Question 1515 Marks
Solve the equation $\begin{vmatrix}x+a&x&x\\x&x+a&x\\x&x&x+a\end{vmatrix}=0,a\neq0$
Answer$\begin{vmatrix}x+a&x&x\\x&x+a&x\\x&x&x+a\end{vmatrix}=0$
Applying $R_1\rightarrow R_1 + R_2 + R_3,$ we get:
$\begin{vmatrix}3x+a&3x+a&3x+a\\x&x+a&x\\x&x&x+a\end{vmatrix}=0$
$\Rightarrow\ \left(3x+a\right)\begin{vmatrix}1&1&1\\x&x+a&x\\x&x&x+a\end{vmatrix}=0$
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$_, we have:
$\left(3x+a\right)\begin{vmatrix}1&0&0\\x&a&0\\x&0&a\end{vmatrix}=0$
Expanding along $R_1$, we have:
$(3x + a)[1 \times a^2] = 0$
$\Rightarrow a^2(3x + a) = 0$
But a $\neq$ 0.
Therefore, we have:
$3x + a = 0$
$\Rightarrow x =-\frac{a}{3}$
View full question & answer→Question 1525 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}\frac{1}{\text{a}}&\text{a}^2&\text{bc}\\\frac{1}{\text{b}}&\text{b}^2&\text{ac}\\\frac{1}{\text{c}}&\text{c}^2&\text{ab} \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^3&\text{abc}\\1&\text{b}^3&\text{abc}\\1&\text{c}^3&\text{abc}\end{vmatrix}$ [Applying $R_1 \rightarrow aR_1, R_2 \rightarrow bR_2$ and $R_3 \rightarrow cR_3$]
$=\text{abc}\begin{vmatrix}1&\text{a}^3&1\\1&\text{b}^3&1\\1&\text{c}^3&1 \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 1535 Marks
Show that $\text{A}=\begin{bmatrix} 6 & 5 \\ 7 & 6 \end{bmatrix}$ satisfies the equation $x^2 - 12x + 1 = 0$. Thus, find $A^{-1}$.
Answer$\text{A}=\begin{bmatrix} 6 & 5 \\ 7 & 6 \end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix} 71 & 60 \\ 84 & 71 \end{bmatrix}$
If $I_2$ is the identity matrix of order 2, then
$\text{A}^2-12\text{A}+\text{I}_2=\begin{bmatrix} 71 & 60 \\ 84 & 71 \end{bmatrix}-12\begin{bmatrix}6 & 5 \\7 & 6 \end{bmatrix}+\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$
$\Rightarrow\ \text{A}^2-12\text{A}+\text{I}_2=\begin{bmatrix}71-72+1 & 60-60+0 \\84-84+0 & 71-72+1 \end{bmatrix}$
$\Rightarrow\ \text{A}^2-12\text{A}+\text{I}_2=0$
Thus, A satisfies $x^2 - 12x + 1 = 0$.
Now,
$A^2 - 12A + I^2 = 0$
$\Rightarrow I^2 = 12A - A^2$
$\Rightarrow A^{-1}I_2 = A^{-1} (12A - A^2)$ [Pre-multiplying both sides by $A^{-1}$]
$\Rightarrow A^{-1} = 12I_2 - A$
$\Rightarrow\ \text{A}^1=12\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-\begin{bmatrix}6 & 5 \\7 & 6 \end{bmatrix}$
$\Rightarrow\ \text{A}^1=\begin{bmatrix}12-6 & 0-5 \\0-7 & 12-6 \end{bmatrix}$
$\Rightarrow\ \text{A}^1=\begin{bmatrix}6 & -5 \\ -7 & 6 \end{bmatrix}$
View full question & answer→Question 1545 Marks
$\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
Answer$\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}=4\text{a}^2\text{b}^2\text{c}^2$
$\text{L.H.S}=\begin{vmatrix}\text{b}^2+\text{c}^2&\text{ab}&\text{ac}\\\text{ba}&\text{c}^2+\text{a}^2&\text{bc}\\\text{ca}&\text{cb}&\text{a}^2+\text{b}^2\end{vmatrix}$
Multiply $R_1, R_2$ and $R_3$ by a, b and c respectively.
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{ab}^2+\text{ac}^2&\text{a}^2\text{b}&\text{a}^2\text{c}\\\text{b}^2\text{a}&\text{bc}^2+\text{ba}^2&\text{b}^2\text{c}\\\text{c}^2\text{a}&\text{c}^2\text{b}&\text{ca}^2+\text{cb}^2\end{vmatrix}$
Take a, b, and c common from $C_1, C_2$ and $C_3$ respectively.
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}\text{b}^2+\text{c}^2&\text{a}^2&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
Now apply $R_1 \rightarrow R_1 + R_2 + R_3$
$=\begin{vmatrix}2(\text{b}^2+\text{c}^2)&2(\text{c}^2+\text{a}^2)&2(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}(\text{b}^2+\text{c}^2)&(\text{c}^2+\text{a}^2)&(\text{a}^2+\text{b}^2)\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\begin{vmatrix}\text{c}^2&0&\text{a}^2\\\text{b}^2&\text{c}^2+\text{a}^2&\text{b}^2\\\text{c}^2&\text{c}^2&\text{a}^2+\text{b}^2\end{vmatrix}$
$=2\big[\text{c}^2\{(\text{c}^2+\text{a}^2)(\text{a}^2+\text{b}^2)-\text{b}^2\text{c}^2\}+\text{a}^2\{\text{b}^2\text{c}^2-(\text{c}^2+\text{a}^2)\text{c}^2\}\big]$
$=4\text{a}^2\text{b}^2\text{c}^2$
$=\text{R.H.S}$
View full question & answer→Question 1555 Marks
Prove that:
$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
Answer$\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$\text{L.H.S}=\begin{vmatrix}\text{b}+\text{c}&\text{a}-\text{b}&\text{a}\\\text{c}+\text{a}&\text{b}-\text{c}&\text{b}\\\text{a}+\text{b}&\text{c}-\text{a}&\text{c}\end{vmatrix}$
$=\begin{vmatrix}\text{b}+\text{c}+\text{a}&-\text{b}&\text{a}\\\text{c}+\text{a}+\text{b}&-\text{c}&\text{b}\\\text{a}+\text{b}+\text{c}&-\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\1&\text{c}&\text{b}\\1&\text{a}&\text{c}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})\begin{vmatrix}1&\text{b}&\text{a}\\0&\text{c}-\text{b}&\text{b}-\text{a}\\0&\text{a}-\text{b}&\text{c}-\text{a}\end{vmatrix}$
$=-(\text{b}+\text{c}+\text{a})[(\text{c}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{a})(\text{a}-\text{b})]$
$=3\text{abc}-\text{a}^3-\text{b}^3-\text{c}^3$
$=\text{R.H.S}$
View full question & answer→Question 1565 Marks
Show that the following systems of linear equations is inconsistent:
3x - y + 2z = 6,
2x - y + z = 2,
3x + 6y + 5z = 20
Answer$\text{D}=\begin{vmatrix}3&-1&2\\2&-1&1\\3&6&5\end{vmatrix}$
$=3(-5-6)+1(10-3)+2(12+3)=4$
Since D is non zero,
$\text{D}_1=\begin{vmatrix}6&-1&2\\2&-1&1\\20&6&5\end{vmatrix}$
$=6(-5-6)+1(10-20)+2(12+20)$
$=-66-10+64=-12$
$\text{D}_2=\begin{vmatrix}3&6&2\\2&2&1\\3&20&5\end{vmatrix}$
$=3(10-20)-6(10-3)+2(40-6)$
$=-30+42+68=-4$
$\text{D}_3=\begin{vmatrix}3&-1&6\\2&-1&2\\3&6&20\end{vmatrix}$
$=3(-20-12)+1(40-6)+6(12+3)$
$=-96+34+90=28$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-12}{4}=-3$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-4}{4}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{28}{4}=7$
$\therefore\text{x}=-3,\text{ y}=-1$ and $\text{z}=7$
View full question & answer→Question 1575 Marks
If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$, then $\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_1&\text{y}_2&1\\\text{x}_2 &\text{y}_3&1\end{vmatrix}^2=\frac{3\text{a}^4}{4}.$
AnswerThe area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$, is given by,
$\triangle=\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_1&\text{y}_2&1\\\text{x}_2 &\text{y}_3&1\end{vmatrix}$
$\Rightarrow\triangle^2=\frac{1}{2}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_1&\text{y}_2&1\\\text{x}_2 &\text{y}_3&1\end{vmatrix}^2\ \ \dots(1)$
We also know that, the area of an equilateral triangle with side a is given by,
$\triangle=\frac{\sqrt{3}}{4}\text{a}^2$
$\Rightarrow\ \triangle^2=\frac{3}{16}\text{a}^4$
From (1) and (2), we get
$\frac{3}{16}\text{a}^4=\frac{1}{4}\begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_1&\text{y}_2&1\\\text{x}_2 &\text{y}_3&1\end{vmatrix}^2$
$\Rightarrow\ \begin{vmatrix}\text{x}_1&\text{y}_1&1\\\text{x}_1&\text{y}_2&1\\\text{x}_2 &\text{y}_3&1\end{vmatrix}^2=\frac{3}{4}\text{a}^4$
Hence proved.
