Question 14 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 24 Marks
Using properties of determinants, prove that:
$\begin{vmatrix} \text{1 + a } & \text{1} & \text{1} \\[0.3em] \text{1} & \text{1 + b} & \text{1} \\[0.3em]\text{1} & 1 &\text{1 + c} \end{vmatrix}= \text{ abc + bc + ca + ab}$
AnswerR1 $\rightarrow$ $\frac{1}{\text{a}}$R1, R2 $\rightarrow$$\frac{1}{\text{b}}$R2, R3$\rightarrow$$\frac{1}{\text{c}}$R3
$\therefore\ \text{ LHS} =\text{abc} \begin{bmatrix} \frac{1}{\text{a}}+1 & \frac{1}{\text{a}}& \frac{1}{\text{a}} \\[0.3em] \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} \\[0.3em] \frac{1}{\text{c}} & \frac{1}{\text{c}} &\frac{1}{\text{c}}+1 \end{bmatrix}$
R1 $\rightarrow$R1 + R2 + R3 ⇒ $\text{ LHS} =\text{abc} \begin{vmatrix} 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}} &1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}& 1+\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}} \\[0.3em] \frac{1}{\text{b}} & \frac{1}{\text{b}}+1&\frac{1}{\text{b}} \\[0.3em] \frac{1}{\text{c}} & \frac{1}{\text{c}} &\frac{1}{\text{c}}+1 \end{vmatrix}$
$=\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 34 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 → C1 — (C2 + C3), 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 44 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 54 Marks
Using properties of determinants, prove that
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AnswerOperating R
2 ; background:rgba(220,220,220,0.5))
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R
2 - 4R
1 and R
3 ![]()
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R3 -8R1, we get
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Expanding along C2,we get
– x (– 5x2 + 4x2) = x3.
View full question & answer→Question 64 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 R1 we get
= 3 (x + y) {1 (2y2 + y2 )}
= 9y2 (x + y) = RHS.
View full question & answer→Question 74 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 R2$\rightarrow$R2-bR3 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 84 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 C2 $\rightarrow$ C2 - 2C1 and C3 $\rightarrow$ C3 - 3C1, 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 R1 $\rightarrow$ R1 + R2 and R3 $\rightarrow$ R3 - 11R2, 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 C2, we get –1[8x – 24 – 20x + 72] = 0 12x = 48 i.e. x = 4
View full question & answer→Question 94 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 104 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 114 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 124 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 134 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 R2 $\rightarrow$ R2 - 2R1, R3 $\rightarrow$ R3 + 2R1, 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 R1 $\rightarrow$ R1 - 2R2, 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 R1 $\rightarrow$ R1 - R3, 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 R2 $\rightarrow$ R2 - R3, 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 144 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 C3, C2 and C1 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$
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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} $
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$\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 + x2 + y2 + z2 = RHS (Expand along C1)
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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 C1
= 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 C1 $\rightarrow$C1 + C2 + C3
$ |\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 R2 and R3
$ |\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 C1
|A|= (1 + x + x2 )(1 - x)2 (1 + x + x2)
= (1 - x3)2
[$\because$ 1 - x3 = (1 - x)(1 + x + x2)].
