Question 514 Marks
If A + B + C = 0, then prove that $\begin{vmatrix}1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&1\end{vmatrix}=0.$
Answer$\text{L.H.S}=\begin{vmatrix}1&\cos\text{C}&\cos\text{B}\\\cos\text{C}&1&\cos\text{A}\\\cos\text{B}&\cos\text{A}&1\end{vmatrix}$
$=(1-\cos^2\text{A})-\cos\text{C}(\cos\text{C}-\cos\text{A}.\cos\text{B})+\cos\text{B}(\cos\text{C}.\cos\text{A}-\cos\text{B})$
$=\sin^2\text{A}-\cos^2\text{C}+\cos\text{A}.\cos\text{B}.\cos\text{C}+\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{B}$
$=\sin^2\text{A}-\cos^2\text{B}+2\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{C}$
$=-\cos(\text{A}+\text{B}).\cos(\text{A}-\text{B})+2\cos\text{A}.\cos\text{B}.\cos\text{C}-\cos^2\text{C}$ $\big[\because\ \cos^2\text{B}-\sin^2\text{A}=\cos(\text{A}+\text{B}).\cos(\text{A}-\text{B}).\big]$
$=-\cos(-\text{C}).\cos(\text{A}-\text{B})+\cos\text{C}(2\cos\text{A}.\cos\text{B}-\cos\text{C})$ $[\because\ \cos(-\theta)=\cos\theta]$
$=-\cos\text{C}.(\cos\text{A}.\cos\text{B}+\sin\text{A}.\sin\text{B}-2\cos\text{A}.\cos\text{B}-\cos\text{C})$
$=\cos\text{C}(\cos\text{A}.\cos\text{B}-\sin\text{A}.\sin\text{B}-\cos\text{C})$
$=\cos\text{C}[\cos(\text{A}+\text{B})-\cos\text{C}]$
$=\cos\text{C}(\cos\text{C}-\cos\text{C})=0$
$=\text{R.H.S}$
Hence proved.
View full question & answer→Question 524 Marks
Solve the following system of equations by matrix method: $\frac{2}{\text{x}}-\frac{3}{\text{y}}+\frac{3}{\text{z}}=10$
$\frac{1}{\text{x}}+\frac{1}{\text{y}}+\frac{1}{\text{z}}=10$
$\frac{3}{\text{x}}-\frac{1}{\text{y}}+\frac{2}{\text{z}}=13$
AnswerLet $\frac{1}{\text{x}}=\text{u},\frac{1}{\text{y}}=\text{v},\frac{1}{\text{z}}=\text{w}$
2u - 3v + 3w = 10
u + v + w = 10
3u - v + 2w = 13
Which can be written as:
$\begin{bmatrix}2&-3&3\\ 1&1&1\\ 3&-2&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ 10\\ 13\end{bmatrix}$
$\text{|A|}=2{(3)}+3{(-1)}+3{(-4)}$
$=6-3-12=-9\neq0$
Hence, the system has a unique solution, given by
$\text{X}=\text{A}^{-1}\times\text{B}$
$\text{C}_{11}=3\ \text{C}_{21}=3\ \text{C}_{31}=-6$
$\text{C}_{12}=1\ \text{C}_{22}=-5\ \text{C}_{32}=1$
$\text{C}_{13}=-4\ \text{C}_{23}=-7\ \text{C}_{33}=5$
$\text{X}=\frac{1}{\text{|A|}}\text{(Adj A)}\times\text{(B)}$
$=\frac{1}{-9}\begin{bmatrix}3&3&-6\\ 1&-5&1\\ -4&-7&5\end{bmatrix}\begin{bmatrix}10\\ 10\\ 13 \end{bmatrix}$
$\frac{1}{-9}\begin{bmatrix}30+30-78\\ 10-50+13\\ -40-70+65\end{bmatrix}$
$\begin{bmatrix}\text{u}\\ \text{v}\\ \text{w}\end{bmatrix}=\frac{-1}{9}\begin{bmatrix}-18\\ -27\\ -45\end{bmatrix}=\begin{bmatrix}2\\ 3\\ 5\end{bmatrix}$
Hence, $\text{x}=\frac{1}{2},\text{y}=\frac{1}{3},\text{z}=\frac{1}{5}$
View full question & answer→Question 534 Marks
Given $\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$ , find BA and use this to solve the system of equations y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17
AnswerHere,
$\text{A}=\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}$
$\text{BA}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}$
$=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}$
$\Rightarrow\text{BA}=6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{BA}=6\text{I}_3$
$\Rightarrow\text{B}\Big(\frac{1}{6}\text{A}\Big)=\text{I}_3$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\text{A}$
$\Rightarrow\text{B}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}$
Now, BX = C
where, $\text{B}=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{C}=\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\therefore$ X = B-1C
$\Rightarrow\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\-6\\24\end{bmatrix}$
$\therefore$ x = 2, y = -1 and z = 4.
View full question & answer→Question 544 Marks
Solve the following for x and y. $\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
AnswerHere,
$\begin{bmatrix}3&-4\\9&2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3\text{x}-4\text{y}\\9\text{x}-2\text{y}\end{bmatrix}=\begin{bmatrix}10\\2\end{bmatrix}$
$\Rightarrow3 \text{x}-4\text{y}=10\ \dots(1)$
$9\text{x}+2 \text{y}=2\ \dots(2)$
Solving both the equation, we get
$\text{x}=\frac{14}{21}$
$=\frac{2}{3}$
Substituting the value of x in eq. (1), we get
$3\times\frac{2}{3}-4\text{y}=10$
$\Rightarrow2-4\text{y}=10$
$\Rightarrow4 \text{y}=-8$
$\Rightarrow\text{y}=-2$
$\therefore\ \text{x}=\frac{2}{3}\text{ and }\text{y}=-2$
View full question & answer→Question 554 Marks
By using properties of determinants, show that:
$\begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix}=1+a^2+b^2+c^2$
Answer$\text{L.H.S.}=\begin{vmatrix}a^2+1&ab&ac\\ab&b^2+1&bc\\ca&cb&c^2+1\end{vmatrix} $
Multiplying C1, C2, C3 by a, b, c respectively and then dividing the determinant by abc,
$=\frac{1}{abc}\begin{vmatrix}a(a^2+1)&ab^2&ac^2\\a^2b&b(b^2+1)&bc^2\\a^2c&b^2c&c(c^2+1)\end{vmatrix}$
$=\frac{abc}{abc}\begin{vmatrix}a^2+1&b^2&c^2\\a^2&b^2+1&c^2\\a^2&b^2&c^2+1\end{vmatrix}$
$=\frac{abc}{abc}\begin{vmatrix}1+a^2+b^2+c^2&b^2&c^2\\1+a^2+b^2+c^2&b^2+1&c^2\\1+a^2+b^2+c^2&b^2&c^2+1\end{vmatrix}\ \ \left[\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\right]$
$=(1+a^2+b^2+c^2)\begin{vmatrix}1&b^2&c^2\\1&b^2+1&c^2\\1&b^2&c^2+1\end{vmatrix}$
$=(1+a^2+b^2+c^2)\begin{vmatrix}1&b^2&c^2\\0&1&0\\0&0&1\end{vmatrix}\ \left[\text{R}_2\rightarrow\text{R}_2-\text{R}_1\text{and R}_3\rightarrow\text{R}_3-\text{R}_1\right]$
$=(1+a^2+b^2+c^2)(1)(1-0)=1+a^2+b^2+c^2=\text{R.H.S.}$ Proved.