View full question & answer→Question 1585 Marks
Find the inverse of the following matrices by using elementry row transformation:
$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}$
Now, A = IA
$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\frac{1}{3}\text{ R}_1$
$\begin{bmatrix} 1 & \frac{10}{7} \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
Applying $R_2 \rightarrow R_2 - 2R_1$
$\begin{bmatrix} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ \frac{-2}{3} & 1 \end{bmatrix}\text{A}$
Applying $R_2 \rightarrow 3R_2$
$\begin{bmatrix}1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ -2 & 3 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\text{R}_1-\frac{10}{3}\text{ R}_3$
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\text{A}$
I = BA
Hence, B is the inverse of A.
View full question & answer→Question 1595 Marks
For the matrix $\text{A}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}.$ Show that $A^{-3} - 6A^2 + 5A + 11I_3 = 0$ Hence, find $A^{-1}$.
Answer$\text{A}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$
$=\begin{bmatrix}1+1+2 & 1+2-1 & 1-3+3 \\1+2-6 & 1+4+3 & 1-6-9 \\ 2-1+6 & 2-2-3 & 2+3+9 \end{bmatrix}$
$=\begin{bmatrix}4 & 2 & 1 \\-3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}$
$\text{A}^3=\text{A}^2\text{A}=\begin{bmatrix}4 & 2 & 1 \\-3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}\begin{bmatrix}1 & 2 & 1 \\1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$
$=\begin{bmatrix} 4+2+2 & 4+4-1 & 4-6+3 \\ -3+8-28 & -3+16+14 & -3-24-42 \\ 7-3+28 & 7-6-14 & 7+9+42 \end{bmatrix}$
$=\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$
$\therefore\ \text{A}^3-6\text{A}^2+5\text{A}+11\text{I}$
$=\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}-6\begin{bmatrix}4 & 2 & 1 \\-3 & 8 & -14 \\ 7 & -3 & 14\end{bmatrix}$
$=+5\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}+11\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}-\begin{bmatrix}24 & 12 & 6 \\-18 & 48 & -84 \\ 42 & -18 & 84\end{bmatrix}$
$=+\begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix}+\begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$
$=\begin{bmatrix}24 & 12 & 6 \\-18 & 48 & -84 \\ 42 & -18 & 84\end{bmatrix}-\begin{bmatrix}24 & 12 & 6 \\-18 & 48 & -84 \\ 42 & -18 & 84\end{bmatrix}$
$=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=0$
Thus, $A^3 - 6A^2 + 5A + 11I = 0$
$\Rightarrow (AAA)A^{-1} - 6(AA)A^{-1} + 5AA^{-1} + 11IA^{-1} = 0$ [Post-multiplying by $A^{-1}$ as $|\text{A}|\neq0$]
$\Rightarrow AA(AA^{-1}) - 6A(AA^{-1}) + 5(AA^{-1}) = -11(IA^{-1})$
$\Rightarrow A^2 - 6A + 5I = -11A^{-1}$
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{11}(\text{A}^2-6\text{A}+5\text{I})\ .....(\text{i})$
Now,
$A^2 - 6A + 5I$
$=\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}-6\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}+5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}-\begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix}+\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$=\begin{bmatrix}9 & 2 & 1 \\ -3 & 13 & -14 \\ 7 & -3 & 19 \end{bmatrix}-\begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix}$
$=\begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$
From equation (i), we have:
$\text{A}^{-1}=-\frac{1}{11}\begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$
$=\frac{1}{11}\begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$
View full question & answer→Question 1605 Marks
If $\text{A}^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}\text{and B}=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix}, $find $(AB)^{-1}$
AnswerGiven: $\text{A}^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix} \text{and B}=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix} $
Since, $(AB)^{-1}= B^{-1}A^{-1}$ [Reversal law] ......(1)
Now $\left|\text{B}\right|=\begin{vmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{vmatrix}=1(3 - 0) -2(- 1 -0) + (-2)(2 - 0)=3+2-4=1\neq0$
Therefore,$ B^{-1}$exists.
$\therefore$
$B_{11} = 3, B_{12} = 1, B_{13} = 2$ and $B_{21} = 2, B_{22} = 1, B_{23}= 2$ and $B_{31} = 6, B_{32} = 2, B_{33} = 5$
$\therefore\text{adj. B}=\begin{bmatrix}3&1&2\\2&1&2\\6&2&5\end{bmatrix}=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}$
$\therefore\text{B}^{-1}=\frac{1}{\text{|B|}}(\text{adj. B})=\frac{1}{1}\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}$
From eq. $(1), (AB)^{-1} =$ $\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
$\Rightarrow $ $(\text{AB})^{-1}=\begin{bmatrix}9-30+30&-3+12-12&3-10+12\\3-15+10&-1+6-4&1-5+4\\6-30+25&-2+12-10&2-10+10\end{bmatrix}=\begin{bmatrix}9&-3&5\\-2&3&1\\1&0&2\end{bmatrix}$
View full question & answer→Question 1615 Marks
Solve the following system of homogeneous linear equations:
x + y - 2z = 0,
2x + y - 3z = 0,
5x + 4y - 9z = 0
AnswerConsider,
x + y - 2z = 0
2x + y - 9z = 0
5x + 4y - 9z = 0
$\text{D}=\begin{vmatrix}1&1&-2\\2&1&-3\\5&4&-9 \end{vmatrix}$
$=1(-9+12)-1(-18+15)-2(8-5)=0$
So, the system has infinitely many solutions, putting z = k in the first two equations,
x + y = 2k
2x + y = 3k
Using cramer's rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{\begin{vmatrix}2\text{k}&1\\3\text{k}&1\end{vmatrix}}{\begin{vmatrix}1&1\\2&1\end{vmatrix}}=\frac{-\text{k}}{-1}=\text{k}$
$ \text{y}=\frac{\text{D}_2}{\text{D}}=\frac{\begin{vmatrix}1&2\text{k}\\1&3\text{k}\end{vmatrix}}{\begin{vmatrix}1&1\\2&1\end{vmatrix}}=\frac{-\text{k}}{-1}=\text{k}$
z = k
Clearly, these value satisfy the third equation.
Thus,
x = y = z - k $[\text{k}\in\text{R}]$
View full question & answer→Question 1625 Marks
Solve the following system of equations by matrix method:
$3x + y = 19$
$3x - y = 23$
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 3&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}19\\ 23\end{bmatrix}$
AX = B
Here,
$\text{A}=\begin{bmatrix}3&1\\ 3&-1\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}19\\ 23\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 3&-1\end{vmatrix}$
$= - 3 - 3$
$=-6\neq0$
So, the given system has a unique solution given by $X = A^{-1}B$.
Let $C_{ij}$ be the co-factors of the elements $a_{ij}$ in $A = [a_{ij}]$. Then,
$\text{C}_{11}=(-1)^{1+1}(-1)=-1,\text{C}_{12}=(-1)^{1+2}(3)=-3$
$\text{C}_{21}=(-1)^{2+1}(1)=-1,\text{C}_{22}=(-1)^{2+2}(3)=3$
$\text{adj A}=\begin{bmatrix}-1&-3\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
View full question & answer→Question 1635 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
$=\frac{\text{xyz}}{\text{xyz}}\begin{vmatrix}0&\text{x}&\text{y}\\-\text{x}&0&\text{z}\\-\text{y}&-\text{z}&0\end{vmatrix}$
$=\frac{1}{\text{xyz}}\begin{vmatrix}0&\text{xz}&\text{yz}\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}$
$=\frac{1}{\text{xyz}}\begin{vmatrix}-2\text{xy}&0&2\text{yz}\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=\frac{1}{\text{xyz}}\begin{vmatrix}0&0&0\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}=0$ [Applying $R_1 \rightarrow R_1 - 2R_2]$
View full question & answer→Question 1645 Marks
Show that the following system of linear equations is consistent and also find solution:
$2x + 2y − 2z = 1$
$4x + 4y − z = 2$
$6x + 6y + 2z = 3$
AnswerThis system can be written as: $\begin{bmatrix}2&2&-2\\ 4&4&-1\\ 6&6&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$ or $\text{AX = B}$
$\text{|A|}=2{(14)}-2(14)-2{(0)}=0$
So, A is singular and the system has either no solution or infinite solutions according as
$\text{(Adj A)}\times\text{(B)}\neq0$ or $\text{(Adj A)}\times\text{(B)}=0$
Let $C_{ij}$ be the co-factors of $a_{ij}i$n A $\text{C}_{11}=14$
$\text{C}_{21}=-16$
$\text{C}_{31}=6$
$\text{C}_{12}=-14$
$\text{C}_{22}=16$
$\text{C}_{32}=-6$
$\text{C}_{1}=0\\ \text{C}_{23}=0\\ \text{C}_{33}=0$
$\text{adj A}=\begin{bmatrix}14&-14&0\\ -16&16&0\\ 0&0&0\end{bmatrix}^\text{T}=\begin{bmatrix}14&-16&6\\ -14&16&-6\\ 0&0&0\end{bmatrix}$
$(\text{adj A})\times\text{B}=\begin{bmatrix}14&-16&0\\ -14&16&-6\\ 0&0&0\end{bmatrix}\begin{bmatrix}1\\ 2\\ 3\end{bmatrix}=\begin{bmatrix}14-32+18\\ -14+32-18\\ 0+0+0\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
So, $\text{AX}=\text{B}$ has infinite solutions.
