Question 1515 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 1525 Marks
Show that the following systems of linear equations has infinite number of solutions and solve:
x + y - z = 0,
x - 2y + z = 0,
3x + 6y - 5z = 0
AnswerUsing the equations we get,
$\text{D}=\begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}$
$=1(10-6)-1(-5-3)-1(6+6)=0$
$\text{D}_1=\begin{vmatrix}1&1&-1\\0&-2&1\\0&6&-5\end{vmatrix}$
$=0(10-6)-1(0-0)-1(0+0)=0$
$\text{D}_2=\begin{vmatrix}1&0&-1\\1&0&1\\3&0&-5\end{vmatrix}$
$=1(0-0)-0(5-3)-1(0-0)=0$
$\text{D}_3=\begin{vmatrix}1&1&0\\1&-2&0\\3&6&0\end{vmatrix}$
$=1(0-0)-1(0-0)+0(6+6)=0$
$\text{D}=\text{D}_1=\text{D}_2$
Thus, the system has infinitely many solution.
View full question & answer→Question 1535 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, $A = IA$
$\begin{bmatrix}3 & 10 \\ 2 & 7 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
Applying $\text{R}_1\rightarrow\frac{1}{3}\text{ R}_1$
$\begin{bmatrix} 1 & \frac{10}{7} \\ 2 & 7 \end{bmatrix}=\begin{bmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
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 $R_2 \rightarrow 3R_2$
$\begin{bmatrix}1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ -2 & 3 \end{bmatrix}\text{A}$
Applying $\text{R}_1\rightarrow\text{R}_1-\frac{10}{3}\text{ R}_3$
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\text{A}$
$I = BA$
Hence$, B$ is the inverse of $A.$
View full question & answer→Question 1545 Marks
For the matrix $\text{A}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}.$ Show that $\ce{A^{-3} - 6A^2 + 5A + 11I_3 = 0}$ 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, $\ce{A^3 - 6A^2 + 5A + 11I = 0}$
$\Rightarrow \ce{(AAA)A^{-1} - 6(AA)A^{-1} + 5AA^{-1} + 11IA^{-1} = 0}[$Post$-$multiplying by $A^{-1}$ as $|\text{A}|\neq0$]
$\Rightarrow \ce{AA(AA^{-1}) - 6A(AA^{-1}) + 5(AA^{-1}) = -11(IA^{-1})}$
$\Rightarrow \ce{A^2 - 6A + 5I = -11A^{-1}}$
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{11}(\text{A}^2-6\text{A}+5\text{I})\ .....(\text{i})$
Now,
$\ce{A^2 - 6A + 5I}$
$=\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}-6\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}+5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}-\begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix}+\begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$=\begin{bmatrix}9 & 2 & 1 \\ -3 & 13 & -14 \\ 7 & -3 & 19 \end{bmatrix}-\begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix}$
$=\begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$
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 1555 Marks
If $\text{A}^{-1}=\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix},$ 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 1565 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 1575 Marks
Solve the following systems of linear equations by cramer's rule:
3x + y = 19,
3x - y = 23
AnswerGiven, 3x + y = 19
3x - y = 23
Using cramer's Rule, we get
$\text{D}=\begin{vmatrix}3&1\\3&-1\end{vmatrix}=-3-3=-6$
$\text{D}_1=\begin{vmatrix}19&1\\23&-1\end{vmatrix}=-19-23=-42$
$\text{D}_2=\begin{vmatrix}3&19\\3&23\end{vmatrix}=(3\times23)-(3\times19)=3\times4=12$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-42}{-6}=7$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{-6}=-2$
$\therefore\text{x}=7$ and $\text{y}=-2$
View full question & answer→Question 1585 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}$ 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 1595 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 1605 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}$ in $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 1615 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,
$\ce{A^2 - 12A + I^2 = 0}$
$\Rightarrow \ce{I^2 = 12A - A^2}$
$\Rightarrow \ce{A^{-1}I_2 = A^{-1} (12A - A^2)}[$Pre$-$multiplying both sides by $A^{-1}]$
$\Rightarrow \ce{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 1625 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 1635 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}$ 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}$
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}$ 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 1645 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 = B$ Where $\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 1655 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 1665 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 \rightarrow C_3 - C_2, C_4 \rightarrow 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 1675 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 1685 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}$ is not possible
View full question & answer→Question 1695 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 1705 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 1715 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 1725 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 1735 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 1745 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 1755 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 1765 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 1775 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 1785 Marks