View full question & answer→Question 564 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
Answer$\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}&\cos\text{x}&\sin\text{y}\\-\cos\text{x}&\sin\text{x}&-\cos\text{y} \end{vmatrix}$
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y})\\\sin\text{x}\sin\text{y}&\cos\text{x}\sin\text{y}&\sin^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
[Applying R2 → siny R2 and R3 → cosy R3]
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\sin\text{x}\sin\text{y}-\cos\text{x}\cos\text{y}&\cos\text{x}\sin\text{y}+\sin\text{x}\sin\text{y}&\sin^2\text{y}-\cos^2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
[Applying R2 → R2 + R3]
$=\frac{1}{\sin\text{y}\cos\text{y}}\begin{vmatrix}\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\\cos(\text{x}+\text{y})&-\sin(\text{x}+\text{y})&\cos2\text{y}\\-\cos\text{x}\cos\text{y}&\sin\text{x}\cos\text{y}&-\cos^2\text{y} \end{vmatrix}$
$=0$
View full question & answer→Question 574 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
Answer$\begin{vmatrix}\sin^2\text{A}&\cot\text{A}&1\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}&\cot\text{C}&1\end{vmatrix}$
$=\begin{vmatrix}\sin^2\text{A}-\sin^2\text{B}&\cot\text{A}-\cot\text{B}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin^2\text{C}-\sin^2\text{B}&\cot\text{C}-\cot\text{B}&0\end{vmatrix}$ [Applying R1 → R1 - R2 and R3 → R3 - R2]
$=\begin{vmatrix}\sin(\text{A}+\text{B})\sin(\text{A}-\text{B})&\frac{\cos\text{A}\sin\text{B}-\cos\text{B}\sin\text{A}}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\text{C}+\text{B})\sin(\text{C}-\text{B})&\frac{\cos\text{C}\sin\text{B}-\cos\text{B}\sin\text{C}}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$
$=\begin{vmatrix}\sin(\pi-\text{C})\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\cot\text{B}&1\\\sin(\pi-\text{A})\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{C}}&0\end{vmatrix}$ $[\because\text{A}+\text{B}+\text{C}=\pi]$
$=\begin{vmatrix}\sin\text{C}\sin(\text{A}-\text{B})&\frac{-\sin(\text{A}-\text{B})}{\sin\text{A}\sin\text{B}}&0\\\sin^2\text{B}&\frac{\cos\text{B}}{\sin\text{B}}&1\\\sin\text{A}\sin(\text{C}-\text{B})&\frac{-\sin(\text{C}-\text{B})}{\sin\text{A}\sin\text{B}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}}\begin{vmatrix}\sin\text{C}&\frac{-1}{\sin\text{A}}&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}&\frac{-1}{\sin\text{C}}&0\end{vmatrix}$
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}\sin\text{C}\sin\text{A}&-1&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix}$ [Applying R1 → sinA R1 and R3 → sinC R3]
$=\frac{\sin(\text{A}-\text{B})\sin(\text{C}-\text{B})}{\sin\text{B}\sin\text{A}\sin\text{C}}\begin{vmatrix}0&0&0\\\sin^2\text{B}&\cos\text{B}&1\\\sin\text{A}\sin\text{C}&-1&0\end{vmatrix}$ [Applying R1 → R1 - R3]
$=0$
View full question & answer→Question 584 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&\text{bc}&\text{ac}+\text{c}^2\\\text{a}^2+\text{ab}&\text{b}^2&\text{ac}\\\text{ab}&\text{b}^2+\text{ac}&\text{c}^2\end{vmatrix}$
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&\text{a}+\text{c}\\\text{a}+\text{b}&\text{b}&\text{a}\\\text{b}&\text{b}+\text{c}&\text{c}\end{vmatrix}$
[Taking out a, b and c common from C1, C2 and C3]
$=\text{abc}\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&-2\text{b}\\\text{b}&\text{b}+\text{c}&-2\text{b}\end{vmatrix}$
[Applying C3 → C3 - C2 - C1]
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}+\text{b}&\text{b}&1\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
[Taking (-2b) common from C3]
$=(\text{abc})(-2\text{b})\begin{vmatrix}\text{a}&\text{c}&0\\\text{a}&-\text{c}&0\\\text{b}&\text{b}+\text{c}&1\end{vmatrix}$
[Applying R2 → R2 - R1]
$=(\text{abc})(-2\text{b})\times1\begin{vmatrix}\text{a}&\text{c}\\\text{a}&-\text{c}\end{vmatrix}$
[expanding along C3]
$=(\text{abc})(-2\text{b})(-2\text{ac})$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 594 Marks
Prove that:
$\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}=(\text{a}^3+\text{b}^3)^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2&2\text{ab}&\text{b}^2\\\text{b}^2&\text{a}^2&2\text{ab}\\2\text{ab}&\text{b}^2&\text{a}^2\end{vmatrix}$
$=\text{a}^2\begin{vmatrix}\text{a}^2&2\text{ab}\\\text{b}^2&\text{a}^2\end{vmatrix}-(2\text{ab})\begin{vmatrix}\text{b}^2&2\text{ab}\\2\text{ab}&\text{a}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}\text{b}^2&\text{a}^2\\2\text{ab}&\text{b}^2\end{vmatrix}$ [Expanding]
$=\text{a}^2(\text{a}^4-2\text{ab}^3)-(2\text{ab})(\text{b}^2\text{a}^2-4\text{a}^2\text{b}^2)+\text{b}^2(\text{b}^4-2\text{a}^3\text{b})$
$=\text{a}^6-2\text{a}^3\text{b}^3-2\text{a}^3\text{b}^3+8\text{a}^3\text{b}^3+\text{b}^6-2\text{a}^3\text{b}^3$
$=\text{a}^6+2\text{a}^3\text{b}^3+(\text{b}^3)^2$
$=(\text{a}^3+\text{b}^3)^2$
$=\text{R.H.S}$
View full question & answer→Question 604 Marks
$$Find the inverse of each of the matrix:
$\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$
Answer$\text{Let A}=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$ $\therefore\ |\text{A}|=\begin{vmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{vmatrix}$
$=1\left(-\cos^2\alpha-\sin^2\alpha\right)-0+0=-\left(\cos^2\alpha+\sin^2\alpha\right)=-1\neq0$
$\text{A}_{11}=+\begin{vmatrix}\cos\alpha&\sin\alpha\\\sin\alpha&-\cos\alpha\end{vmatrix}=+\left(-\cos^2\alpha-\sin^2\alpha\right)=-\left(\cos^2\alpha+\sin^2\alpha\right)=-1$
$\text{A}_{12}=-\begin{vmatrix}0&\sin\alpha\\0&-\cos\alpha\end{vmatrix}=-(0-0)=0,$
$\text{A}_{13}=+\begin{vmatrix}0&\cos\alpha\\0&\sin\alpha\end{vmatrix}=+(0-0)=0,$
$\text{A}_{21}=-\begin{vmatrix}0&0\\\sin\alpha&-\cos\alpha\end{vmatrix}=-(0-0)=0,$
$\text{A}_{22}=+\begin{vmatrix}1&0\\0&-\cos\alpha\end{vmatrix}=+(-\cos\alpha-0)=-\cos\alpha,$
$\text{A}_{23}=-\begin{vmatrix}1&0\\0&\sin\alpha\end{vmatrix}=(-\sin\alpha-0)=\sin\alpha,$
$\text{A}_{31}=+\begin{vmatrix}0&0\\\cos\alpha&\sin\alpha\end{vmatrix}=(0-0)=0,$
$\text{A}_{32}=-\begin{vmatrix}1&0\\0&\sin\alpha\end{vmatrix}=-(\sin\alpha-0)=-\sin\alpha,$
$\text{A}_{33}=+\begin{vmatrix}1&0\\0&\cos\alpha\end{vmatrix}=+(\cos\alpha-0)=\cos\alpha$
$\therefore\ \text{adj.A}=\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj. A}=-\begin{bmatrix}-1&0&0\\0&-\cos\alpha&-\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}1&0&0\\0&\cos\alpha&\sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}$
View full question & answer→Question 614 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$
Answer$\begin{vmatrix}49&1&6\\39&7&4\\26&2&3 \end{vmatrix}$
Applying: C1 → C1 + (-8)C3
$=\begin{vmatrix}1&1&6\\7&7&4\\2&2&3 \end{vmatrix}=0$
$\because\text{C}_1=\text{C}_2$
View full question & answer→Question 624 Marks
Solve the following system of homogeneous linear equations:
3x + y + z = 0,
x - 4y + 3z = 0,
2x + 5y - 2z = 0
AnswerGiven,
3x + y + z = 0,
x - 4y + 3z = 0,
2x + 5y - 2z = 0
$\text{D}=\begin{vmatrix}3&1&1\\1&-4&3\\2&5&-2\end{vmatrix}=0$
The system has infinitely many solution Puting z = k in the first two equation we get
3x + y = -k
x - 4y = -3k
Solving these equations by Cramer's rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{\begin{vmatrix}-\text{k}&1\\-3\text{k}&-4\end{vmatrix}}{\begin{vmatrix}3&1\\1&-4\end{vmatrix}}=-\frac{7\text{k}}{13}$
$\text{y}=\frac{\text{D}_2}{\text{D}} =\frac{\begin{vmatrix}3&-\text{k}\\1&-3\text{k}\end{vmatrix}}{\begin{vmatrix}3&1\\1&-4\end{vmatrix}}=\frac{8\text{k}}{13} \text{z}=\text{k}$
$\text{z}=\text{k}$
$\Rightarrow\text{x}=-\frac{7\text{k}}{13},\text{ y}=\frac{8\text{k}}{13}$ and $\text{z}=\text{x}$
Or x = -7k, y = 8k and z = 13k
Clearly, these value satisfy the thried equation.
Thus, x = -7k, y = 8k, z = 13k $[\text{k}\in\text{R}]$
View full question & answer→Question 634 Marks
If a, b, c are real numbers such that $\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}=0,$ then show that either a + b + c = 0 or a = b= c.