Now, let $z = k So, 2x + 2y = 1 + 2k 4x + 4y = 2 + k$
which can be written as: $\begin{bmatrix}2&2\\ 4&4\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}1+2\text{k}\\ 2+\text{k}\end{bmatrix}$.
or $\text{AX = B}$|A| = 0, z = 0
Again, $2\text{x}+2\text{y}=1$
$4\text{x}+4\text{y}=2$ Let $\text{y = k}$
$2\text{x}=1-2\text{k}$
$\text{x}=\frac{1}{2}-\text{k}$
Hence, $\text{x}=\frac{1}{2}-\text{k}$
$\text{y}=\text{k}$
$\text{z}=0$
View full question & answer→Question 1655 Marks
Prove the following identities:
$\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^3&2&\text{a}\\\text{b}^3&2&\text{b}\\\text{c}^3&2&\text{c}\end{vmatrix}$
$=2\begin{vmatrix}\text{a}^3&1&\text{a}\\\text{b}^3&1&\text{b}\\\text{c}^3&1&\text{c}\end{vmatrix}$
$=2\{\text{a}^3(\text{c}-\text{d})-1(\text{b}^3\text{c}-\text{bc}^3)+\text{a}(\text{b}^3-\text{c}^3)\}$
$=2\{\text{a}^3(\text{c}-\text{b})-\text{bc}(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=(\text{b}-\text{c})\{-\text{a}^3-\text{bc}(\text{b}+\text{c})+\text{a}(\text{b}^2+\text{bc}+\text{c}^2)\}$
$=2(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}+\text{b}+\text{c})$
$=\text{R.H.S}$
View full question & answer→Question 1665 Marks
If $\text{A} = \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},$ find $A^{-1}.$ Using $A^{-1}$ solve the system of equations:
$2x - 3y + 5z = 11$
$3x + 2y - 4z = -5$
$x + y - 2z = -3$
Answer$\text{Given:}\ \text{Matrix A}=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{vmatrix}$
$\Rightarrow\ \text{|A|}=2(-4+4)-3(-3)(-6+4)+5(3-2)=0-6+5=-1\neq0$
$\therefore\ \text{A}^{-1}\ \text{exists and A}^{-1}=\frac{1}{\text{|A|}}\text{(adj. A)} \dots\dots(1)$
Now, $A_{11}= 0, A_{12} = 2, A_{13} = 1$ and $A_{21} = -1, A_{22} = -9, A_{23} = -5$ and $A_{31} = 2, A_{32} = 23, A_{13} = 13$
$\therefore\ \text{adj.A}=\begin{bmatrix}0&2&1\\-1&-9&-5\\2&23&13\end{bmatrix}=\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}$ $\therefore$ From eq. (1),
$\text{A}^{-1}=\frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}$
Now, Matrix form of given equations is AX = B
$ \Rightarrow\ \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$\text{Here}\ \text{A}=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$ Therefore, solution is unique and
$X = A^{-1}B$ $\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
$=\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$
Therefore, $x = 1, y = 2$ and $ z = 3$
View full question & answer→Question 1675 Marks
Solve the following system of equations by matrix method:
$3x + 7y = 4$
$x + 2y = -1$
AnswerThe above system can be written in matrix form as:$\begin{bmatrix}3&7\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Or AX = BWhere $\text{A}=\begin{bmatrix}3&7\\ 1&2\end{bmatrix},\text{x}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{ and }\text{B}=\begin{bmatrix}4\\ -1\end{bmatrix}$
Now,$\text{|A|}=-1\neq0$
So, the above system has a unique solution, given by $X = A^{-1}$ B Now, let $C_{ij}$ be the co-factors of $a_{ij}$ in A $\text{C}_{11} = 2,\text{C}_{12} = -1$$\text{C}_{21} = -7,\text{C}_{22} = 3$
$\text{Adj A}=\begin{bmatrix}2&-1\\ -7&3\end{bmatrix}^\text{T}=\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
$\therefore\text{A}^{-1}=\frac{1}{\text{|A|}}.\text{adj A}=\frac{1}{(-1)}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}$
Now, $X = A^{-1}$ B$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}2&-7\\ -1&3\end{bmatrix}\begin{bmatrix}4\\ -1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}15\\ -7\end{bmatrix}=\begin{bmatrix}-15\\ 7\end{bmatrix}$
Hence, $x = -15 y = 7$
View full question & answer→Question 1685 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Answer$\begin{vmatrix}1^2&2^2&3^2&4^2\\2^2&3^2&4^2&5^2\\3^2&4^2&5^2&6^2\\4^2&5^2&6^2&7^2\end{vmatrix}$
Applying: $C_3 → C_3 - C_2, C_4 → C_4 - C_1$
$=\begin{vmatrix}1^1&2^2&3^2-2^2&4^2-1^2\\2^2&3^2&4^2-3^2&5^2-2^2\\3^3&4^2&5^2-4^2&6^2-3^2\\4^2&5^2&6^2-5^2&7^2-4^2 \end{vmatrix}$
$=\begin{vmatrix}1^1&2^2&5&15\\2^2&3^2&7&21\\3^3&4^2&9&27\\4^2&5^2&11&33 \end{vmatrix}$
Take 3 common from $C_4$
$=0$
$\because\text{C}_3=\text{C}_4$
View full question & answer→Question 1695 Marks
Evaluate the following:
$\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$\triangle=\begin{vmatrix}0&\text{xy}^2&\text{xz}^2\\\text{x}^2\text{y}&0&\text{yz}^2\\\text{x}^2\text{z}&\text{zy}^2&0\end{vmatrix}$
$=\text{x}^2\text{y}^2\text{z}^2\begin{vmatrix}0&\text{x}&\text{x}\\\text{y}&0&\text{y}\\\text{z}&\text{z}&0\end{vmatrix}$
[Taking $x^2$ common from from $C_1, y^2$ common from $C_2$ and $z^2$ common from $C_3$]
$=\text{x}^3\text{y}^3\text{z}^3\begin{vmatrix}0&0&1\\1&-1&1\\1&1&0\end{vmatrix}$
[Applying $C_2 → C_2 - C_3$]
$=\text{x}^3\text{y}^3\text{z}^3(1+1)$ [Expanding along first row]
$=2\text{x}^3\text{y}^3\text{z}^3$
$\therefore\ \triangle=2\text{x}^3\text{y}^3\text{z}^3$
View full question & answer→Question 1705 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
Answer$\begin{vmatrix}\sqrt{23}+\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}+\sqrt{46}&5&\sqrt{10}\\3+\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\begin{vmatrix}\sqrt{3}&\sqrt{5}&\sqrt{5}\\\sqrt{15}&5&\sqrt{10}\\3&\sqrt{15}&5\end{vmatrix}+\begin{vmatrix}\sqrt{23}&\sqrt{5}&\sqrt{5}\\\sqrt{46}&5&\sqrt{10}\\\sqrt{115}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{15}&5\end{vmatrix}+\sqrt{23}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{2}&5&\sqrt{10}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=\sqrt{3}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&\sqrt{5}\\\sqrt{5}&5&\sqrt{10}\\\sqrt{3}&\sqrt{3}&5\end{vmatrix}+\sqrt{23}\times\sqrt{5}\begin{vmatrix}1&\sqrt{5}&1\\\sqrt{2}&5&\sqrt{2}\\\sqrt{5}&\sqrt{15}&5\end{vmatrix}$
$=0+0$
$=0$
View full question & answer→Question 1715 Marks
Find the value of $\theta$ satisfying $\begin{bmatrix}1&1&\sin3\theta\\-4&3&\cos2\theta\\7&-7&-2\end{bmatrix}=0.$
AnswerWe have, $\begin{vmatrix}1&1&\sin3\theta\\-4&3&\cos2\theta\\7&-7&-2\end{vmatrix}=0$
Expandiug along $C_3$, we get
$\sin3\theta\times(28-21)-\cos2\theta\times(-7-7)-2(3+4)=0$
$\Rightarrow\ 7\sin3\theta+14\cos2\theta-14=0$
$\Rightarrow\ \sin3\theta+2\cos2\theta-2=0$
$\Rightarrow\ (3\sin\theta-4\sin^3\theta)+2(1-2\sin^2\theta)-2=0$
$\Rightarrow\ 4\sin^3\theta-4\sin^2\theta+3\sin\theta=0$
$\Rightarrow\ \sin\theta(4\sin^2\theta-4\sin\theta+3)=0$
$\Rightarrow\ \sin\theta(2\sin\theta+1)(2\sin\theta-3)=0$
$\Rightarrow\ \sin\theta=0\text{ or }\sin\theta=-\frac{1}{2}\text{or }\sin\theta=\frac{3}{2}$
$\Rightarrow\ \theta=\text{n}\pi\text{ or }\theta=\text{m}\pi+(-1)^\text{n}\Big(-\frac{\pi}{6}\Big);\text{ m, n}\in\text{Z}$
$\sin\theta=\frac{-3}{2}\text{ is not possible}$
View full question & answer→Question 1725 Marks
Find the inverse of the following matrices by using elementry row transformation:
$\begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix}$
Answer$\text{A}=\begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix}$
We have A = IA
$\Rightarrow\begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & -1 \\ 1 & -1 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix} 1 & 2 & 0 \\ 0 & -1 & -1 \\ 0 & -3 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-2\text{R}_1\text{ and R}_3\rightarrow\text{R}_3-\text{R}_1\big]$
$\Rightarrow\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & -3 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & -1 & 0 \\ -1 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow-\text{R}_2\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 6 \end{bmatrix}=\begin{bmatrix} -3 & 2 & 0 \\ 2 & -1 & 0 \\ 5 & -3 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1-2\text{R}_2\text{ and R}_3\rightarrow\text{R}_3+3\text{R}_2\big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -3 & 2 & 0 \\ 2 & -1 & 0 \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_3\rightarrow\frac{1}{6}\text{ R}_3\Big]$
$\Rightarrow\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{3} & -\frac{1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{bmatrix}\text{A}$
$\big[\text{Applying R}_1\rightarrow\text{R}_1+2\text{R}_1\text{ and R}_2\rightarrow\text{R}_2-\text{R}_3\big]$
$\therefore\ \text{A}^{-1}=\begin{bmatrix} -\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{3} & -\frac{1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6} \end{bmatrix}$
View full question & answer→Question 1735 Marks
Find the inverse of each of the matrix:
$\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}$
Answer$\text{Let A}=\begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix}$
$\therefore\ |\text{A}|=\begin{vmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{vmatrix}=1(8-6)-(-1)(0+9)+2(0-6)=-1\neq0$
$\text{A}_{11}=+\begin{vmatrix}2&-3\\-2&4\end{vmatrix}=+(8-6)=2,$
$\text{A}_{12}=-\begin{vmatrix}0&-3\\3&4\end{vmatrix}=-(0+9)=-9,$
$\text{A}_{13}=+\begin{vmatrix}0&2\\3&-2\end{vmatrix}=+(0-6)=-6,$
$\text{A}_{21}=-\begin{vmatrix}-1&2\\-2&4\end{vmatrix}=-(-4+4)=0,$
$\text{A}_{22}=+\begin{vmatrix}2&2\\3&4\end{vmatrix}=+(4-6)=-2,$
$\text{A}_{23}=-\begin{vmatrix}1&-1\\3&-2\end{vmatrix}=-(-2+3)=-1,$
$\text{A}_{31}=+\begin{vmatrix}-1&2\\2&-3\end{vmatrix}=+(3-4)=-1,$
$\text{A}_{32}=-\begin{vmatrix}1&2\\0&-3\end{vmatrix}=-(-3-0)=3,$
$\text{A}_{33}=+\begin{vmatrix}1&-1\\0&2\end{vmatrix}=+(2-0)=2$
$\therefore\ \text{adj. A}=\begin{bmatrix}2&-9&-6\\0&-2&-1\\-1&3&2\end{bmatrix}= \begin{bmatrix}\text{A}_{11}&\text{A}_{21}&\text{A}_{31}\\\text{A}_{12}&\text{A}_{22}&\text{A}_{32}\\\text{A}_{13}&\text{A}_{23}&\text{A}_{33}\end{bmatrix} =\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj. A}=\frac{1}{-1}=\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}=\begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}$
View full question & answer→Question 1745 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 1755 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$
$=\begin{vmatrix}3\text{a}+3\text{b}&3\text{a}+3\text{b}&3\text{a}+3\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Applying $R_1 → R_2 + R_2 + R_3$]
$=3(\text{a}+\text{b})\begin{vmatrix}1&1&1\\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Taking out 3(a + b) common from $R_1$]
$=3(\text{a}+\text{b})\begin{vmatrix}0&0&1\\2\text{b}&-\text{b}&\text{a}+\text{b}\\-\text{b}&2\text{b}&\text{a} \end{vmatrix}$ [Applying $C_1 → C_1 - C_2$ and $C_2 → C_2 - C_3$]
$=3(\text{a}+\text{b})\text{b}^2\begin{vmatrix}0&0&1\\2&-1&\text{a}+\text{b}\\-1&2&\text{a} \end{vmatrix}$ [Taking out b common from $C_1$ and $C_2$]
$=3(\text{a}+\text{b})\text{b}^2\times3$
$=9(\text{a}+\text{b})\text{b}^2$
$=\text{R.H.S}$
View full question & answer→Question 1765 Marks
$\begin{vmatrix}0&\text{b}^2\text{a}&\text{c}^2\text{a}\\\text{a}^2\text{b}&0&\text{c}^2\text{b}\\\text{a}^2\text{c}&\text{b}^2\text{c}&0\end{vmatrix}=2\text{a}^3\text{b}^3\text{c}^3$
Answer$\text{L.H.S}=\begin{vmatrix}0&\text{b}^2\text{a}&\text{c}^2\text{a}\\\text{a}^2\text{b}&0&\text{c}^2\text{b}\\\text{a}^2\text{c}&\text{b}^2\text{c}&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&\text{b}^3\text{a}&\text{c}^3\text{a}\\\text{a}^3\text{b}&0&\text{c}^3\text{b}\\\text{a}^3\text{c}&\text{b}^3\text{c}&0\end{vmatrix}$ [Multiplying the three columns by a, b, and c]
$=\frac{\text{abc}}{\text{abc}}\begin{vmatrix}0&\text{b}^3&\text{c}^3\\\text{a}^3&0&\text{c}^3\\\text{a}^3&\text{b}^3&0\end{vmatrix}$ [Taking out a, b and c common from the three rows]
$=\text{b}^3\begin{vmatrix}\text{b}^3&\text{c}^3\\\text{a}^3&0\end{vmatrix}+\text{c}^3\begin{vmatrix}\text{a}^3&0\\\text{a}^3&\text{b}^3\end{vmatrix}$ [Expanding along $R_1$]
$=2\text{a}^3\text{b}^3\text{c}^3$
View full question & answer→Question 1775 Marks
Show that $\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}=0,$ where $\alpha,\beta,\gamma$ are in A.P.
AnswerSince, $\alpha,\beta,\gamma$ are in A.P, $2\beta=\alpha+\gamma$
$\text{L.H.S}=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\\text{x}-2&\text{x}-3&\text{x}-\beta\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$\text{R}_2\rightarrow\text{R}_2-\frac{\text{R}_1}{2}-\frac{\text{R}_3}{2}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\$\text{x}-2)-\frac{\text{x}-3}{2}-\frac{\text{x}-1}{2}&(\text{x}-3)-\frac{\text{x}-4}{2}-\frac{\text{x}-2}{2}&(\text{x}-\beta)-\frac{\text{x}-\alpha}{2}-\frac{\text{x}-\gamma}{2}\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$
$=\begin{vmatrix}\text{x}-3&\text{x}-4&\text{x}-\alpha\\0&0&0\\\text{x}-1&\text{x}-2&\text{x}-\gamma\end{vmatrix}$ $[\because2\beta=\alpha+\gamma]$
$=0$
$=\text{R.H.S}$
View full question & answer→Question 1785 Marks
Show that $\text{A}=\begin{bmatrix} 5 & 3 \\-1 & -2 \end{bmatrix}$ satisfies the equation $x^2 - 3x - 7 = 0$. Thus, find $A^{-1}.$
Answer$\text{A}=\begin{bmatrix} 5 & 3 \\-1 & -2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix} 22 & 9 \\-3 & 1 \end{bmatrix}$
If $I_2$ is the identity matrix of order 2, then
$\text{A}^2-3\text{A}-7\text{I}_2$
$=\begin{bmatrix}22 & 9 \\-3 & 1 \end{bmatrix}-3\begin{bmatrix}5 & 3 \\ -1 & -2 \end{bmatrix}-7\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}$
$\Rightarrow\ \text{A}^2-3\text{A}-7\text{I}_2=\begin{bmatrix}22-15-7 & 9-9-0 \\-3+3+0 & 1+6-7 \end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}=0$
$\Rightarrow\ \text{A}^2-3\text{A}-7\text{I}_2=0$
Thus, A satisfies $x^2 - 3x - 7 = 0$
Now,
$A^2 - 3A - 7I_2 = 0$
$\Rightarrow A^2 - 3A = 7I_2$
$\Rightarrow A^{-1} (A^2 - 3A) = A^{-1} \times 7I_{2$} [Pre-multiplying both sides by $A^{-1}]$
$\Rightarrow A - 3I_2 = 7A^{-1}$