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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 R1, R2, R3 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 c1, c2, c3 to get
LHS = a2b2c2 $\begin{vmatrix} \text{-1} & \text{1} & \text{1} \\ \text{1} & \text{-1} & \text{1} \\ \text{1} & \text{1} & \text{-1} \end{vmatrix}$
Applying R2 $\rightarrow$ R2+R1, R3 $\rightarrow$ R3+ R1, to get
LHS = a2b2c2 $\begin{vmatrix} \text{-1} & \text{1} & \text{1} \\ \text{0} & \text{0} & \text{2} \\ \text{0} & \text{2} & \text{0} \end{vmatrix}$
= a2b2c2 . (-1) (-4) = 4 a2b2c2 = 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 284 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 294 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 eq
uation 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 304 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 C1 → C1 + C2 + C3
$=\begin{vmatrix}\text{a}+\text{b}+\text{c} & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\2(\text{a}+\text{b}+\text{c}) & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}$
Taking (a + b + c) common from C1
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1 & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\2 & \text{c}+\text{a} & \text{a}+\text{b} \end{vmatrix}$
Apply: R3 → R3 - 2R1
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1 & \text{b} & \text{c} \\0 & \text{b}-\text{c} & \text{c}-\text{a}\\0 & \text{c}+\text{a} -\text{2b}& \text{a}+\text{b} -2\text{c}\end{vmatrix}$
$=(\text{a}+\text{b}+\text{c})[(\text{b}-\text{c})(\text{a}+\text{b}-2\text{c})-(\text{c}-\text{a})(\text{c}+\text{a}-2\text{b})]$
$=\text{a}^3+\text{b}^3+\text{c}^{3}-3\text{abc}$
$=\text{RHS}$
View full question & answer→Question 314 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 324 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 334 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 344 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 C3
$=1\big[(1-\text{a}^2)(2-2\text{a})-(1-\text{a})(3-\text{a}^2-2\text{a})\big]$
$=2(1-\text{a})(1-\text{a})(1+\text{a})-(1-\text{a})(3-\text{a}^2-2\text{a})$
$=(1-\text{a})\big[2(1-\text{a}^2)-3+\text{a}^2+2\text{a}\big]$
$=(1-\text{a})(2\text{a}-\text{a}^2-1)$
$=(\text{a}-1)^3$
$=\text{RHS}$
View full question & answer→Question 354 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 R2 → R2 → R1]
$\text{A}^{-1}=\begin{bmatrix}1 & 0 & 0\\1 & 1 & 2\\0 & 0 & 1 \end{bmatrix}\begin{bmatrix}1 & 2 & -2\\0 & 1 & 0\\0 & -2 & 1 \end{bmatrix}$ [Applying R2 → R2 → R3]
$\text{A}^{-1}=\begin{bmatrix}-1 & -2 & -4\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & -2\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$ [Applying R1 → R1 + (-2)R2, R3 → R3 + 2R2]
$\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$ [Applying R2 → R2 → R3]
Hence, $\text{A}^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5 \end{bmatrix}$
View full question & answer→Question 364 Marks
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1,600. School B wants to spend ₹ 2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹ 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
AnswerLet the award money given for sincerity, truthfulness and helpfulness be ₹ x, ₹ y and ₹ z respectively. Since, the total cash award is ₹ 900. $\therefore$ x + y + z = 900 .....(1) Award money given by school A is ₹ 1,600. $\therefore$ 3x + 2y + z = 1600 .....(2) Award money given by school B is ₹ 2,300. $\therefore$ 4x + y + 3z = 2300 .....(3) The above system of equations can be written in matrix form CX = D as: $\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}900\\1600\\2300\end{bmatrix}$ Where, $\text{C}=\begin{bmatrix}1&1&1\\3&2&1\\4&1&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{D}=\begin{bmatrix}900\\1600\\2300\end{bmatrix}$ Now, $|\text{C}|=\begin{vmatrix}1&1&1\\3&2&1\\4&1&3\end{vmatrix}$ $=1(6-1)-1(9-4)+1(3-8)$ $=5-5-5$ $=-5$ Let Cij be the co-factors of elements cij in C = [cij]. Then,
$\text{C}_{11}=(-1)^{1+1}\begin{vmatrix}2&1\\1&3\end{vmatrix}=5,$ $\text{C}_{12}=(-1)^{1+2}\begin{vmatrix}3&1\\4&3\end{vmatrix}=-5,$ $\text{C}_{13}=(-1)^{1+3}\begin{vmatrix}3&2\\4&1\end{vmatrix}=-5$ $\text{C}_{21}=(-1)^{2+1}\begin{vmatrix}1&1\\1&3\end{vmatrix}=-2,$ $\text{C}_{22}=(-1)^{2+2}\begin{vmatrix}1&1\\4&3\end{vmatrix}=-1,$ $\text{C}_{23}=(-1)^{2+3}\begin{vmatrix}1&1\\1&1\end{vmatrix}=3$ $\text{C}_{31}=(-1)^{3+1}\begin{vmatrix}1&1\\2&1\end{vmatrix}=-1,$ $\text{C}_{32}=(-1)^{3+2}\begin{vmatrix}1&1\\3&1\end{vmatrix}=2,$ $\text{C}_{33}=(-1)^{3+3}\begin{vmatrix}1&1\\3&2\end{vmatrix}=-1$ $\text{adj }\text{C}=\begin{bmatrix}5&-5&-5\\-2&-1&3\\-1&2&-1\end{bmatrix}^\text{T}$ $=\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}$ $\Rightarrow\text{C}^{-1}=\frac{1}{|\text{C}|}\text{adj }\text{C}$ $=\frac{1}{-5}\begin{bmatrix}5&-2&-1\\-5&-1&2\\-5&3&-1\end{bmatrix}$ $\text{X}=\text{C}^{-1}\text{D}$ $\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}4500-3200-2300\\-4500-4600+4600\\-4500+4800-2300\end{bmatrix}$ $\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=-\frac{1}{5}\begin{bmatrix}-1000\\-1500\\-2000\end{bmatrix}$ $\Rightarrow\text{x}=\frac{-1000}{-5},\text{y}=\frac{-1500}{-5}\ \text{and }\text{z}=\frac{-2000}{-5}$ $\therefore$ x = 200, y = 300 and z = 400. Hence, the award money for each value of sincerity, truthfulness and helpfulness is ₹ 200, ₹ 300 and ₹ 400.