Solve the following determinant equations:
$\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}=0,\text{a}\neq0$
AnswerLet $\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+\text{a}&\text{x}&\text{x}\\3\text{x}+\text{a}&\text{x}+\text{a}&\text{x}\\3\text{x}+\text{a}&\text{x}&\text{x}+\text{a}\end{vmatrix} [$Applying $C_1 \rightarrow C_1 + C_2 + C_3]$
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\1&\text{x}+\text{a}&\text{x}\\1&\text{x}&\text{x}+\text{a}\end{vmatrix}$
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&\text{x}&\text{x}+\text{a}\end{vmatrix} [$Applying $R_2 \rightarrow R_2 - R_1]$
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&0&\text{a}\end{vmatrix} [$Applying $R_3 \rightarrow R_3 - R_1]$
$=(3\text{x}+\text{a})=(\text{a}^2-0)=0$
$\text{x}=\frac{-\text{a}}{3}$
View full question & answer→Question 1795 Marks
Solve the following determinant equations:
$\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}1&1&\text{x}\\\text{p}+1&\text{p}+1&\text{p}+\text{x}\\3&\text{x}+1&\text{x}+2\end{vmatrix}=0$
$=\begin{vmatrix}1&1&\text{x}\\\text{p}&\text{p}&\text{p}\\3&\text{x}+1&\text{x}+2\end{vmatrix} [$Applying $R_{2 }\rightarrow R_2 - R_1]$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\3&\text{x}+1&\text{x}+2\end{vmatrix}$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\2&\text{x}&2\end{vmatrix}$
$=\text{p}\begin{vmatrix}0&1&\text{x}\\0&1&1\\2-\text{x}&\text{x}&2\end{vmatrix} [$Applying $C_1 \rightarrow C_2 - C_1]$
$=\text{p}\left\{(2-\text{x})\times\begin{vmatrix}1&\text{x}\\1&1 \end{vmatrix}\right\} [$Expanding along $C_1]$
$=\text{p}(2-\text{x})(1-\text{x})=0$
$\text{x}=1,2$
View full question & answer→Question 1805 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 1815 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 1825 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 1835 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 1845 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 1855 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
$\ce{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
$\ce{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$
$\ce{\Rightarrow (A^{-1} A) X (BB^{-1}) = A^{-1}CB^{-1}}$
$\ce{\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 1865 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 1875 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 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 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 \rightarrow C_1 - C_2, C_3 \rightarrow 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 1905 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 1915 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 1925 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 1935 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 1945 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.
$\ce{A_{11} = (-1)^{1+1} (-1) = -1; A_{12} = (-1)^{1+2} (-1) = 1; A_{13} = (-1)^{1+3} (1) = 1}$
$\ce{A_{21} = (-1)^{2+1} (-1) = 1; A_{22} = (-1)^{2+2} (-1) = -1; A_{23} = (-1)^{2+3} (-1) = 1}$
$\ce{A_{31} = (-1)^{3+1} (1) = 1; A_{32} = (-1)^{3+2} (-1) = 1; A_{33} = (-1)^{3+3} (-1) = -1}$
$\text{adj A}=\begin{bmatrix}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}^\text{T}$
$=\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\frac{1}{2}\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}\ .....\text{(i)}$
$\text{A}^2-3\text{I}=\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$
$=-3\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\text{A}^2-3\text{I}=\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}-=\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$
$\text{A}^2-3\text{I}=\begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}\ .....\text{(ii)}$
From $(i)$ and $(ii)$ we can see that,
$\text{A}^{-1}=\frac{1}{2}(\text{A}^2-3\text{I})$
View full question & answer→Question 1955 Marks
Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?
AnswerLet x, y and z be the prize amount per person for Resourcefulness, Competence and Determination respectively.
As per the data in the question, we get
4x + 3y + 2z = 37000
5x + 3y + 4z = 47000
x + y + z = 12000
The above three simultaneous equations can be written in matrix form as
$\begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}37000\\47000\\12000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}37000\\47000\\12000\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}4&3&2\\5&3&4\\1&1&1\end{bmatrix}$
$|\text{A}|=4(-1)-3(1)+2(2)=-3$
$\text{cof }\text{A}=\begin{bmatrix}-1&-1&2\\-1&2&-1\\6&-6&-3\end{bmatrix}$
$\text{adj }\text{A}=(\text{cof }\text{A})^\text{T}=\begin{bmatrix}-1&-1&6\\-1&2&-6\\2&-1&-3\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{1}{-3}\begin{bmatrix}-1&-1&2\\-1&2&-1\\6&-6&-3\end{bmatrix}$
From (1)
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-1&-1&2\\-1&2&-1\\6&-6&-3\end{bmatrix}\begin{bmatrix}37000\\47000\\12000\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-12000\\-15000\\-9000\end{bmatrix}=\begin{bmatrix}4000\\5000\\3000\end{bmatrix}$
The values of x, y and z describe the amount of prizes per person of Resourcefulness, Competence and Determination.