$\Rightarrow\ \begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}-3\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=7\text{A}^{-1}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{7}\begin{bmatrix}5-3 & 3-0 \\-1-0 & -2-3 \end{bmatrix}$
$=\frac{1}{7}\begin{bmatrix}2 & 3 \\-1 & -5 \end{bmatrix}$
View full question & answer→Question 1795 Marks
Using the proprties of determinants in Exercise:
$\begin{vmatrix}\text{y}^2\text{z}^2&\text{yz}&\text{y}+\text{z}\\\text{z}^2\text{x}^2&\text{zx}&\text{z}+\text{x}\\\text{x}^2\text{y}^2&\text{xy}&\text{x}+\text{y}\end{vmatrix}=0$
AnswerWe have, $\begin{vmatrix}\text{y}^2\text{z}^2&\text{yz}&\text{y}+\text{z}\\\text{z}^2\text{x}^2&\text{zx}&\text{z}+\text{x}\\\text{x}^2\text{y}^2&\text{xy}&\text{x}+\text{y}\end{vmatrix}$
$[\text{Multiplying R}_1, \text{R}_2, \text{R}_3\text{ by x, y, z respectively}]$
$=\frac{1}{\text{xyz}}\begin{vmatrix}\text{xy}^2\text{z}^2&\text{xyz}&\text{xy}+\text{xz}\\\text{x}^2\text{yz}^2&\text{xyz}&\text{yz}+\text{xy}\\\text{x}^2\text{y}^2\text{z}&\text{xyz}&\text{xz}+\text{yz}\end{vmatrix}$
$\big[\text{Taking (xyz) common from C}_1\text{ and C}_2\big]$
$=\frac{1}{\text{xyz}}(\text{xyz})^2\begin{vmatrix}\text{yz}&1&\text{xy}+\text{xz}\\\text{xz}&1&\text{yz}+\text{xy}\\\text{xy}&1&\text{xz}+\text{yz}\end{vmatrix}$
$[\text{Applying C}_3\rightarrow\text{C}_3+\text{C}_1]$
$=\text{xyz}\begin{bmatrix}\text{yz}&1&\text{xy}+\text{yz}+\text{zx}\\\text{xz}&1&\text{xy}+\text{yz}+\text{zx}\\\text{xy}&1&\text{xy}+\text{yz}+\text{zx}\end{bmatrix}$
$\big[\text{Taking (xy}+\text{yz}+\text{zx})\text{ common from }\text{C}_3\big]$
$=\text{xyz (yz}+\text{yz}+\text{zx})\begin{vmatrix}\text{yz}&1&1\\\text{xz}&1&1\\\text{xy}&1&1\end{vmatrix}$
$=0$
$\big[\because\text{C}_2\text{ and C}_3\text{ are identical}\big]$
View full question & answer→Question 1805 Marks
Find $A^{-1}$ if $\text{A}=\begin{vmatrix}0&1&1\\1&0&1\\1&1&0\end{vmatrix}$ and show that $\text{A}^{-4}=\frac{\text{A}^2-3\text{I}}{2}.$
AnswerWe have, $\text{A}=\begin{vmatrix}0&1&1\\1&0&1\\1&1&0\end{vmatrix}$ Cofactors are: $\text{A}_{11}=-1,\text{A}_{12}=1,\text{A}_{13}=1,$ $\text{A}_{21}=1,\text{A}_{22}=-1,\text{A}_{23}=1,$ $\text{A}_{31}=1,\text{A}_{31}=1,\text{A}_{32}=1\text{ A}_{33}=-1$ $\therefore\ \ \text{adj A}=\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}^\text{T}=\begin{vmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{vmatrix}$ and $|\text{A}|=-1(-1)+1.1=2$ $\therefore\ \ \text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\frac{1}{2}\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}\ \ \dots(\text{i})$ And $\text{A}^2=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}\ \ \dots(\text{ii})$ $\therefore\ \ \frac{\text{A}^2-3\text{I}}{2}=\frac{1}{2}\begin{Bmatrix}\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}-\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}\end{Bmatrix}$ $=\frac{1}{2}\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}=\text{A}^{-1}$ $= A^{-1}$ [Using Eq.(i)] Hence proved.
View full question & answer→Question 1815 Marks
Solve the following systems of linear equations by cramer's rule:
2y - 3z = 0,
x + 3y = -4,
3x + 4y = 3
AnswerThese equations can be written as
0x + 2y - 3z = 0
x + 3y + 0z = -4
3x + 4y + 0z= 3
$\text{D}=\begin{vmatrix}0&2&-3\\1&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-2(0-0)-3(4-9)$
$=15$
$\text{D}_1=\begin{vmatrix}0&2&-3\\-4&3&0\\3&4&0 \end{vmatrix}$
$=0(0-0)-0(0-0)-3(3+12)$
$=-45$
$\text{D}_3=\begin{vmatrix}0&2&0\\1&3&-4\\3&4&3 \end{vmatrix}$
$=0(9+16)-2(3+12)-0(4-9)$
$=-30$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{75}{15}=5$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-45}{15}=-3$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-30}{12}=-2$
$\therefore\text{x}=5,\text{y}=-3$ and $\text{z}=-2$
View full question & answer→Question 1825 Marks
$\text{A}=\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$, find AB. Hence, solve the system of equations:
x - 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
Answer$\text{A}=\begin{bmatrix}1&-2&0\\2&1&3\\0&-2&1\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}$
$\text{A}\times\text{B}=\begin{bmatrix}11&0&0\\ 0&11&0\\\ 0&0&11\end{bmatrix}$
AB = 11I, where I is a 3 × 3 unit matrix
$\text{A}^{-1}=\frac{1}{11}\text{B}$ [By def. of inverse]Or
Or $\frac{1}{11}=\begin{bmatrix}7&2&-6\\ -2&1&-3\\ -4&2&5\end{bmatrix}$
Now, the given system of equations can be written as:
$\begin{bmatrix}1&-2&0\\ 2&1&3\\ 0&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$
Or $\text{AX = B}$
$\text{X = A}^{-1}\text{B}$
Or $=\frac{1}{11}\begin{bmatrix}7&2&-6\\-2&1&-3\\-4&2&5\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{11}\begin{bmatrix}44\\ -33\\ 11\end{bmatrix}=\begin{bmatrix}4\\ -3\\ 1\end{bmatrix}$
Hence, x = 4, y = -3, z = 1
View full question & answer→Question 1835 Marks
Show that the following system of linear equation is inconsistent:
x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4
AnswerThe above system can be written as:
$\begin{bmatrix}1&1&-2\\ 1&-2&1\\ -2&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}5\\ -2\\ 4\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1{(-3)}-1{(-3)}=-3-3+6=0$
So, A is singular. Now the system can be inconsistent, if
$(\text{adj A})\times\text{B}\neq0$
$\text{C}_{11}=-3\\ \text{C}_{21}=-3\\ \text{C}_{31}=-3$
$\text{C}_{12}=-3\\ \text{C}_{22}=-3\\ \text{C}_{32}=-3$
$\text{C}_{13}=-3\\ \text{C}_{23}=-3\\ \text{C}_{33}=-3$
$(\text{adj A})=\begin{bmatrix}-3&-3&-3\\ -3&-3&-3\\ -3&-3&-3\end{bmatrix}^\text{T}=\begin{bmatrix}-3&-3&-3\\ -3&-3&-3\\ -3&-3&-3\end{bmatrix}$
$(\text{adj A})\times\text{(B)}=\begin{bmatrix}-3&-3&-3\\ -3&-3&-3\\ -3&-3&-3\end{bmatrix}\begin{bmatrix}5\\ -2\\ 4\end{bmatrix}=\begin{bmatrix}-15+6-12\\ -15+6-12\\ -15+6-12\end{bmatrix}$
$=\begin{bmatrix}-21\\ -21\\ -12\end{bmatrix}$
$\neq0$
Hence, the given system is inconsistent.
View full question & answer→Question 1845 Marks
Show that the following system of linear equation is inconsistent:
$2x + 5y = 7$
$6x + 15y = 13$
AnswerThe given system of equations can be expresed as follows:
$\text{AX = B}$
Here,
$\text{A}=\begin{bmatrix}2&5\\ 6&15\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}7\\ 13\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}2&5\\ 6&15\end{vmatrix}$
$={(30-30)}$
$=0$
Let $C_{ij}$ be the co-factors of the elemennts $a_{ij} in A = [a_{ij}].$ Then,
$\text{C}_{11}=-1^{1+1}{(15)}=15,\\ \text{C}_{12}=-1^{1+2}{(6)}=-6\\ $
$\text{C}_{21}=-1^{2+1}{(5)}=-5,\\ \text{C}_{22}=-1^{2+2}{(6)}=2\\ $
$\text{adj A}=\begin{bmatrix}15&-6\\ -5&2\end{bmatrix}^\text{T}$
$=\begin{bmatrix}15&-5\\ -6&2\end{bmatrix}$
$(\text{adj A) B}=\begin{bmatrix}15&-5\\ -6&2\end{bmatrix}\begin{bmatrix}7\\ 13\end{bmatrix}$
$=\begin{bmatrix}105-65\\ -42+26\end{bmatrix}$
$=\begin{bmatrix}40\\ -16\end{bmatrix}\neq0$
Hence, the given system of equations is inconsitent.