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})&2(\text{a}+\text{b}+\text{c})\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$ [Applying R1 → R1 + R2 + R3]
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{c}+\text{a}&\text{a}+\text{b}&\text{b}+\text{c}\\\text{a}+\text{b}&\text{b}+\text{c}&\text{c}+\text{a}\end{vmatrix}$
$=2(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&0&0\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}-\text{a}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}$ [Applying C2 → C2 - C1 and C3 → C3 - C1]
$=2(\text{a}+\text{b}+\text{c})\left\{1\begin{vmatrix}\text{b}-\text{c}&\text{b}-\text{a}\\\text{c}-\text{a}&\text{c}-\text{b}\end{vmatrix}\right\}$
$=2(\text{a}+\text{b}+\text{c})\{(\text{b}-\text{c})(\text{c}-\text{b})-(\text{b}-\text{a})(\text{c}-\text{a})\}$
$=-2(\text{a}+\text{b}+\text{c})\{\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\}$
$=-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}$
But $\triangle=0$ [Given]
$\Rightarrow-(\text{a}+\text{b}+\text{c})\{(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\}=0$
Either,
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2=0$
$\Rightarrow(\text{a}+\text{b}+\text{c})=0$ or $\text{a}=\text{b}=\text{c}$
Hence proved.
View full question & answer→Question 644 Marks
Find the adjoint of the following matrices: $\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$ Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
AnswerHere, $\text{A}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$
cofactors of A are:
C11 = d
C12 = -d
C21 = -b
C22 = a
$\therefore\ \text{adjoint A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} \\ \text{C}_{21} & \text{C}_{22} \end{bmatrix}^\text{T}$
$\therefore\ \text{adjoint A}=\begin{bmatrix}\text{d} & -\text{C} \\ -\text{b} & \text{a} \end{bmatrix}$
$=\begin{bmatrix}\text{d} & -\text{b} \\ -\text{C} & \text{a} \end{bmatrix}$
Now, $\text{adjoint A}=\begin{bmatrix}\text{d} & -\text{b} \\ -\text{C} & \text{a} \end{bmatrix}\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & \text{bd}-\text{bd} \\ -\text{ac}+\text{ac} & \text{ad}-\text{bc} \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & 0 \\ 0 & \text{ad}-\text{bc} \end{bmatrix}$
And, $|\text{A}|\text{I}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$=(\text{ad}-\text{bc})\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & 0 \\ 0 & \text{ad}-\text{bc} \end{bmatrix}$
Also, $\text{A(adjoint A)}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \text{d} \end{bmatrix}\begin{bmatrix} \text{d} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
$=\begin{bmatrix}\text{ad}-\text{bc} & 0 \\ 0 & \text{ad}-\text{bc} \end{bmatrix}$
$\therefore\ \text{(adjoint A)A}=|\text{A}|\text{I}=\text{A(adjoint A)}$
View full question & answer→Question 654 Marks
Solve the following systems of linear equations by cramer's rule:
5x + 7y = -2
4x + 6y = -3
AnswerLet $\text{D}=\begin{vmatrix}5&7\\4&6\end{vmatrix}=-2$
$\text{D}_1=\begin{vmatrix}-2&7\\-3&6\end{vmatrix}=9$
$\text{D}_2=\begin{vmatrix}5&-2\\4&-3\end{vmatrix}=-7$
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{9}{2}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-7}{2}$
View full question & answer→Question 664 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}2&3&7\\13&17&5\\15&20&12 \end{vmatrix}$
Answer$\begin{vmatrix}2&3&7\\13&17&5\\15&20&12 \end{vmatrix}$
Use: R3 → R3 - R2
$=\begin{vmatrix}2&3&7\\13&17&5\\2&3&7 \end{vmatrix}$
$=0$
$\because\text{R}_3=\text{R}_1$
View full question & answer→Question 674 Marks
For what value of x the matrix A is singular?
$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
Answer$\text{A}=\begin{vmatrix}1+\text{x}&7\\3-\text{x}&8 \end{vmatrix}$
A matrix A is called singular if |A| = 0
Now expanding along the first row |A|
$=(\text{x}-1)\begin{vmatrix}\text{x}-1&1\\1&\text{x}-1 \end{vmatrix}-1\begin{vmatrix}1&1\\1&\text{x}-1 \end{vmatrix}+1\begin{vmatrix}1&1\\\text{x}-1&-1 \end{vmatrix}$
$=(\text{x}-1)\big[(\text{x}-1)^2-1\big]-1[\text{x}-1-1]+1[1-\text{x}+1]$
$=(\text{x}-1)(\text{x}^2+1-2\text{x}-1)-1(\text{x}-2)+1(2-\text{x})$
$=(\text{x}-1)(\text{x}^2-2\text{x})-\text{x}+2+2-\text{x}$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+(4-2\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)+2(2-\text{x})$
$=(\text{x}-1)\times(\text{x})\times(\text{x}-2)-2(\text{x}-2)$
$=(\text{x}-2)[\text{x}(\text{x}-1)-2]$ (Taking (x - 2) common)
Since A is a singular matrix, so |A| = 0
i.e., $(\text{x}-2)(\text{x}^2-\text{x}-2)=0$
either $(\text{x}-2)=0$ or $\text{x}^2-\text{x}-2=0$
$\text{x}=2$ or $\text{x}^2-2\text{x}+\text{x}-2=0$
$\text{x}(\text{x}-2)+1(\text{x}-2)=0$
$(\text{x}-2)(\text{x}+1)=0$
$\text{x}=2,-1$
$\text{x}=2\text{ or}-1$
View full question & answer→Question 684 Marks
Find the inverse of the matrix $\text{A}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$ ans show that aA-1 = (a2 + bc + 1) I - aA.
Answer$\text{A}=\begin{bmatrix}\text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$
$\Rightarrow\ |\text{A}|=(1+\text{bc})-\text{bc}=1\neq0$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{1}{1}\text{A}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
Now aA-1 = (a2 + bc + 1)I - aA
$\text{L.H.S}:\text{aA}^{-1}=\text{a}\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
$=\begin{bmatrix} 1+\text{bc} & -\text{ab} \\ -\text{ac} & \text{a}^2 \end{bmatrix}$
$\text{R.H.S}:(\text{a}^2+\text{bc}+1)\text{I}-\text{aA}$
$=\begin{bmatrix}\text{a}^2+\text{bc}+1 & 0 \\0 & \text{a}^2+\text{bc}+1 \end{bmatrix}-\begin{bmatrix}\text{a}^2 & \text{ab} \\ \text{ac} & 1+\text{bc} \end{bmatrix}$
$=\begin{bmatrix}1+\text{bc} & -\text{ab} \\ -\text{ac} & \text{a}^2 \end{bmatrix}$
Since, L.H.S = R.H.S
Hence proved.
View full question & answer→Question 694 Marks
Find the inverse of the following matrices and verify that A-1 A = I3.