One more value which should be considered for award hardwork.
View full question & answer→Question 374 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 384 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 C1 → C1 - C2, C2 → C2 - C3]
$=\begin{vmatrix}\lambda&0&\text{x}\\-\lambda&0&2\text{x}+\lambda\\0&-\lambda&\text{x}+\lambda\end{vmatrix}$ [Applying R1 → R2 + R3]
$=\lambda\begin{vmatrix}0&2\text{x}+\lambda\\-\lambda&\text{x}+\lambda\end{vmatrix}+\text{x}\begin{vmatrix}-\lambda&0\\0&-\lambda\end{vmatrix}$
$=\lambda\big[\lambda(2\text{x}+\lambda)\big]+\text{x}\lambda^2$
$=\lambda^2(2\text{x}+\lambda+\lambda^2\text{x})$
$=3\lambda^2\text{x}+\lambda^3$
$=\lambda^2(3\text{x}+\lambda)$
View full question & answer→Question 394 Marks
Solve the following systems of linear equations by cramer's rule:
2x + 3y = 10,
x + 6y = 4
AnswerGiven, 2x + 3y = 10
x + 6y = 4
Using Cramer's Rule, we get
$\text{D}=\begin{vmatrix}2&3\\1&6\end{vmatrix}=12-3=9$
$\text{D}_1=\begin{vmatrix}10&3\\4&6\end{vmatrix}=60-12=48$
$\text{D}_2=\begin{vmatrix}2&10\\1&4\end{vmatrix}=8-10=-2$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{48}{9}=\frac{16}{3}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-2}{9}$
$\therefore\text{x}=\frac{16}{3}$ and $\text{y}=\frac{-2}{9}$
View full question & answer→Question 404 Marks
Solve the following systems of homogeneous linear equations by matrix method:
3x - y + 2z = 0
4x + 3y + 3z = 0
5x + 7y + 4z =0
AnswerHere, 3x - y + 2z = 0 .....(1)
4x + 3y + 3z = 0 .....(2)
5x + 7y + 4z =0 .....(3)
The given system of homogeneous equaions can be written in matrix form as follows:
$\begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
$\text{AX}=\text{O}$
Here,
$\text{A}=\begin{bmatrix}3&-1&2\\4&3&3\\5&7&4\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix} $
Now,
$|\text{A}|=\begin{vmatrix}3&-1&2\\4&3&3\\5&7&4\end{vmatrix}$
$=3(12-21)+1(16-15)+2(28-15)$
$=0$
$\therefore\ |\text{A}|\neq0$
So, the given system of homogeneous equations has non-trivial solutions.
Substituting z = k in eq. (1) & eq. (2), we get
3x - y = -2k and 4x + 3y = -3k
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}3&-1\\4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix} $
$\Rightarrow\begin{bmatrix}3&-1\\4&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix} $
$|\text{A}|=\begin{vmatrix}3&-1\\4&3\end{vmatrix}$
$=(3\times3+4\times1)$
$=13$
So, A-1 exists.
We have
$\text{adj }\text{A}=\begin{bmatrix}3&1\\-4&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$\Rightarrow\text{A}^{-1}=\frac{1}{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{13}\begin{bmatrix}3&1\\-4&3\end{bmatrix}\begin{bmatrix}-2\text{k}\\-3\text{k}\end{bmatrix}$
$=\frac{1}{13}\begin{bmatrix}-6\text{k}-3\text{k}\\8\text{k}-9\text{k}\end{bmatrix} $
Thus, $\text{x}=\frac{-9\text{k}}{13}, \text{y}=\frac{-\text{k}}{13}\text{ and }\text{z}=\text{k}$ (where k is any real number) satisfy the given system of equations.