View full question & answer→Question 1855 Marks
Using the proprties of determinants in Exercise:
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
AnswerWe have to prove,
$\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}=4\text{xyz}$
$\therefore\ \text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{z}&\text{y}\\\text{z}&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$=\begin{vmatrix}\text{y}+\text{z}+\text{z}+\text{y}&\text{z}&\text{y}\\\text{z}+\text{z}+\text{x}+\text{x}&\text{z}+\text{x}&\text{x}\\\text{y}+\text{x}+\text{x}+\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$ $\big[\because\text{C}_1\rightarrow\text{C}_1+\text{C}_2+\text{C}_3\big]$
$=2\begin{vmatrix}(\text{y}+\text{z})&\text{z}&\text{y}\$\text{z}+\text{x})&\text{z}+\text{x}&\text{x}\$\text{x}+\text{y})&\text{x}&\text{x}+\text{y}\end{vmatrix}$
$\big[\text{Taking 2 common from C}_1\big]$
$=2\begin{vmatrix}\text{y}&\text{z}&\text{y}\\0&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$ $[\because\ \text{C}_1\rightarrow\text{C}_1-\text{C}_2]$
$=2\begin{vmatrix}0&\text{z}-\text{x}&-\text{x}\\0&\text{z}+\text{x}&\text{x}\\\text{y}&\text{x}&\text{x}+\text{y}\end{vmatrix}$ $[\because\ \text{R}_1\rightarrow\text{R}_1-\text{R}_3]$
$=2\big[\text{y}(\text{xz}-\text{x}^2+\text{xz}+\text{x}^2)\big]$
$=4\text{xyz}=\text{R.H.S}$
Hence proved.
View full question & answer→Question 1865 Marks
A shopkeeper has $3$ varieties of pens $'A', 'B'$ and $'C'$. Meenu purchased $1$ pen of each variety for a total of Rs $21$. Jeevan purchased $4$ pens of $'A'$ variety $3$ pens of $'B'$ variety and $2$ pens of $'C'$ variety for Rs $60$. While Shikha purchased $6$ pens of $'A'$ variety, $2$ pens of $'B'$ variety and $3$ pens of $'C'$ variety for Rs $70$. Using matrix method, find cost of each variety of pen.
AnswerAs there are 3 varieties of pen A, B and C
Meenu purchased 1 pen of each variety which costs her Rs. 21
Therefore,
A + B + C = 2
Similarly,
For Jeevan
4A + 3B + 2C = 60
For Shikha
6A + 2B + 3C = 70
$\begin{bmatrix}1&1&1\\4&3&2\\6&2&3\end{bmatrix}\begin{bmatrix}\text{A}\\\text{B}\\\text{C}\end{bmatrix}=\begin{bmatrix}21\\60\\70\end{bmatrix}$
where, $\text{P}=\begin{bmatrix}1&1&1\\4&3&2\\6&2&3\end{bmatrix},\text{Q}=\begin{bmatrix}21\\60\\70\end{bmatrix}$
$|\text{P}|=1(9-4)-1(12-12)+1(8-18)$
$=-5\neq0$
$\therefore P^{-1}$ exists
$\text{X}=\text{P}^{-1}\text{Q}$
$\begin{matrix}\text{C}_{11}=5&\text{C}_{12}=0&\text{C}_{13}=-10\\\text{C}_{21}=-1&\text{C}_{22}=-3&\text{C}_{23}=4\\\text{C}_{31}=-1&\text{C}_{32}=2&\text{C}_{33}=-1\end{matrix}$
$\text{adj }\text{P}=\begin{bmatrix}5&0&-10\\-1&-3&4\\-1&2&-1\end{bmatrix}^\text{T}=\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}$
$\text{P}^{-1}=\frac{1}{-5}\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}$
$\text{X}=\text{P}^{-1}\text{Q}$
$=\frac{1}{-5}\begin{bmatrix}5&-1&-1\\0&-3&2\\-10&4&-1\end{bmatrix}\begin{bmatrix}21\\60\\70\end{bmatrix}$
$=\frac{1}{-5}\begin{bmatrix}105-60-70\\0-180+140\\-210+240-70\end{bmatrix}$
$=\frac{1}{-5}\begin{bmatrix}-25\\-40\\-40\end{bmatrix}$
$\therefore\ \text{X} = \begin{bmatrix}5\\8\\8\end{bmatrix}$
Therefore, cost of A variety of pens = Rs. 5
Cost of B variety of pens = Rs. 8
Cost of C variety of pens = Rs. 8
View full question & answer→Question 1875 Marks
Let $\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$
$\text{and 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)\text{G}(\beta)\big]^{-1}=\text{G}(-\beta)\text{F}(-\alpha).$
Answer$\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\ .....\text{(i)}$
$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(ii)}$
$\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}\ .....(\text{iii})$
$\text{G }(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
$\Rightarrow\ \text{G}(-\beta)=\begin{bmatrix} \cos(-\beta) & 0 & \sin(-\beta) \\ 0 & 1 & 0 \\ -\sin(-\beta) & 0 & \cos(-\beta) \end{bmatrix}$
$=\begin{bmatrix} \cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(iv)}$
$\big[\text{F}(\alpha)\text{G}(\beta)\big]^{-1}=\big[\text{G}(\beta)\big]^{-1}\big[\text{F}(\alpha)\big]^{-1}$
$\begin{bmatrix} \cos\beta & 0 & -\sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix}=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$
$[\text{Using eqn (i) and (ii)}]$
$=\text{G}(-\beta)\text{F}(-\alpha)\ [\text{Using eqn (iii) and (iv)}]$
View full question & answer→Question 1885 Marks
Show that $\text{A}=\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$ sastisfies the equation $A^2 + 4A - 42I = 0$. Hence find $A^{-1}.$
Answer$\text{A}=\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$
Now $A^2 + 4A - 42I = 0$
For this $\text{A}^2=\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}\begin{bmatrix}-8 & 5 \\2 & 4 \end{bmatrix}=\begin{bmatrix}74 & -20 \\-8 & 26 \end{bmatrix}$
Hence,
$\Rightarrow A^2 + 4A - 42I$
$=\begin{bmatrix}74 & -20 \\-8 & 26 \end{bmatrix}+\begin{bmatrix} -32 & 20 \\8 & 16 \end{bmatrix}-\begin{bmatrix}42 & 0 \\0 & 42 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
Hence proved.
Now, $A^2 + 4A - 42I = 0$
$\Rightarrow A^{-1}AA + 4A^{-1}A - 42A^{-1}I = 0$
$\Rightarrow IA + 4I - 42A^{-1} = 0$
$\Rightarrow 42A^{-1} = A + 4I$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{42}\big[\text{A}+4\text{I}\big]=\frac{1}{42}\left\{\begin{bmatrix}-8 & 5 \\2 & 4 \end{bmatrix}+\begin{bmatrix}4 & 0 \\0 & 4 \end{bmatrix}\right\}$
$=\frac{1}{42}\begin{bmatrix}-4 & 5 \\2 & 8 \end{bmatrix}$
View full question & answer→Question 1895 Marks
Solve the following system of equations by matrix method:
$3x + 4y + 7z = 14$
$2x - y + 3z = 4$
$x + 2y - 3z = 0$
AnswerThe given system of equations can be written in matrix form as follows:
$\begin{bmatrix}3&1\\ 3&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}19\\ 23\end{bmatrix}$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}3&1\\ 3&-1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}\text {and }\text{B}=\begin{bmatrix}19\\ 23\end{bmatrix}$
Now,
$\text{|A|}=\begin{vmatrix}3&1\\ 3&-1\end{vmatrix}$
$=-3-3$
$=-6\neq0$
So, the given system has a unique solution given by $X = A^{-1}B.$
Let $c_{ij}$be the co-factors of the elements $a_{ij}$in $\text{A}=\text{[a}_\text{ij}]$. Then,
$\text{C}_{11}{(-1)}^{1+1}{(-1)}=-1,\text{C}_{12}={(-1)}^{1+2}{(3)}=-3$
$\text{C}_{21}={(-1)}^{2+1}{(1)}=-1,\text{C}_{22}={(-1)}^{2+2}{(3)}=3$
$\text{adj}\ \text{A}=\begin{bmatrix}-1&-3\\ -1&3\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-6}\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$=\frac{1}{-6}\begin{bmatrix}-1&-1\\ -3&3\end{bmatrix}\begin{bmatrix}19\\ 23\end{bmatrix}$
$=\frac{1}{-6}\begin{bmatrix}-19-23\\ -57+69\end{bmatrix}$
$=\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{-42}{-6}\\ \frac{12}{-6}\end{bmatrix}$
View full question & answer→Question 1905 Marks
Find the matrix X for which:
$\begin{bmatrix}3 & 2 \\ 7 & 5 \end{bmatrix}\text{X}\begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix}=\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix}$
AnswerLet $\text{A}=\begin{bmatrix}3 & 2 \\ 7 & 5 \end{bmatrix},\text{B}=\begin{bmatrix} -1 & 1 \\ -2 & 1 \end{bmatrix}\text{and C}=\begin{bmatrix} 2 & -1 \\ 0 & 4 \end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}3 & 2 \\ 7 & 5 \end{vmatrix}=15-14=1$
$|\text{B}|=\begin{vmatrix} -1 & 1 \\ -2 & 1 \end{vmatrix}=-1+2=1$
Since, $|\text{A}|\neq0\text{ and }|\text{B}|\neq0$
Hence, A & B are invertible, so $A^{-1} $and $B^{-1}$^ exist.