$\begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{bmatrix}4 & 3 \\ 3 & 4 \end{bmatrix}=7,\ \text{C}_{12}=-\begin{bmatrix}1 & 3 \\ 1 & 4 \end{bmatrix}=-1$
$\text{and C}_{13}=\begin{bmatrix}1 & 4 \\ 1 & 3 \end{bmatrix}=-1$
$\text{C}_{21}=-\begin{bmatrix}3 & 3 \\ 3 & 4 \end{bmatrix}=-3,\ \text{C}_{22}=\begin{bmatrix}1 & 3 \\ 1 & 4 \end{bmatrix}=1$
$\text{and C}_{23}=-\begin{bmatrix}1 & 3 \\ 1 & 3 \end{bmatrix}=0$
$\text{C}_{31}=\begin{bmatrix}3 & 3 \\ 4 & 3 \end{bmatrix}=-3,\ \text{C}_{32}=-\begin{bmatrix}1 & 3 \\ 1 & 3 \end{bmatrix}=0$
$\text{and C}_{33}=\begin{bmatrix}1 & 3 \\ 1 & 4 \end{bmatrix}=1$
$\text{adj A}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -3 & 0 & 1\end{bmatrix}^\text{T}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{bmatrix}$
and |A| = 1
$\text{A}^{-1}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{bmatrix}$
Now, $\text{A}^{-1}\text{A}=\begin{bmatrix}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\text{I}_3$
View full question & answer→Question 704 Marks
Let $\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$ and
$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that
$\big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$
Answer$\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\ \text{F}(-\alpha)=\begin{bmatrix}\cos(-\alpha) & -\sin(-\alpha) & 0 \\ \sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\-\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{vmatrix}\cos\alpha & 0 \\0 & 1 \end{vmatrix}=\cos\alpha,\ \text{C}_{12}=-\begin{vmatrix}\sin\alpha & 0 \\ 0 & 1 \end{vmatrix}=-\sin\alpha$
$\text{and C}_{13}=\begin{vmatrix}\sin\alpha & \cos\alpha \\0 & 0 \end{vmatrix}=0$
$\text{C}_{21}=-\begin{vmatrix}-\sin\alpha & 0 \\0 & 1 \end{vmatrix}=\sin\alpha,\ \text{C}_{22}=\begin{vmatrix}\cos\alpha & 0 \\0 & 1 \end{vmatrix}=\cos\alpha$
$\text{and C}_{23}=-\begin{vmatrix}\cos\alpha & -\sin\alpha \\0 & 0 \end{vmatrix}=0$
$\text{C}_{31}=\begin{bmatrix} -\sin\alpha & 0 \\ \cos\alpha & 0 \end{bmatrix}=0, \text{C}_{32}=-\begin{bmatrix}\cos\alpha & 0 \\ \sin\alpha & 0 \end{bmatrix}=0$
$\text{and C}_{33}=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ 0 & 0 \end{bmatrix}=0$
$\Rightarrow\ \text{adj}\left\{\text{F}(\alpha)\right\}=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\ |\text{F}(\alpha)|=1$
$\therefore\ \big[\text{F}(\alpha)\big]^{-1}=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\ .....(\text{i})$
$\Rightarrow\ \big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$
View full question & answer→Question 714 Marks
Using properties of determinants, prove that:
$\begin{vmatrix}3\text{a}&-\text{a+b}&-\text{a+c}\\-\text{b+a}&3\text{b}&-\text{b+c}\\-\text{c+a}&\text{c+b}&3\text{c}\end{vmatrix}=3 (\text{a + b + c}) (\text{ab + bc + ca})$
Answer$\triangle=\begin{vmatrix}3\text{a}&-\text{a + b}&-\text{a + c}\\-\text{b + a}&3\text{b}&-\text{b + c}\\-\text{c + a}&\text{c + b}&3\text{c}\end{vmatrix}$
Applying C1 → C1 + C2 + C3, we have:
$\triangle=\begin{vmatrix}\text{a + b + c}&-\text{a + b}&-\text{a + c}\\\text{a + b + c}&3\text{b}&-\text{b + c}\\\text{a + b + c}&-\text{c + b}&3\text{c}\end{vmatrix}$
$= (\text{a + b + c})\begin{vmatrix}1&-\text{a + b}&-\text{a + c}\\1&3\text{b}&-\text{b + c}\\1&-\text{c + b}&3\text{c}\end{vmatrix}$
Applying R2 → R2 - R1 and R3 → R3 - R1, we have:
$\triangle=(\text{a + b + c})\begin{vmatrix}1&-\text{a + b}&-\text{a + c}\\0&2\text{b + a}&\text{a}-\text{b}\\0&\text{a}-\text{c}&2\text{c + a}\end{vmatrix}$
Expanding along C1, we have:
$\triangle$ = (a + b + c) [(2b + a) (2c + a) - (a - b) (a - c)]
= (a + b + c) [4bc + 2ab + 2ac + a2 - a2 + ac + ba - bc]
= (a + b + c) (3ab + 3bc + 3ac)
= 3(a + b + c) (ab + bc + ca)
Hence, the given result is proved.
View full question & answer→Question 724 Marks
For the following pairs of matrices verify that (AB)-1 = B-1 A-1:
$\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\text{ and B}=\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\therefore|\text{A}|=1\neq0$
$\text{adj A}=\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}\Rightarrow\ \text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\frac{1}{1}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$\text{B}=\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}\therefore|\text{B}|=-10\neq0$
$\text{adj B}=\begin{bmatrix}2 & -6 \\-3 & 4 \end{bmatrix} \Rightarrow\ \text{B}^{-1}=\frac{\text{adj B}}{|\text{B}|}=\frac{1}{-10}\begin{bmatrix}2 & -6 \\-3 & 4 \end{bmatrix}$
Also, $\text{AB}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}=\begin{bmatrix}18 & 22 \\43 & 52 \end{bmatrix}$
$|\text{AB}|=936-946=-10\neq0$
$\text{adj(AB)}=\begin{bmatrix}52 & -22 \\-43 & 18 \end{bmatrix}$
$(\text{AB})^{-1}=\frac{\text{adj(AB)}}{|\text{AB}|}=\frac{1}{|\text{Ab}|}\begin{bmatrix}52 & -22 \\-43 & 18 \end{bmatrix}$
$=\frac{1}{-10}\begin{bmatrix}+52 & -22 \\-43 & +18 \end{bmatrix}=\frac{1}{10}\begin{bmatrix}-52 & 22 \\43 & -18 \end{bmatrix}$
$\text{B}^{-1}\text{A}^{-1}=\frac{-1}{10}\begin{bmatrix}2 & -6 \\-3 & 4 \end{bmatrix}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$=\frac{-1}{10}\begin{bmatrix}52 & -22 \\-43 & 18 \end{bmatrix}=\frac{1}{10}\begin{bmatrix}-52 & 22 \\43 & -18 \end{bmatrix}$
Hence, (AB)-1 = B-1A-1
View full question & answer→Question 734 Marks
If $\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0,$ then using properties of determinants, find the value of $\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}},$ where $\text{x},\text{y},\text{z}\neq0.$
Answer$\Rightarrow\begin{vmatrix}\text{a}&\text{b}-\text{y}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
R1 → R1 - R2
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\\text{a}-\text{x}&\text{b}&\text{c}-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
R2 → R2 - R3
$\Rightarrow\begin{vmatrix}\text{x}&-\text{y}&0\\0&\text{y}&-\text{z}\\\text{a}-\text{x}&\text{b}-\text{y}&\text{c}\end{vmatrix}=0$
Expanding along first row, we get
$\text{x}(\text{yc}+\text{zb}-\text{zy})+\text{y}(0-\text{za}+\text{zx})=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
$\Rightarrow\text{xyc}+\text{xzb}-2\text{xyz}+\text{zya}=0$
Dividing by xyz, we get
$\frac{\text{c}}{\text{z}}+\frac{\text{b}}{\text{y}}-2+\frac{\text{a}}{\text{x}}=0$
$\therefore\frac{\text{a}}{\text{x}}+\frac{\text{b}}{\text{y}}+\frac{\text{c}}{\text{z}}=2$
View full question & answer→Question 744 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = 17,
3x + 5y = 6
AnswerGiven, 2x - y = 17
3x + 5y = 6
Using cramers Rule, we get
$\text{D}=\begin{vmatrix}2&1\\3&5\end{vmatrix}=10+3=13$
$\text{D}_1=\begin{vmatrix}17&-1\\6&5\end{vmatrix}=85+6=91$
$\text{D}_2=\begin{vmatrix}2&17\\3&6\end{vmatrix}=12-51=-39$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{13}=7$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-39}{13}=-3$
$\therefore\text{x}=7$ and $\text{y}=-3$
View full question & answer→Question 754 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = 1,
7x - 2y = -7
AnswerLet $\text{D}=\begin{vmatrix}2&-1\\7&-2 \end{vmatrix}=-4+7=3$
$\text{D}_1=\begin{vmatrix}1&-1\\-7&-2 \end{vmatrix}=-9$
$\text{D}_2=\begin{vmatrix}2&1\\7&-7\end{vmatrix}=-21$
Now, $\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-9}{3}=-3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-21}{3}=-7$
Hence x = -3, y = -7
View full question & answer→Question 764 Marks
Show that the matrix $\text{A}=\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$ satisfies the equation, A3 - A2 - 3A - I3 = 0. Hence, find A-1.
AnswerWe have, $\text{A}=\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$
$\Rightarrow\ |\text{A}|=\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}=1(-9)+0-2(-8)=-9+16=7$
Since, $|\text{A}|\neq0$
Hence, A-1 exists.