View full question & answer→Question 1965 Marks
If $\text{A}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$, find $A^{-1}$ and hence solve the system of linear equations:
$2x - 3y + 5z = 11, 3x + 2y - 4z = -5, x + y + 2z = -3$
AnswerHere,
$\text{A}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$
$\text{|A|}=\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}$
$=2{(-4+4)}+3{(-6+4)}+5{(3-2)}$
$=0-6+5$
$=-1$
Let $C_{ij}$ be the co$-$factors of the elements $a_{ij}$ in $A [a_{ij}].$ Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}2&-4\\ 1&-2\end{vmatrix}=0,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}3&-3\\ 1&-2\end{vmatrix}=2,\\ \text{C}_{13}{(-1)}^{1+3}\begin{vmatrix}3&2\\ 1&1\end{vmatrix}=1$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}-3&5\\ 1&-2\end{vmatrix}=-1,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}2&5\\ 1&-2\end{vmatrix}=-9,\\ \text{C}_{23}{(-1)}^{2+3}\begin{vmatrix}2&-3\\ 1&1\end{vmatrix}=-5$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}-3&5\\ 2&-4\end{vmatrix}=2,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}2&5\\ 3&-4\end{vmatrix}=23,\\ \text{C}_{23}{(-1)}^{3+3}\begin{vmatrix}2&-3\\ 3&2\end{vmatrix}=13$
$\text{adj A}=\begin{bmatrix}0&2&1\\ -1&-9&-5\\ 2&23&13\end{bmatrix}^\text{T}$
$=\begin{bmatrix}0&-1&2\\ 2&-9&23\\ 1&-5&13\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-1}\begin{bmatrix}0&-1&2\\ 2&-9&23\\ 1&-5&13\end{bmatrix}$
The given system of equations can be wriiten in matrix form as follows:
$\begin{bmatrix}2&-3&5\\ 3&2&-4\\ 1&1&-2\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}0&-1&2\\ 2&-9&23\\ 1&-5&13\end{bmatrix}\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}0+5-6\\ 22+45-69\\ 11+25-39\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-1}\begin{bmatrix}-1\\ -2\\ -3\end{bmatrix}$
$\Rightarrow\text{x}=\frac{-1}{-1},\text{y}=\frac{-2}{-1}$ and $\text{z}=\frac{-3}{-1}$
$\therefore x = 1, y = 2$ and $z = 3$
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
Prove that $|A \ adj \ A| = |A|^n.$
AnswerLet $A = [a_{ij}]$ be a square matrix of order $n \times n.$
If $C_{ij}$ is a cofactor of $a_{ij}$ in $A,$ then $adj \ A = [C_{ij}]^T = [C_{ij}].$
Also, it is a matrix of order $n \times n.$
Because $A$ and ahj $A$ are matrices of order $n \times n, A \times (adj \ A)$ exists and is of order $n \times n.$
$\Rightarrow\left\{\text{A}\times(\text{adj A})\right\}_{\text{ij}}=\sum\limits_{\text{r}-1}^\text{n}\text{A}_{\text{ir }}(\text{adj A})_\text{rj}$
$=\sum\limits_{\text{n}}^{\text{r}-1}\text{a}_{\text{i r}}\text{C}_{\text{r j}}=\begin{cases}|\text{A}| \text{ if i}=\text{j}\\ 0 \text{ otherwise}\end{cases}$
Thus, each diagonal element of $A \times (adj \ A)$ is $|A|.$
Also, the non$-$diagonal elements are zero.
$\Rightarrow\ \text{A}\times(\text{adj A})=\begin{bmatrix}|\text{A}| & 0 & 0 & ......... & 0 \\ 0 & |\text{A}| & 0 & .......... & 0 \\ 0 & 0 & |\text{A}| & 0 ....... & 0 \\ . \\ . \\ . \\ 0 & 0 & 0 & .......... & |\text{A}|\end{bmatrix} $
$\Rightarrow\ |\text{A}\times(\text{adj A})|=\begin{bmatrix}|\text{A}| & 0 & 0 & ......... & 0 \\ 0 & |\text{A}| & 0 & .......... & 0 \\ 0 & 0 & |\text{A}| & 0 ....... & 0 \\ . \\ . \\ . \\ 0 & 0 & 0 & .......... & |\text{A}|\end{bmatrix} $
$=|\text{A}|^\text{n}\begin{bmatrix} 1 & 0 & 0 & ......... & 0 \\ 0 & 1 & 0 & .......... & 0 \\ 0 & 0 & 1 & 0 ....... & 0 \\ . \\ . \\ . \\ 0 & 0 & 0 & .......... & 1\end{bmatrix} $
$=|\text{A}|^\text{n}\text{I}_\text{n}=|\text{A}|^\text{n}$
$\Rightarrow\ |\text{A}\times(\text{adj A})|$
$=|\text{A}|^\text{n}$
Hence proved.
