=0$
$\Rightarrow\text{x}=3$
View full question & answer→Question 475 Marks
Evaluate:
$\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\text{x}+\lambda&\text{x}&\text{x}\\\text{x}&\text{x}+\lambda&\text{x}\\\text{x}&\text{x}&\text{x}+\lambda\end{vmatrix}$
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&\lambda&\text{x}\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ [Applying $C_1 → C_1 - C_2, C_2 → C_2 - C_3$]
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&0&2\text{x}+\lambda\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ [Applying $R_1 → R_2 + R_3$]
$=\lambda\begin{vmatrix}0&2\text{x}+\lambda\\-\lambda&\text{x}+\lambda\end{vmatrix}+\text{x}\begin{vmatrix}-\lambda&0\\0&-\lambda\end{vmatrix}$
$=\lambda\big[\lambda(2\text{x}+\lambda)\big]+\text{x}\lambda^2$
$=\lambda^2(2\text{x}+\lambda+\lambda^2\text{x})$
$=3\lambda^2\text{x}+\lambda^3$
$=\lambda^2(3\text{x}+\lambda)$
View full question & answer→Question 485 Marks
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
AnswerGiven,
$\triangle=\begin{vmatrix}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\-\sin\beta&\cos\beta&0\\\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha \end{vmatrix}$
$\Rightarrow\triangle=(-1)^{1+1}\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)\\+(-1)^{1+2}\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\+(-1)^{1+3}(-\sin\alpha)(-\sin^2\beta\sin\alpha-\sin\alpha\cos^2\beta)$ [Expanding along $R_1$]
$=\cos\alpha\cos\beta(\cos\alpha\cos\beta-0)-\cos\alpha\sin\beta(-\sin\beta\cos\alpha-0)\\-\sin\alpha(\sin^2\beta\sin\alpha-\sin\alpha-\sin\alpha\cos^2\beta)$
$=\cos^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+\sin^2\alpha\sin^2\beta+\sin^2\alpha\cos^2\beta$
$=\cos^2\alpha(\cos^2\beta+\sin^\beta)+\sin^2\alpha(\sin^2\beta+\cos^2\beta)$
$\Rightarrow\triangle=\cos^2\alpha+\sin^2\alpha$ $\big[\because\sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\triangle=1$
View full question & answer→Question 495 Marks
If $\text{A}=\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix},$ show that $A^{-1} = A^3$.
AnswerWe have, $\text{A}=\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}\begin{bmatrix}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{bmatrix}$
$=\begin{bmatrix} 9-6+0 & -9+9-4 & 12-12+4 \\ 6-6+0 & -6+9-4 & 8-12+4 \\ 0-2+0 & 0+3-1 & 0-4+1 \end{bmatrix}$
$=\begin{bmatrix}3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}$
Now, $\text{A}^3=\text{A}^2\times\text{A}=\begin{bmatrix} 3 & -4 & 4 \\ 0 & -1 & 0 \\ -2 & 2 & -3 \end{bmatrix}\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
$=\begin{bmatrix}9-8 & -9+12-4 & 12-16+4 \\ 0-2+0 & 0+3+0 & -4 \\ -6+4+0 & 6-6+3 & -8+8-3 \end{bmatrix}$
$= \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ -2 & 3 & -3 \end{bmatrix}$
Again, $\text{A}^3\times\text{A}=\begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ -2 & 3 & -3\end{bmatrix}\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$
$=\begin{bmatrix}-3-2+0 & -3+3+0 & 4-4+0 \\ -6+6 & 6-9+4 & -8+12-4 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\text{I}_3 \ \big[\text{Identity matrix of order }3\big]$
$\Rightarrow\ \text{A}^3\times\text{A}=\text{I}_3$
$\Rightarrow\ \text{A}^3=\text{A}^{-1}$
View full question & answer→Question 505 Marks
Without expanding, prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
Answer$=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}$ $\text{R}_2\leftrightarrow\text{R}_3$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
$\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
$=\begin{vmatrix}\text{y}&\text{x}&\text{z}\\\text{b}&\text{a}&\text{c}\\\text{q}&\text{p}&\text{r}\end{vmatrix}$ $\text{C}_1\leftrightarrow\text{C}_2$
$=-\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$ $\text{R}_1\leftrightarrow\text{R}_2$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}$
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