Now,
$\text{A}^2=\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & -2 \\-2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$
$=\begin{bmatrix}1+0-6 & 0+0-8 & -2+0-2 \\ -2+2+6 & 4-2+2 & 4-2+2 \\ 3-8+3 & 0-4+4 & -6+8+1\end{bmatrix}$
$=\begin{bmatrix} -5 & -8 & -4 \\ -6 & 9 & 4 \\ -2 & 0 & 3 \end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}=\begin{bmatrix} -5 & -8 & 2 \\ 6 & 9 & 4 \\ -2 & 0 &3 \end{bmatrix}\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ -2 & -1 & 2 \end{bmatrix}$
$=\begin{bmatrix} 5+16-12 & 0+8+16 & 10-16-4 \\ 6-18+12 & 0-9+16 & -12+18+4 \\ -2+0+9 & 0+0+12 & 4+0+3 \end{bmatrix}$
$=\begin{bmatrix}-1 & -8 & -10 \\0 & 7 & 10 \\ 7 & 12 & 7 \end{bmatrix}$
Now, $\text{A}^3-\text{A}^2-3\text{A}-\text{I}_3=\begin{bmatrix}-1 & -8 & -10 \\0 & 7 & 10 \\ 7 & 12 & 7 \end{bmatrix}-\begin{bmatrix} -5 & -8 & -4 \\ 6 & 9 & 4 \\ -2 & 0 & 3\end{bmatrix}$
$=-3\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ -2 & -1 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}-1+5-3-1 & -8+8+0+0 & -10+4+6-0 \\ 0-6+6-0 & 7-9+3-1 & 10-4-6-0 \\ 7+2-9-0 & 12+0-12-0 & 7-3-3-1\end{bmatrix}$
$=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}=0$
Hence proved.
Now, A3 - A2 - 3A - I3 = 0(Null matrix)
⇒ A-1(A3 - A2 - 3A - I3) = A-1 0 (Pre - multiplying by A-1)
⇒ A2 - A1 - 3I3 = A-1
$\Rightarrow\ \begin{bmatrix}-5 & -8 & -4\\6 & 9 & 4\\ -2 & 0 & 4 \end{bmatrix}-\begin{bmatrix}1 & 0 & 2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}-3\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\text{A}^{-1}$
$\Rightarrow\ \begin{bmatrix}-5-1-3 & -8-0-0 & -4+2+0\\ 6+2+0 & 9+1-3 & 4-2 \\ -2-3-0 & 0-4-0 & 3-1-3 \end{bmatrix}$
$=\begin{bmatrix} -9 & -8 & -2\\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix}=\text{A}^{-1}$
$\Rightarrow\ \text{A}^{-1}=\begin{bmatrix} -9 & -8 & -2\\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix}$
View full question & answer→Question 774 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}1&\text{a}&\text{bc}\\1&\text{b}&\text{ca}\\1&\text{c}&\text{ab} \end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}\text{b} &\text{ca}\\ \text{c}&\text{ab} \end{vmatrix}=\text{ab}^{2}-\text{c}^2\text{a}=\text{a}(\text{b}^2-\text{c}^2)$
$\text{M}_{21}=\begin{vmatrix}\text{a} &\text{bc}\\ \text{c}&\text{ab} \end{vmatrix}=\text{a}^{2}\text{b}-\text{c}^2\text{b}=\text{b}(\text{a}^2-\text{c}^2)$
$\text{M}_{31}=\begin{vmatrix}\text{a} &\text{bc}\\ \text{b}&\text{ca} \end{vmatrix}=\text{a}^{2}\text{c}-\text{b}^2\text{c}=\text{c}(\text{a}^2-\text{b}^2)$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=\text{a}(\text{b}^2-\text{c}^2)$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{21}=-\text{b}(\text{a}^2-\text{c}^2)$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=\text{c}(\text{a}^2-\text{b}^2)$
$\text{D}=1\text{a}(\text{b}^2-\text{c}^2)-\text{a}(\text{ab}-\text{ca})+\text{b}(\text{c}-\text{b})$
$=\text{ab}^2-\text{ac}^2-\text{a}^2\text{b}+\text{a}^2\text{c}+\text{c}^2\text{b}-\text{b}^2\text{c}$
$=\text{a}^{2}(\text{c}-\text{b})+\text{b}^{2}(\text{a}-\text{c})+\text{c}^{2}(\text{b}-\text{a})$
View full question & answer→Question 784 Marks
Evaluate the following:
$\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$\triangle=\begin{vmatrix}\text{x}&1&1\\1&\text{x}&1\\1&1&\text{x}\end{vmatrix}$
$=\begin{vmatrix}\text{x}-1&1-\text{x}&0\\1&\text{x}&1\\0&1-\text{x}&\text{x}-1\end{vmatrix}$ [Applying R1 → R1 - R2 and R3 → R3 - R2]
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}&1\\0&-1&1\end{vmatrix}$
$=(\text{x}-1)^2\begin{vmatrix}1&-1&0\\1&\text{x}+1&1\\0&0&1\end{vmatrix}$ [Applying C2 → C2 + C3]
$=(\text{x}-1)^2(\text{x}+1+1)$ [Expanding along last row]
$=(\text{x}-1)^2(\text{x}+2)$
$\therefore\triangle=(\text{x}-1)^2(\text{x}+2)$
View full question & answer→Question 794 Marks
Solve the follwing system of equations by matrix method:
3x + 4y - 5 = 0
x - y + 3 = 0
AnswerThe given system of equations can be written in matrix form as follow:
$\begin{bmatrix}3&4\\ 1&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ -3\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&4\\ 1&-1\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{ B}=\begin{bmatrix}5\\ -3\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&4\\ 1&-1\end{vmatrix}$
$=-3-4$
$=-7\neq0$
So, the given system has a unique solution given by X = A-1 B.
Let Cij be the cofactors of the elements aij in A = [aij]. Then,
$\text{C}_{11}=(-1)^{1+1}(-1)=-1,\text{C}_{12}=(-1)^{1+2}(1)=-1$
$\text{C}_{21}=(-1)^{2+1(4)}=-4,\text{C}_{22}=-{(-1)}^{2+2}{(3)}={(3)}$
$\text{adj A}=\begin{bmatrix}-1&-1\\ -4&3\end{bmatrix}^{T}=\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-7}\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}$
X = A-1 B
$=\frac{1}{-7}\begin{bmatrix}-1&-4\\ -1&3\end{bmatrix}\begin{bmatrix}5\\ -3\end{bmatrix}$
$=\frac{1}{-7}\begin{bmatrix}-5+12\\ -5-9\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{7}{-7}\\ \frac{-14}{-7}\end{bmatrix}$
$\therefore$ x = -1 and y = 2
View full question & answer→Question 804 Marks
Show that the following system of linear equation is inconsistent:
4x − 2y = 3
6x − 3y = 5
AnswerThis system can be written as:
$\begin{bmatrix}4&-2\\ 6&-3\end{bmatrix}\begin{bmatrix} \text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 5\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=-12+12=0$
So, A is singular, Now system will be inconsisent, if
$(\text{adj A})\times\text{B}\neq0$
$\text{C}_{11}=-3$
$\text{C}_{12}=-6$
$\text{C}_{21}=2$
$\text{C}_{22}=4$
$\text{adj A}=\begin{bmatrix}-3&-6\\ 2&4\end{bmatrix}^\text{T}\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}$
$(\text{adj A})\times(\text{B})=\begin{bmatrix}-3&2\\ -6&4\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$
$=\begin{bmatrix}-9+10\\ -18+20\end{bmatrix}$
$=\begin{bmatrix}1\\ 2\end{bmatrix}$
$\neq0$
Hence, the above system is inconsisent.
View full question & answer→Question 814 Marks
Solve the following systems of homogeneous linear equations by matrix method:
2x + 3y - z = 0
x - y - 2z = 0
3x + y + 3z = 0
Answer2x + 3y - z = 0 x - y - 2z = 0 3x + y + 3z = 0 Hence, $\text{A}=\begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix}$ $|\text{A}|=\begin{bmatrix}2&3&-1\\1&-1&-2\\3&1&3\end{bmatrix}$ $=2(-3+2)-3(3+6)-1(4)$ $=-2-27-4$
$\neq0$
Hence, the system has only trivial solutions given by x = y = z = 0 View full question & answer→Question 824 Marks
Solve the follwing system of equations by matrix method:
5x + 2y = 3
3x + 2y = 5
AnswerThe above system can be written in matrix form as: $\begin{bmatrix}5&2\\ 3&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}3\\ 2\end{bmatrix}$ Or $\text{AX = B}$ Where, $\text{A}=\begin{bmatrix}5&2\\ 3&2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix},\text{B}=\begin{bmatrix}3\\ 5\end{bmatrix}$ Now, $\text{|A|}=10-6=4\neq0$
So, the above system has a unique solution, given by $\text{X}=\text{A}^{-1}\text{B}$ Let Cij be the co factor of aij in A, then $\text{C}_{11} = 2,\text{C}_{12} = -3$
$\text{C}_{21} = -2,\text{C}_{22} = 5$
Also, $\text{Adj A}=\begin{bmatrix}2&-3\\ -2&5\end{bmatrix}^\text{T}=\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}$ Now, X = A-1 B $=\frac{1}{4}\begin{bmatrix}2&-2\\ -3&5\end{bmatrix}\begin{bmatrix}3\\ 5\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}-4\\ 16\end{bmatrix}$ $\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-1\\ 4\end{bmatrix}$ Hence, x = -1 y = 4 View full question & answer→Question 834 Marks
Show that the following system of linear equations is consistent and also find solution:
6x + 4y = 2
9x + 6y =3
AnswerHere,
$6\text{x}+4\text{y}=2\ \dots(1)$
$9\text{x}+6\text{y}\ \dots(2)$
AX = B
Here,
$\text{A}=\begin{bmatrix}6&4\\ 9&6\end{bmatrix},\text{X}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\begin{bmatrix}6&4\\ 9&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}2\\ 3\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}6&4\\ 9&6\end{vmatrix}$
$= 36 -36$
$= 0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because (adj A)B ≠ 0 or (adj A) = 0.