View full question & answer→Question 414 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 R2 → R2 - R1 and R3 → R3 - R1, 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 R3 → R3 - R2, 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 R3, we have:
$\triangle$ = (x - y) (y - z) (z - x) [(-1) (p) (xy2 + x3 + x2y) + 1 + px3 + p(x + y + z) (xy)]
= (x - y) (y - z) (z - x) [-pxy2 - px3 - px2y + 1 + px3 + px2y + pxy2 + pxyz]
= (x - y) (y - z) (z - x) (1 + pxyz)
Hence, the given result is proved.
View full question & answer→Question 424 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 C1 → C1 + C2]
$=\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=(-1)\begin{vmatrix}1&1&-1\\-1&-\sin^223^{\circ}&1\\-\cos^223^{\circ}&\sin^223^{\circ}&\cos^223^{\circ} \end{vmatrix}$
$=0$
View full question & answer→Question 434 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 × 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-1B .....(i)
$|\text{A}|=1\neq0$
Cofactors of A are:
C11 = 1, C12 = -1
C21 = -4, C22 = 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 444 Marks
Using the properties of determinants:
$\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
Answer We have, $\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$ $=\begin{vmatrix}2\text{x}+4&2\text{x}+4&2\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$ $\big[\text{R}_1\rightarrow\text{R}_1+\text{R}_2\big]$ $\begin{vmatrix}2\text{x}&2\text{x}&2\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}+\begin{vmatrix}4&4&4\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$=2\text{x}\begin{vmatrix}1&1&1\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}+4\begin{vmatrix}1&1&0\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$ [Taking 2x and 4 common from R1]
$=2\text{x}\begin{vmatrix}0&0&1\\0&4&\text{x}\\-4&-4&\text{x}+4\end{vmatrix}+4\begin{vmatrix}0&1&0\\-4&\text{x}+4&\text{x}\\0&\text{x}&\text{x}+4\end{vmatrix}$ [C1 → C1 - C3 and C2 → C2 - C3 in first and C1 → C1 - C2 in second determinant]
Expanding along C1, we get
$2\text{x}[-4(-4)]+4[4(\text{x}+4-0)]$
$=32\text{x}+16\text{x}+64$
$=16(3\text{x}+4)$
View full question & answer→Question 454 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 × 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 (AT)-1.
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^\text{T})^{-1}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=(\text{A}^{-1})^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\text{B}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$ $[\text{Using}\ (1)]$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}^\text{T}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-2&9&6\\0&2&1\\1&-3&-2\end{bmatrix}\begin{bmatrix}9\\4\\-3\end{bmatrix}$
$=\begin{bmatrix}-18+36-18\\0+8-3\\9-12+6\end{bmatrix}$
$=\begin{bmatrix}0\\5\\3\end{bmatrix}$
Hence, x = 0, y = 5 and z = 3.
View full question & answer→Question 464 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 Cij be the co-factors of aij 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 474 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-1B
Let Cij be the co-factor of aij 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 484 Marks
For the matrix $\text{A}=\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}$ Show that A3 - 6A2 + 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 A3 - 6A2 + 5A + 11I = 0 by A-1
⇒ A3A-1 - 6A2A-1 + 5AA-1 + 11IA-1 = 0.A-1
⇒ A2 - 6A + 5I + 11A-1 = 0 ⇒ 11A-1 = 6A - 5I - A2
$\Rightarrow\ \ 11\text{A}^{-1}=6\begin{bmatrix}1&1&1\\1&2&-3\\2&-1&3\end{bmatrix}-5\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}-\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}$
$\Rightarrow\ \ 11\text{A}^{-1}=\begin{bmatrix}6&6&6\\6&12&-18\\12&-6&18\end{bmatrix}-\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}-\begin{bmatrix}4&2&1\\-3&8&-14\\7&-3&14\end{bmatrix}$
$\Rightarrow \ 11\text{A}^{-1}\begin{bmatrix}6-5-4&6-2&6-1\\6+3&12-5-8&-18+14\\12-7&-6+3&18-5-14\end{bmatrix}=\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{11}\begin{bmatrix}-3&4&5\\9&-1&-4\\5&-3&-1\end{bmatrix}$
View full question & answer→Question 494 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 Cij be the co-factors of the elements aij in A = [aij]. 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 504 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 Cij be the co-factors of aij 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→