Cofactors of matrix A are
$A_{11} = 5, A_{12} = -7, A_{21} = -2, A_{22} = 3$
Now, $\text{adj A}\begin{bmatrix}5 & -7 \\ -2 & 3 \end{bmatrix}^\text{T}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}$
Cofactors of matrix B are
$B_{11} = 1, B_{12} = 2, B_{21} = -1, B_{22} = -1$
Now, $\text{adj B}=\begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}^\text{T}=\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$
$\text{B}^{-1}=\frac{1}{|\text{B}|}\text{ adj B}=\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$
The given equation Becomes$ AXB = C$
$\Rightarrow (A^{-1} A) X (BB^{-1}) = A^{-1}CB^{-1}$
$\Rightarrow (I) X (I) = A^{-1}CB^{-1}$
$\Rightarrow\ \text{X}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}\begin{bmatrix}2 & -1 \\0 & 4 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}\begin{bmatrix}2-2 & -2+1 \\0+8 & 0-4 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix}5 & -2 \\ -7 & 3 \end{bmatrix}\begin{bmatrix} 0 & -1 \\8 & -4 \end{bmatrix}$
$\Rightarrow\ \text{X}=\begin{bmatrix}-16 & 3 \\ 24 & -5 \end{bmatrix}$
View full question & answer→Question 1915 Marks
Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P 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 ₹ 2,200. School Q wants to spend ₹ 3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹ 1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.
Answerx, y and z be prize amount per student for Tolerence, Kindness and Leadership respectively.
As per the data in the question, we get
3x + 2y + z = 2200
4x + y + 3z = 3100
x + y + z = 1200
The above three simultaneous equations can be written in matrix form as
$\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2200\\3100\\1200\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}2200\\3100\\1200\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}3&2&1\\4&1&3\\1&1&1\end{bmatrix}$
$|\text{A}|=3(-2)-2(1)+1(3)=-5$
$\text{cof }\text{A}=\begin{bmatrix}-2&-1&3\\-1&2&-1\\5&-5&-5\end{bmatrix}$
$\text{adj }\text{A}=(\text{cof }\text{A})^\text{T}=\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}}{-5}$
From (1)
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}}{-5}\begin{bmatrix}2200\\3100\\1200\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}-2&-1&5\\-1&2&-5\\3&-1&-5\end{bmatrix}\begin{bmatrix}-440\\-620\\-240\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}300\\400\\500\end{bmatrix}$
Excellence in extra-curricular activities should be another value considered for an award.
View full question & answer→Question 1925 Marks
A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission.
|
Month
|
Sale of units
|
Total commission drawn (in Rs.)
|
|
|
A
|
B
|
C
|
|
|
Jan
|
90
|
100
|
20
|
800
|
|
Feb
|
130
|
50
|
40
|
900
|
|
March
|
60
|
100
|
30
|
850
|
Find out the rates of commission on items A, B and C by using determinant method.
AnswerLet the rates of commissions on iteams A, B and C be x, y and z respectively.
Then we can express the given modal as system of linear equations
90x + 100y + 20z = 800
130x + 50y + 40z = 900
60x + 100y + 30z = 850
We will solve this using the Cramer's rule
Here,
$\text{D}=\begin{vmatrix}90&100&20\\130&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-200&0&-50\end{vmatrix}$
$=50(8500-12000)=-175000$
$\text{D}_1=\begin{vmatrix}800&100&20\\900&50&40\\60&100&30\end{vmatrix}=\begin{vmatrix}-170&0&-60\\130&50&40\\-950&0&-50\end{vmatrix}$
$=50(50000-57000)=-350000$
$\text{D}_2=\begin{vmatrix}90&800&20\\130&900&40\\60&850&30\end{vmatrix}=\begin{vmatrix}90&800&20\\-50&-700&0\\-75&-350&0\end{vmatrix}$
$=20(17500-52500)=-700000$
$\text{D}_3=\begin{vmatrix}90&100&800\\130&50&900\\60&100&850\end{vmatrix}=\begin{vmatrix}-170&0&-1000\\130&50&900\\-200&0&-950\end{vmatrix}$
$=50(161500-200000)=-1925000$
$\therefore\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-350000}{-175000}=2$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-700000}{-175000}=4$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-1925000}{-175000}=11$
$\therefore$ The rates of commission of iteam A, B and C are 2%, 4% and 11% respectively.
View full question & answer→Question 1935 Marks
Solve the following systems of linear equations by cramer's rule:
$x + y + z + 1 = 0,$
$ax + by + cz + d = 0,$
$a^2x + b^2y + x^2z + d^2 = 0$
AnswerThese equations can be written as
$x + y + z + 1 = 0$
$ax + by + cz + d = 0$
$a^2x + b^2y + x^2z + d^2 = 0$
$\text{D}=\begin{vmatrix}1&1&1\\\text{a}&\text{b}&\text{c}\\\text{a}^2&\text{b}^2&\text{c}^2 \end{vmatrix}$
$=\begin{vmatrix}1&0&0\\\text{a}&\text{a}-\text{b}&\text{b}-\text{c}\\\text{a}^2&\text{a}^2-\text{b}^2&\text{b}^2-\text{c}^2 \end{vmatrix} $ [Applying $C_2 → C_1 - C_2, C_3 → C_2 - C_3$]
Taking (b - a) and (c - a) common from $C_1$ and $C_2$, respectively, we get
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}1&0&0\\\text{a}&1&1\\\text{a}^2&\text{a}+\text{b}&\text{b}+\text{c}\end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})\ ....(\text{i})$
$\text{D}_1=\begin{vmatrix}-1&1&1\\-\text{d}&\text{b}&\text{c}\\-\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{d}&\text{b}&\text{c}\\\text{d}^2&\text{b}^2&\text{c}^2\end{vmatrix}$
$\text{D}_1=-(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})$ [Replacing a by d in eq. (i)]
$\text{D}_2=\begin{vmatrix}1&-1&1\\\text{a}&-\text{d}&\text{c}\\\text{a}^2&-\text{d}^2&\text{c}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{c}\\\text{a}^2&\text{d}^2&\text{c}^2\end{vmatrix}$
$\text{D}_2=-(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})$
$\text{D}_3=\begin{vmatrix}1&1&-1\\\text{a}&\text{b}&-\text{d}\\\text{a}^2&\text{b}^2&-\text{d}^2\end{vmatrix}=-\begin{vmatrix}1&1&1\\\text{a}&\text{d}&\text{d}\\\text{a}^2&\text{b}^2&\text{d}^2\end{vmatrix}$
$\text{D}_3=-(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})$
Thus,
$\text{x}=\frac{\text{D}_1}{\text{D}}=-\frac{(\text{d}-\text{b})(\text{b}-\text{c})(\text{c}-\text{d})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=-\frac{(\text{a}-\text{d})(\text{d}-\text{c})(\text{c}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{(\text{a}-\text{b})(\text{b}-\text{d})(\text{d}-\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
View full question & answer→Question 1945 Marks
Find $A^{-1}$, If $\text{A}=\begin{bmatrix}1&2&5\\ 1&-1&-1\\ 2&3&-1\end{bmatrix}$.
Hence solve the follwing system of linear equations:
$x + 2y +5z = 10, x- y - z = - 2, 2x + 3y - z = - 11$
Answer$\text{A}=\begin{bmatrix}1&2&5\\ 1&-1&-1\\ 2&3&-1\end{bmatrix}$
$\text{|A|}=1{(1+3)}-2{(-1+2)}+5{(5)}=4-2+25=27\neq0$ $\text{C}_{11}=4\\ \text{C}_{12}=-1\\ \text{C}_{13}=5$ $\text{C}_{31}=3\\ \text{C}_{32}=6\\ \text{C}_{33}=-3$ $\text{C}_{21}=17\\ \text{C}_{22}=-11\\ \text{C}_{23}=1$ $\text{A}^{-1}=\frac{1}{\text{|A|}}\times\text{adj A}=\frac{1}{27}\begin{bmatrix}4&17&3\\ -1&-11&6\\ 5&1&-3\end{bmatrix}$
Now, the given set of equations can be represented as: $\text{x}+2\text{y}+5\text{z}=10$
$\text{x}-\text{y}-\text{z}=-2$ $2\text{x}+3\text{y}-\text{z}=-11$ or $\begin{bmatrix}1&2&5\\ 1&-1&-1\\ 2&3&-1\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}10\\ -2\\ -11\end{bmatrix}$ or $\text{X = A}^{-1}\times\text{B}$
$=\frac{1}{27}\begin{bmatrix}4&17&3\\ -1&-11&6\\ 5&1&-3\end{bmatrix}\begin{bmatrix}10\\ -2\\ -11\end{bmatrix}$
$=\frac{1}{27}\begin{bmatrix}40-34-33\\ -10+22-66\\ 50-2+33\end{bmatrix}=\frac{1}{27}\begin{bmatrix}-27\\ -54\\ 81\end{bmatrix}=\begin{bmatrix}-1\\ -2\\ 3\end{bmatrix}$
Hence, x = -1, y = -2, z = 3
View full question & answer→Question 1955 Marks
Solve the following systems of linear equations by cramer's rule:
$6x + y - 3z = 5,$
$x + 3y - 2z = 5,$
$2x + y + 4z = 8$
AnswerLet $\text{D}=\begin{vmatrix}6&1&-3\\1&3&-2\\2&1&4\end{vmatrix}$
Expanding along $R_1$
$=6(14)-1(8)-3(-5)$
$=84-8+15=91$
Also, $\text{D}_1=\begin{vmatrix}5&1&-3\\5&-3&-2\\8&1&4\end{vmatrix}$
Expanding along $R_1$
$=5(14)-1(36)-3(-19)$
$=70-36+57=91$
Again $\text{D}_2=\begin{vmatrix}6&5&-3\\1&5&-2\\2&8&4\end{vmatrix}$
Expanding along $R_1$
$=6(36)-5(8)-3(-2)$
$=216-40+6=182$
Also $\text{D}_3=\begin{vmatrix}6&1&5\\1&3&5\\2&1&8\end{vmatrix}$
Expanding along $R_1$
$=6(19)-1(-2)+5(-5)$
$=114+2-25=91$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{91}{91}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{182}{91}=2$
Also $\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{91}{91}=1$
Hence, x = 1, y = 2, z = 1
View full question & answer→Question 1965 Marks
Show that the following system of linear equations is consistent and also find solution:
$x + y + z = 6$
$x + 2y + 3z = 14$
$x + 4y + 7z = 30$
AnswerThis system can be written as:
$\begin{bmatrix}1&1&1\\ 1&2&3\\ 1&4&7\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}6\\ 14\\ 30\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1{(2)}-1{(4)}+1{(2)}$
$=2-4+2$
$=0$
So, A is singular, thus the given system has either no solutions or infinite solutions depending on as
$(\text{Adj A})\times\text{(B)}\neq0$ or $(\text{Adj A})\times\text{(B)}=0$
Let $C_{ij}$ be the co-factors of $a_{ij}$ in A
$\text{C}_{11}=2\\ \text{C}_{21}=-3\\ \text{C}_{31}=1$
$\text{C}_{12}=-4\\ \text{C}_{22}=6\\ \text{C}_{32}=-2$
$\text{C}_{13}=2\\ \text{C}_{23}=-3\\ \text{C}_{33}=1$
$\text{adj A}=\begin{bmatrix}2&-4&2\\ -3&6&-3\\ 1&-2&1\end{bmatrix}^\text{T}=\begin{bmatrix}2&-3&1\\ -4&6&-2\\ 2&-3&1\end{bmatrix}$
$(\text{adj A})\times\text{B}=\begin{bmatrix}2&-4&2\\ -3&6&-3\\ 1&-2&1\end{bmatrix}\begin{bmatrix}6\\ 14\\ 30\end{bmatrix}\begin{bmatrix}12-42+30\\ -24+84-60\\ 12-42+30\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
So, $\text{AX = B}$ has infinite solutions.