View full question & answer→Question 1995 Marks
Solve the following system of equations by matrix method: $5x +3y + z = 16 , 2x + y +3z = 19 ,x + 2y + 4z = 25$
AnswerHere,
$\text{A}=\begin{bmatrix}5&3&1\\ 2&1&3\\ 1&2&4\end{bmatrix}$
$\text{|A|}=\begin{vmatrix}5&3&1\\ 2&1&3\\ 1&2&4\end{vmatrix}$
$=5{(4-6)}-3{(8-3)}+1{(4-1)}$
$=-10-15+3$
$=-22$
Let $C_{ij}$be the co$-$factors of the elements aij in $A [a_{ij}].$ Then,
$\text{C}_{11}={(-1)}^{1+1}\begin{vmatrix}1&2\\ 2&4\end{vmatrix}=-2,\\ \text{C}_{12}={(-1)}^{1+2}\begin{vmatrix}2&3\\ 1&4\end{vmatrix}=-5,\\ \text{C}_{13} ={(-1)}^{1+3}\begin{vmatrix}2&1\\ 1&2\end{vmatrix}=3$
$\text{C}_{21}={(-1)}^{2+1}\begin{vmatrix}3&1\\ 2&4\end{vmatrix}=-10,\\ \text{C}_{22}={(-1)}^{2+2}\begin{vmatrix}5&1\\ 1&4\end{vmatrix}=19,\\ \text{C}_{23} ={(-1)}^{2+3}\begin{vmatrix}5&3\\ 1&2\end{vmatrix}=-7$
$\text{C}_{31}={(-1)}^{3+1}\begin{vmatrix}3&1\\ 1&3\end{vmatrix}=-8,\\ \text{C}_{32}={(-1)}^{3+2}\begin{vmatrix}5&1\\ 2&3\end{vmatrix}=-13,\\ \text{C}_{33} ={(-1)}^{3+3}\begin{vmatrix}5&3\\ 2&1\end{vmatrix}=-1$
$\text{adj A}=\begin{bmatrix}-2&-5&3\\ -10&19&-7\\ 8&-13&-1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}-2&-10&8\\ -5&19&-13\\ 3&-7&-1\end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj A}$
$=\frac{1}{-22}\begin{bmatrix}-2&-10&8\\ -5&19&-13\\ 3&-7&-1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-2&-10&8\\ -5&19&-13\\ 3&-7&-1\end{bmatrix}\begin{bmatrix}16\\ 19\\ 25\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-32-190+200\\ -80+361-325\\ 48-133-25\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\\ \text{z}\end{bmatrix}=\frac{1}{-22}\begin{bmatrix}-22\\ -44\\ -110\end{bmatrix}$
$\Rightarrow\text{x}=\frac{-22}{-22},\text{y}=\frac{-44}{-22}$ and $\text{z}=\frac{-110}{-22}$
Hence, $x = 1, y = 2, z = 5$
View full question & answer→Question 2005 Marks
Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping clam in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honuor respectively 2, 4 and 3 employees with a total prize money of ₹ 29000. The second factory decided to honuor respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then
- Represent the above situation by matrix equation and form linear equation using matrix multiplication.
- Solve this equation by matrix method.
- Which values are reflected in the questions?
AnswerLet x, y and z be the prize amount per person for adaptibility, carefulness and calmness respectively.
as per the given data, we get
2x + 4y + 3z = 29000
5x + 2y + 3z = 30500
x + y + z = 9500
The above three simultaneous equations can be written in the matrix form as
$\begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}29000\\30500\\9500\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix}^{-1}\begin{bmatrix}29000\\30500\\9500\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}2&4&3\\5&2&3\\1&1&1\end{bmatrix}$
$|\text{A}|=2(-1)-4(2)+3(3)=-1$
$\text{cof }\text{A}=\begin{bmatrix}-1&-2&3\\-1&-1&2\\6&9&-16\end{bmatrix}$
$\text{adj }\text{A}=(\text{cof }\text{A})^\text{T}=\begin{bmatrix}-1&-2&3\\-1&-1&2\\6&9&-16\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{=\begin{bmatrix}-1&-1&6\\-2&-1&9\\3&2&-16\end{bmatrix}}{-1}$$=\begin{bmatrix}1&1&-6\\2&1&-9\\-3&-2&16\end{bmatrix}$
From (1)
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1&1&-6\\2&1&-9\\-3&-2&16\end{bmatrix}\begin{bmatrix}29000\\30500\\9500\end{bmatrix}\ \dots(1)$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}2500\\3000\\4000\end{bmatrix}$
Keeping calm in a tense situation is more rewarding than carefulness, and carefulness is more rewarding than adaptability.
View full question & answer→