Let Cij be the co-factors of the elements aij in A [aij]. Then,
$\text{C}_{11}=6,\\ \text{C}_{12}=-9,\\ \text{C}_{21}=-4,\\ \text{C}_{22}=6$
$\text{adj A}=\begin{bmatrix}6&-9\\ -4&6\end{bmatrix}^\text{T}$
$=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}$
$\text{(adj A) B}=\begin{bmatrix}6&-4\\ -9&6\end{bmatrix}\begin{bmatrix}2\\ 3\end{bmatrix}$
$=\begin{bmatrix}12-12\\ -18+18\end{bmatrix}$
$=\begin{bmatrix}0\\ 0\end{bmatrix}$
If |A| = 0 and (adj A) B = 0, then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions.
Substituting y = k in the eq. (1), we get
$6\text{x}+4\text{k}=2$
$\Rightarrow6\text{x}=2-4\text{k}$
$\Rightarrow\text{x}=\frac{2-\text{4k}}{6}$
$\Rightarrow\text{x}=\frac{1-\text{2k}}{3}$
$\therefore \text{x}=\frac{1-2\text{k}}{3}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\text{x}=\frac{1-2\text{k}}{3}$ and y = k ( where k is a real number ) satisfy the given system of equations.
View full question & answer→Question 844 Marks
If $\text{a}+\text{b}+\text{c}\neq0\text{ and}\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}=0,$ then prove that a = b = c.
AnswerLet $\Delta=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3\big]$
$\Delta=\begin{vmatrix}\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}&\text{a}+\text{b}+\text{c}\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$ $=(\text{a}+\text{b}+\text{c})\begin{vmatrix}1&1&1\\\text{b}&\text{c}&\text{a}\\\text{c}&\text{a}&\text{b}\end{vmatrix}$
$\big[\text{Applying C}_1\rightarrow\text{C}_1-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2-\text{C}_1\big]$
$\Delta=(\text{a}+\text{b}+\text{c})\begin{vmatrix}0&0&1\\\text{b}-\text{a}&\text{c}-\text{a}&\text{a}\\\text{c}-\text{b}&\text{a}-\text{b}&\text{b}\end{vmatrix}$
[Expanding along R1]
$=(\text{a}+\text{b}+\text{c})[(\text{b}-\text{a})(\text{a}-\text{b})-(\text{c}-\text{a})(\text{c}-\text{b})]$
$=(\text{a}+\text{b}+\text{c})(\text{ba}-\text{b}^2-\text{a}^2+\text{ab}-\text{c}^2+\text{cb}+\text{ac}-\text{ab})$
$=-(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[2\text{a}^2+2\text{b}^2-2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}\big]$
$=-\frac{1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}^2+\text{b}^2-2\text{ab})+(\text{b}^2+\text{c}^2-2\text{bc})+(\text{c}^2+\text{a}^2-2\text{ac})\big]$
$-\frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\big]$
Given $\Delta=0$
$\Rightarrow\ \frac{-1}{2}(\text{a}+\text{b}+\text{c})\big[(\text{a}-\text{b})^2+(\text{b}-\text{c})^2+(\text{c}-\text{a})^2\big]=0$
$\Rightarrow\ (\text{a}-\text{b})^2+(\text{b}-\text{c}^2)+(\text{c}-\text{a})^2=0$ $[\because\ \text{a}+\text{b}+\text{c}\neq0,\text{ given}]$
$\Rightarrow\ \text{a}-\text{b}=\text{b}-\text{c}=\text{c}-\text{a}=0$
$\Rightarrow\ \text{a}=\text{b}=\text{c}$
View full question & answer→Question 854 Marks
Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 5&3\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}7\\ 12\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&1\\ 5&3\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}7\\ 12\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 5&3\end{vmatrix}$
$=9-5$
$=4\neq0$
So, the given system has a unique solution given by X = A-1 B.
Let Cij be the co-factors of the elements aij in A = [aij]. then,
$\text{C}_{11}=-{(-1)}^{1+1}{(3)}=3,\text{C}_{12}={(-1)}^{1+2}{(5)}=-5$
$\text{C}_{21}={(-1)}^{2+1}{(1)}=-1,\text{C}_{22}=-{(-1)}^{2+2}{(3)}=3$
$\text{adj A}=\begin{bmatrix}3&-5\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}3&-1\\-5&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj A}$
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}$
X = A-1B
$=\frac{1}{4}\begin{bmatrix}3&-1\\ -5&3\end{bmatrix}\begin{bmatrix}7\\ 12\end{bmatrix}$
$=\frac{1}{4}\begin{bmatrix}21-12\\ -35+36\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{4}\\ \frac{1}{4}\end{bmatrix}$
$\therefore\ \text{x}=\frac{9}{4}\text{ and }\text{y}=\frac{1}{4}$
View full question & answer→Question 864 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}=0,\text{a}\neq\text{b}$
AnswerLet $\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{a}&\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\1&\text{b}&\text{b}^2\end{vmatrix}$ [Applying R2 → R1 - R2]
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{x}-\text{a}&\text{x}^2-\text{a}^2\\0&\text{x}-\text{b}&\text{x}^2-\text{b}^2\end{vmatrix}$ [Applying R3 → R1 - R3]
$=(\text{x}-\text{a})(\text{x}-\text{b})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{x}+\text{a}\\0&1&\text{x}+\text{b}\end{vmatrix}$
$=(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}+\text{b}-\text{x}-\text{a})=0$
$\text{x}=\text{a},\text{b}$
View full question & answer→Question 874 Marks
If $\text{A}=\begin{bmatrix}4 & 3 \\ 2 & 5 \end{bmatrix},$ find x and y such that A2 = zA + yI = 0. Hence, evaluate A-1.
Answer$\text{A}=\begin{bmatrix}4 & 3 \\2 & 5 \end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix}22 & 27 \\ 18 & 31 \end{bmatrix}$
Now, A2 - xA + yI = 0
$\Rightarrow\ \begin{bmatrix}22 & 27 \\ 18 & 31 \end{bmatrix}-\begin{bmatrix}4\text{x} & 3\text{x} \\2\text{x} & 5\text{x} \end{bmatrix}+\begin{bmatrix}\text{y} & 0 \\0 & \text{y} \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}22-4\text{x}-\text{y} & 27-3\text{x} \\ 18-2\text{x} & 31-5\text{x}-\text{y} \end{bmatrix}=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
Thus, we have
22 - 4x + y = 0, 37 - 3x = 0, 18 - 2x = 0 and 31 - 5x + y = 0
⇒ -3x = -27
⇒ x = 9
On putting x = 9 in 22 - 4x + y = 0, we get
22 - 36 + y = 0
⇒ -14 = -y
⇒ y = 14
Now,
A2 - 9A + 14I = 0
⇒ A2 - 9A = -14I
⇒ A-1A2 - 9AA-1 = -14IA-1 [Pre-multiplying both sides by A-1]
⇒ A - 9I = -14A-1
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{14}(\text{A}-9\text{I})$
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{14}\left\{\begin{bmatrix}4 & 3 \\2 & 5 \end{bmatrix}-\begin{bmatrix}9 & 0 \\0 & 9 \end{bmatrix}\right\}$
$=\frac{1}{14}\begin{bmatrix}5 & -3 \\-2 & 4 \end{bmatrix}$
View full question & answer→Question 884 Marks
$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)$
$\text{L.H.S}=\begin{vmatrix}-\text{a}(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^3&2\text{c}^3\\2\text{a}^3&-\text{b}(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^3\\2\text{a}^3&2\text{b}^3&-\text{c}(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
Tkae a, b and c common from C1, C2 and C3 respectively.