Now, let z = k
So, x + y = 6 - k
x + 2y = 14 - 3k
which can be written as:
$\begin{bmatrix}1&1\\ 1&2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}6-\text{k}\\ 14-\text{3k}\end{bmatrix}$
or $\text{AX = B}$
$\text{|A|}=1\neq0$
$\text{adj A}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}^\text{T}=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$
$\text{X = A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{adj A}\times\text{B}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\frac{1}{1}\begin{bmatrix}2&-1\\ -1&1\end{bmatrix}\begin{bmatrix}6-\text{k}\\ 14-3\text{k}\end{bmatrix}$
$=\begin{bmatrix}12-2\text{k}-14+3\text{k}\\ -6+\text{k}+14-3\text{k}\end{bmatrix}$
$=\begin{bmatrix}12-2\text{k}-14+3\text{k}\\ -6+\text{k}+14-3\text{k}\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2+\text{k}\\ 8-2\text{k}\end{bmatrix}$
Hence, x = k - 2
y = 8 - 2k
z = k
View full question & answer→Question 1975 Marks
Using the property of determinants and without expanding, prove that:
$\begin{vmatrix}b+c&q+r&y+z\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}=2\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}$
Answer$\text{L.H.S.}= \begin{vmatrix}(b+c)&q+r&y+z\$c+a)&r+p&z+x\$a+b)&p+q&x+y\end{vmatrix}\ \text{operating}\ \text{R}_1\rightarrow \text{R}_1+\text{R}_2+\text{R}_3$
$=\begin{vmatrix}b+c+c+a+a+b&q+r+r+p+p+q&y+z+z+x+x+y\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}$
$=\begin{vmatrix}2(a+b+c)&2(p+q+r)&2(x+y+z)\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}$
$=2\begin{vmatrix}(a+b+c)&(p+q+r)&(x+y+z)\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}$
$=2\begin{vmatrix}b&q&y\\c+a&r+p&z+x\\a+b&p+q&x+y\end{vmatrix}\ \left[\text{operating}\ \text{R}_1\rightarrow \text{R}_1-\text{R}_2\right]$
$=2\begin{vmatrix}b&q&y\\c+a&r+p&z+x\\a&p&x\end{vmatrix}\ \left[\text{operating}\ \text{R}_3\rightarrow \text{R}_3-\text{R}_1\right]$
$=2\begin{vmatrix}b&q&y\\c&r&z\\a&p&x\end{vmatrix}\ \left[\text{operating}\ \text{R}_2\rightarrow \text{R}_2-\text{R}_3\right]$
$=-2\begin{vmatrix}b&q&y\\a&p&x\\c&r&z\end{vmatrix}\ \left[\text{interchanging}\ \text{R}_2\ \text{and} \ \text{R}_3\right]$
$=-(-2)\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}\ \left[\text{interchanging}\ \text{R}_1\ \text{and} \ \text{R}_2\right]$
$=2\begin{vmatrix}a&p&x\\b&q&y\\c&r&z\end{vmatrix}=\text{R.H.S.}$
View full question & answer→Question 1985 Marks
Solve the following systems of linear equations by cramer's rule:
5x - 7y + z = 11,
6x - 8y - z = 15,
3x + 2y - 6z = 7
Answer$\text{D}=\begin{vmatrix}5&-7&1\\6&-8&-1\\3&2&-6 \end{vmatrix}$
$=5(48+2)+7(-36+3)+1(12+24)$
$=5(50)+7(-33)+1(36)=55$
$\text{D}_1=\begin{vmatrix}11&-7&1\\15&-8&-1\\7&2&-6 \end{vmatrix}$
$=11(48+2)+7(-90+7)+1(30+36)$
$=11(50)+7(-83)+1(86)=55$
$\text{D}_2=\begin{vmatrix}5&11&1\\6&15&-1\\3&7&-6 \end{vmatrix}$
$=5(-90+7)-11(-36+3)+1(42-45)$
$=5(-83)-11(-33)+1(-3)=-55$
$\text{D}_3=\begin{vmatrix}5&-7&11\\6&-8&15\\3&2&7 \end{vmatrix}$
$=5(-56-30)+7(42-45)+11(12+24)$
$=5(-86)+7(-3)+11(36)=-55$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{55}{55}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-55}{55}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-55}{55}=-1$
$\therefore\text{x}=1,\text{ y}=-1$ and $\text{z}=-1$
View full question & answer→Question 1995 Marks
If $\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$ are two square matrices, find AB and hence solve the system of linear equations:
x - y = 3, 2x + 3y + 4z = 17, y + 2z = 7
AnswerHere,
$\text{A}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\text{and}$$\text{B}=\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}1&-1&0\\ 2&3&4\\ 0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}6&0&0\\ 0&6&0\\ 0&0&6\end{bmatrix}$
$\Rightarrow\text{AB}=6\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$
$\Rightarrow\text{AB}=6\text{I}_{3}$
$\Rightarrow\frac{1}{6}\text{AB}=\text{I}_{3}$
$\Rightarrow\big(\frac{1}{6}\text{B}\big)\text{A}=\text{I}_{3}\ (\because\text{AB}=\text{AB})$
$\Rightarrow\text{A}^{-1}=\frac{1}{6}\text{B}$
$\Rightarrow\text{A}^{-1}=\frac{1}{6}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\text{X}=\frac{1}{6}\begin{bmatrix}2&2&-4\\ -4&2&-4\\ 2&-1&5\end{bmatrix}\begin{bmatrix}3\\ 17\\ 7\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}6+34-28\\ -12+34-28\\ 6-17+35\end{bmatrix}$
$\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{6}\begin{bmatrix}12\\ -6\\ 24\end{bmatrix}$
$\therefore$ x = 2, y = -1 and z = 4
View full question & answer→Question 2005 Marks
If $\text{A}=\begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix},$ find $A^{-1}$ and show that $\text{A}^{-1}=\frac{1}{2}(\text{A}^2-3\text{I}).$
Answer$|\text{A}|=\begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$
$=0-1(0-1)+1(1-0)$
$= 0+1+1=2\neq0$
So, A is invertuble matrix.
$A_{11} = (-1)^{1+1} (-1) = -1; A_{12} = (-1)^{1+2} (-1) = 1; A_{13} = (-1)^{1+3} (1) = 1$
$A_{21} = (-1)^{2+1} (-1) = 1; A_{22} = (-1)^{2+2} (-1) = -1; A_{23} = (-1)^{2+3} (-1) = 1$
$A_{31} = (-1)^{3+1} (1) = 1; A_{32} = (-1)^{3+2} (-1) = 1; A_{33} = (-1)^{3+3} (-1) = -1$
$\text{adj A}=\begin{bmatrix}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}^\text{T}$
$=\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{1}{2}\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}\ .....\text{(i)}$
$\text{A}^2-3\text{I}=\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$
$=-3\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\text{A}^2-3\text{I}=\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}-=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$
$\text{A}^2-3\text{I}=\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}\ .....\text{(ii)}$
From (i) and (ii) we can see that,
$\text{A}^{-1}=\frac{1}{2}(\text{A}^2-3\text{I})$
View full question & answer→