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)&2\text{b}^2&2\text{c}^2\\2\text{a}^2&-(\text{c}^2+\text{a}^2-\text{b}^2)&2\text{c}^2\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
Apply: R1 → R1 - R3, R2 → R2 - R3
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)-2\text{a}^2&0&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\0&-(\text{c}^2+\text{a}^2-\text{b}^2)-2\text{b}^2&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}\begin{vmatrix}-(\text{b}^2+\text{c}^2-\text{a}^2)-2\text{a}^2&0&(\text{a}^2+\text{b}^2-\text{c}^2)\\0&-(\text{c}^2+\text{a}^2-\text{b}^2)-2\text{b}^2&2\text{c}^2+(\text{a}^2+\text{b}^2-\text{c}^2)\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-(\text{a}^2+\text{b}^2-\text{c}^2)+2\text{a}^2\end{vmatrix}$
$=\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)^2\begin{vmatrix}-1&0&1\\0&-1&1\\2\text{a}^2&2\text{b}^2&-\text{b}^2+\text{c}^2+\text{a}^2\end{vmatrix}$
$=-\text{abc}(\text{b}^2+\text{c}^2+\text{a}^2)[(-1)(-\text{b}^2+\text{c}^2+\text{a}^2)-(1)(2\text{b}^2)]$
$=\text{abc}(\text{a}^2+\text{b}^2+\text{c}^2)^3$
$=\text{R.H.S}$
View full question & answer→Question 894 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 R2 → R2 - R1]
$=\begin{vmatrix}\text{x}&-2&1\\5-\text{x}&4&0\\8-\text{x}&10&0\end{vmatrix}$ [Applying R3 → R3 - R1]
$=\begin{vmatrix}5-\text{x}&4\\8-\text{x}&10\end{vmatrix}$
$=50-10\text{x}-32+4\text{x}$
$=18-6\text{x}=0$
$\Rightarrow\text{x}=3$
View full question & answer→Question 904 Marks
Prove that:
$\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
Answer$\text{L.H.S}=\begin{vmatrix}-\text{bc}&\text{b}^2+\text{bc}&\text{c}^2+\text{bc}\\\text{a}^2+\text{ac}&-\text{ac}&\text{c}^2+\text{ac}\\\text{a}^2+\text{ab}&\text{b}^2+\text{ab}&-\text{ab}\end{vmatrix}$
Multiply R1, R2 and R3 by a, b and c respectively.
$=\frac{1}{\text{abc}}\begin{vmatrix}-\text{abc}&\text{ab}^2+\text{abc}&\text{ac}^2+\text{abc}\\\text{a}^2\text{b}+\text{abc}&-\text{abc}&\text{bc}^2+\text{abc}\\\text{a}^2\text{c}+\text{abc}&\text{b}^2\text{c}+\text{abc}&-\text{abc}\end{vmatrix}$
Take a, b and c common from C1, C2 and C3 respectively.
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}-\text{bc}&\text{ab}+\text{ac}&\text{ac}+\text{ab}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
Apply: R1 → R1 + R2 + R3
$=\begin{vmatrix}\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}&\text{ab}+\text{bc}+\text{ca}\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}1&1&1\\\text{a}\text{b}+\text{bc}&-\text{ac}&\text{bc}+\text{ab}\\\text{a}\text{c}+\text{bc}&\text{b}\text{c}+\text{ac}&-\text{ab}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})\begin{vmatrix}0&1&0\\\text{a}\text{b}+\text{bc}+\text{ac}&-\text{ac}&\text{bc}+\text{ab}+\text{ac}\\0&\text{b}\text{c}+\text{ac}&-\text{ab}-\text{bc}-\text{ac}\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3\begin{vmatrix}0&1&0\\0&-\text{ac}&1\\0&\text{b}\text{c}+\text{ac}&1\end{vmatrix}$
$=(\text{ab}+\text{bc}+\text{ca})^3$
$=\text{R.H.S}$
View full question & answer→Question 914 Marks
Using Cofactors of elements of third column, evaluate $\triangle=\begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}.$
AnswerThe given determinant is $\begin{vmatrix}1&x&yz\\1&y&zx\\1&z&xy\end{vmatrix}.$
We have:
M13 = $\begin{vmatrix}1&y\\1&z\end{vmatrix}=z-y$
M23 = $\begin{vmatrix}1&x\\1&z\end{vmatrix}=z-x$
M33 = $\begin{vmatrix}1&x\\1&y\end{vmatrix}=y-x$
$\therefore$ A13 = cofactor of a13 = (-1)1+3 M13 = (z - y)
A23 = cofactors of a23 = (-1)2+3 M23 = - (z - x) = (x - z)
A33 = cofactors of a33 = (-1)3+3 M33 = (y - x)
We know that $\triangle$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
$\therefore\triangle$ = a13A13 + a23A23 + a33A33
= yz(z - y) + zx (x - z) + xy (y - x)
= yz2 - y2z + x2z - xz2 + xy2 - x2y
=(x2z - y2z) + (yz2 - xz2) + (xy2 - x2y)
= z(x2 - y2) + z2(y - x) + xy(y - x)
=z(x - y)(x + y) + z2(y - x) + xy(y - x)
=(x- y)[zx + zy - z2 - xy]
=(x - y)[z(x - z) + y(z - x)]
=(x - y)(z - x)[-z + y]
=(x - y)(y - z)(z - x)
View full question & answer→Question 924 Marks
$\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{a}&\text{a}\\1&1&1+\text{a}\end{vmatrix}=\text{a}^3+3\text{a}^2$
Answer$\text{L.H.S}=\begin{vmatrix}1+\text{a}&1&1\\1&1+\text{a}&\text{a}\\1&1&1+\text{a}\end{vmatrix}$
$=1+\text{a}\begin{vmatrix}1+\text{a}&1\\1&1+\text{a}\end{vmatrix}-1\begin{vmatrix}1&1\\1&1+\text{a}\end{vmatrix}+1\begin{vmatrix}1&1+\text{a}\\1&1\end{vmatrix}$
$=(1+\text{a})[(1+\text{a})^2-1]-1(1+\text{a}-1)+(1-1-\text{a})$
$=(1+\text{a})[1+\text{a}^2+2\text{a}-1]-\text{a}-\text{a}$
$=1+\text{a}+\text{a}^2+\text{a}^3+2\text{a}+2\text{a}^2-2\text{a}$
$=\text{a}^3+3\text{a}^2$
$=\text{R.H.S}$
View full question & answer→Question 934 Marks
Show that the following system of linear equations is consistent and also find solution:
2x + 3y = 5
6x + 9y = 15
Answer$2\text{x}+3\text{y}=5\dots(1)$
$6\text{x}+9\text{y}=15\dots(2)$
Or , AX = B
Where,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\begin{bmatrix}2&3\\ 6&9\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}5\\ 15\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}2&3\\6&9\end{vmatrix}$
$=18-18$
$=0$
So, A is singular. Thus, the given system of equations is either inconsistent or it is consistent with infinitely many solutions because $(\text{adj A})\text{B}\neq0\text{ or }(\text{adj A})=0$.
C11 = 9, C12 = -6, C21 = -3 and C22 = 2
$\therefore\ \text{adj A}=\begin{bmatrix}9&-6\\-3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}2&-3\\-6&9\end{bmatrix}$
$\Rightarrow(\text{adj A})\text{B}=\begin{bmatrix}9&-3\\-6&2\end{bmatrix}\begin{bmatrix}5\\15\end{bmatrix}$
$=\begin{bmatrix}45-45\\-30+30\end{bmatrix}$
$=\begin{bmatrix}0\\0\end{bmatrix}$
If |A| = 0 and (adj A) B = 0, then the system is consistent and has infinitely many solutions.
Thus, AX = B has infinitely many solutions.
Substituting y = k in eq. (1), we get
$2\text{x} + 3\text{k}=5$
$\Rightarrow2\text{x}=5-3\text{k}$
$\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$
These values of x and y satisfy the third equation.
Thus, $\Rightarrow\text{x}=\frac{5-3\text{k}}{2}\text{ and }\text{y}=\text{k}$ (where k is a real number) satisfy the given system of equations.
View full question & answer→Question 944 Marks
Using properties of determinants, prove that:
$\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$
Answer$\triangle=\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta&\beta^2&\gamma+\alpha\\\gamma&\gamma^2&\alpha+\beta\end{vmatrix}$
Applying R2 → R2 - R1 and R3 → R3 - R1, we have:
$\triangle=\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\\beta-\alpha&\beta^2-\alpha^2&\alpha-\beta\\\gamma-\alpha&\gamma^2-\alpha^2&\alpha-\gamma\end{vmatrix}$
$=(\beta-\alpha)(\gamma-\alpha)\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\1&\beta+\alpha&-1\\0&\gamma+\alpha&-1\end{vmatrix}$
Applying R3 → R3 - R2, we have:
$\triangle=(\beta-\alpha)(\gamma-\alpha)\begin{vmatrix}\alpha&\alpha^2&\beta+\gamma\\1&\beta+\alpha&-1\\1&\gamma-\beta&0\end{vmatrix}$
Expanding along R3, we have:
$\triangle=(\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$
$=(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)$
$=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$
Hence, the given result is proved.
View full question & answer→Question 954 Marks
Show that $\triangle\text{ABC}$ is an isosceles triangle, if the determinant $\triangle=\begin{vmatrix}1&1&1\\1+\cos\text{A}&1+\cos\text{B}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}&\cos^2\text{B}+\cos\text{B}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0.$
AnswerWe have, $\triangle=\begin{vmatrix}1&1&1\\1+\cos\text{A}&1+\cos\text{B}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}&\cos^2\text{B}+\cos\text{B}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0$
$ \triangle=\begin{vmatrix}0&0&0\\\cos\text{A}-\cos\text{C}&\cos\text{B}-\cos\text{C}&1+\cos\text{C}\\\cos^2\text{A}+\cos\text{A}-\cos^2\text{C}-\cos\text{C}&\cos^2\text{B}+\cos\text{B}-\cos^2\text{C}-\cos\text{C}&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0.$
$[\text{C}_1\rightarrow\text{C}_1-\text{C}_3\text{ and C}_2\rightarrow\text{C}_2+\text{C}_3]$
$=\begin{vmatrix}0&0&1\\1 & 1&1+\cos\text{C}\\\cos\text{A}+\cos\text{C}+1&\cos\text{B}+\cos\text{C}+1&\cos^2\text{C}+\cos\text{C}\end{vmatrix}=0$
$\big[\text{Taking }(\cos\text{A}\cos\text{C})\text{ common from C}_1\text{ and }(\cos\text{B}-\cos\text{C})\text{ common from C}_2\big]$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})\big[(\cos\text{B}+\cos\text{C}+1)-(\cos\text{A}+\cos\text{C}+1)\big]=0$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})(\cos\text{B}+\cos\text{C}+1-\cos\text{A}-\cos\text{C}-1)=0$
$\Rightarrow\ (\cos\text{A}-\cos\text{C}).(\cos\text{B}-\cos\text{C})(\cos\text{B}-\cos\text{A})=0$
$\text{i.e., }\cos\text{A}=\cos\text{C or }\cos\text{B}=\cos\text{C or}\cos\text{B}=\cos\text{A}$
$\Rightarrow\ \text{A = C or B = C or B = A}$
Hence, ABC is an isosceles triangle.
View full question & answer→Question 964 Marks
If $\begin{bmatrix}4-\text{x}&4+\text{x}&4+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0,$ then find values of x.
AnswerGiven, $\begin{bmatrix}4-\text{x}&4+\text{x}&4+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}12+\text{x}&12+\text{x}&12+\text{x}\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$ $[\because\ \text{R}_1\rightarrow\text{R}_1+\text{R}_2+\text{R}_3]$
$\Rightarrow\ (12+\text{x})\begin{bmatrix}1&1&1\\4+\text{x}&4-\text{x}&4+\text{x}\\4+\text{x}&4+\text{x}&4-\text{x}\end{bmatrix}=0$ [Taking (12 + x) common from R1]
$\Rightarrow\ (12+\text{x})\begin{vmatrix}0&0&0\\0&8&4+\text{x}\\2\text{x}&8&4-\text{x}\end{vmatrix}=0$ $\big[\because\text{C}_1\rightarrow\text{C}_ 1-\text{C}_ 3\text{ and }\text{C}_ 2\rightarrow\text{C}_ 2+\text{C}_ 3]$
$\Rightarrow\ (12+\text{x})\big[1.(-16\text{x})\big]=0$
$\Rightarrow\ (12+\text{x})(-16\text{x})=0$
$\therefore\ \text{x}=-12,0$
View full question & answer→Question 974 Marks
Solve the following systems of homogeneous linear equations by matrix method:
2x - y + z = 0
3x + 2y - z = 0
x + 4y + 3z = 0
AnswerThe given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}2&-1&1\\3&2&-1\\1&4&3\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$|\text{A}|=\begin{vmatrix}2&-1&1\\3&2&-1\\1&4&3\end{vmatrix}$
$=2(6+4)+1(9+1)+1(12-2)$
$=40$
$\therefore\ |\text{A}|\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0
View full question & answer→Question 984 Marks
Show that the following system of linear equation is inconsistent:
2x + 3y = 5
6x + 9y = 10
AnswerThe given system of equations can be expresesed as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}2&3\\ 6&9\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}5\\ 10\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}2&3\\ 6&9\end{vmatrix}$
$={(18-18)}$
$=0$
Let Cij be the co-factors of the elements aij in A = [aij]. Then,
$\text{C}_{11}={(-1)}^{1+1}{(9)}=9,\\ \text{C}_{12}={(-1)}^{1+2}{(6)}=-6$
$\text{C}_{21}={(-1)}^{2+1}{(3)}=-3,\\ \text{C}_{22}={(-1)}^{2+2}{(6)}=2$
$\text{adj A}=\begin{bmatrix}9&-6\\ -3&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}$
$\text{(adj A) = B}=\begin{bmatrix}9&-3\\ -6&2\end{bmatrix}\begin{bmatrix}5\\ 10\end{bmatrix}$
$=\begin{bmatrix}45-30\\ -30+30\end{bmatrix}$
$=\begin{bmatrix}15\\ -10\end{bmatrix}\neq0$
Hence, the given system of equations is inconsistent.
View full question & answer→Question 994 Marks
If $\triangle=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix},$ $\triangle_1=\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix},$ then prove that $\triangle+\triangle_1=0$
Answer$\triangle+\triangle_1=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&1&1\\\text{yz}&\text{zx}&\text{xy}\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}+\begin{vmatrix}1&\text{yz}&\text{x}\\1&\text{zx}&\text{y}\\1&\text{xy}&\text{z} \end{vmatrix}$
$[$Interchanging rows and coloumns in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\1&\text{y}&\text{y}^2\\1&\text{z}&\text{z}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\1&\text{y}&\text{zx}\\1&\text{z}&\text{xy} \end{vmatrix}$
$[$Applying $\text{C}_2\leftrightarrow\text{C}_3$ in $\triangle_1]$
$=\begin{vmatrix}1&\text{x}&\text{x}^2\\0&\text{y}-\text{x}&\text{y}^2-\text{x}^2\\0&\text{z}-\text{x}&\text{z}^2-\text{x}^2\end{vmatrix}-\begin{vmatrix}1&\text{x}&\text{yz}\\0&\text{y}-\text{x}&\text{zx}-\text{yz}\\0&\text{z}-\text{x}&\text{xy}-\text{yz}\end{vmatrix}$
[Applying R2 → R2 - R1 and R3 → R3 - R1]
$=(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{x}^2\\0&1&\text{y}+\text{x}\\0&1&\text{z}+\text{x}\end{vmatrix}-(\text{y}-\text{x})(\text{z}-\text{x})\begin{vmatrix}1&\text{x}&\text{yz}\\0&1&-\text{z}\\0&1&-\text{y}\end{vmatrix}$
[Taking (y - x) common from R2 and (z - x) common from R3]
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}+\text{x}-\text{y}-\text{x})-(\text{y}-\text{x})(\text{z}-\text{x})(-\text{y}+\text{z})$
[Expanding along first column]
$=(\text{y}-\text{x})(\text{z}-\text{x})(\text{z}-\text{y})(1-1)$
$=0$
$\therefore\ \triangle+\triangle_1=0$
View full question & answer→Question 1004 Marks
If $\text{A}=\begin{bmatrix}2 & 3 \\ 1 & 2 \end{bmatrix},$ verify that A2 - 4A + I = 0, where $\text{I}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{ and O}\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}.$ Hence find A-1.
Answer$\text{A}=\begin{bmatrix}2 & 3 \\ 1 & 2 \end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$
and $\text{A}^2-4\text{A}+\text{I}$
$=\begin{bmatrix}7 & 12 \\4 & 7 \end{bmatrix}-\begin{bmatrix}8 & 12 \\4 & 8 \end{bmatrix}+\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$
$\Rightarrow\ \text{A}^2-4\text{A}+\text{I}$
$=\begin{bmatrix}7-8+1 & 12-12+0 \\4-4+0 & 7-8+1 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}=0$
⇒ A2 - 4A + I
⇒ A2 - 4A = -I
⇒ A-1A2 - 4AA-1 = -IA-1 [Pre-multiplying both sides by A-1]
⇒ A - 4I = -A-1
⇒ A-1 = 4I - A
$\Rightarrow\ \text{A}^{-1}=\left\{\begin{bmatrix}4 & 0 \\0 & 4 \end{bmatrix}-\begin{bmatrix}2 & 3 \\ 1 & 2 \end{bmatrix}\right\}$
$=\begin{bmatrix}2 & -3 \\-1 & 2 \end{bmatrix}$
View full question & answer→