Question 1014 Marks
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.
AnswerLet ₹ x, ₹ y, ₹ z per kg be the prices of onion, wheat and rice respectively. $\therefore$ According to given data, we have three equations, 4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
Matrix form of given equations is AX = B $\Rightarrow\ \begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\90\\70\end{bmatrix}$ $\text{Here}\ \text{A}=\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}60\\90\\70\end{bmatrix}$ $\therefore\ \left|\text{A}\right|=\begin{vmatrix}4&3&2\\2&4&6\\6&2&3\end{vmatrix}=4(12-12)-3(6-36)+2(4-24)=0+90-40=50\neq0$ Therefore, solution is unique and $\text{X=A}^{-1}\text{B}=\frac{1}{\text{|A|}}\text{(adj. A)B} \dots\dots(1)$ Now, A11 = 0, A12 = 30, A13 = -20 A21 = -5, A22 = 0, A23 = 10 A31 = 10, A32 = -20, A33 = 10 $\therefore\ \text{adj. A}=\begin{bmatrix}0&30&-20\\-5&0&10\\10&-20&10\end{bmatrix}=\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}$ ⇒ From eq. (1), $\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}\begin{bmatrix}60\\90\\70\end{bmatrix}$ $=\frac{1}{50}\begin{bmatrix}-450+700\\1800-1400\\-1200+900+700\end{bmatrix}=\frac{1}{50}\begin{bmatrix}250\\400\\400\end{bmatrix}=\begin{bmatrix}5\\8\\8\end{bmatrix}$
Therefore, x = 5, y = 8 and z = 8 Hence, the cost of onion, wheat and rice are ₹ 5, ₹ 8 and ₹ 8 per kg. View full question & answer→Question 1024 Marks
Evaluate the following determinant:
$\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&3&9&27\\3&9&27&1\\9&27&1&3\\27&1&3&9 \end{vmatrix}$
$=1\begin{vmatrix}9&27&1\\27&1&3\\1&3&9 \end{vmatrix}-3\begin{vmatrix}3&27&1\\9&1&3\\27&3&6\end{vmatrix}\\+9\begin{vmatrix}3&9&1\\9&27&3\\27&1&9 \end{vmatrix}-27\begin{vmatrix}3&9&27\\9&27&1\\27&1&3 \end{vmatrix}$
$=1\{9(9-9)-27(243-3)+1(81-1)\}-3\{3(9-9)-27(81-81)+1(27-27)\}\\+9\{3(243-3)-9(81-81)+1(9-729)\}-27\{(81-1)-9(27-27)+27(9-729)\}$
$=1\{0-6480+80\}-3\{0-0+0\}+9\{720-0-720\}-27\{80-0-19440\}$
$=-6400+522720$
$=516320$
View full question & answer→Question 1034 Marks
Evaluate $\begin{vmatrix}2&3&-5\\7&1&-2\\-3&4&1\end{vmatrix}$ by two methods.
AnswerWe will evaluate the given determinant: - Along the first row.
$|\text{A}|=2\begin{vmatrix}1&-2\\4&1 \end{vmatrix}-3\begin{vmatrix}7&-2\\-3&1\end{vmatrix}-3\begin{vmatrix}7&1\\-3&4\end{vmatrix}$
$=2(1+8)-7(3+20)-3(-6+5)$
$=18-7(23)-3(-1)$
$=21-161$
$=-140$
- Along the first column.
$|\text{A}|=\begin{vmatrix}1&-2\\4&1 \end{vmatrix}-7\begin{vmatrix}3&-5\\4&1\end{vmatrix}-3\begin{vmatrix}3&-5\\1&-2\end{vmatrix}$
$=2(1+8)-7(3+20)-3(-6+5)$
$=18-7(23)-3(-1)$
$=18-161+3$
$=21-161$
$=-140$
We can see, the answer is same with both the methods.
View full question & answer→Question 1044 Marks
Find values of k, if area of triangle is 4 square units whose vertices are:
(-2, 0), (0, 4), (0, k)
Answer$4=\frac{1}{2}\begin{vmatrix}-2&0&1\\0&4&1\\0&\text{k}&1 \end{vmatrix}$
$\pm8=\begin{vmatrix}-2&0&1\\0&4&1\\0&\text{k}&1 \end{vmatrix}$
Expanding along R1
$\pm8=-2(4-\text{k})-0(0-0)+1(0)$
$\pm=8=-8+2\text{k}$
Taking positive (+) sign
$\pm=8=-8+2\text{k}$
Taking positive (-) sing
-8 = -8 + 2k or k = 0
Hence k = 0, 8
View full question & answer→Question 1054 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{b}+\text{c}&\text{c}+\text{a}&\text{a}+\text{b}\end{vmatrix}$
$=\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\\text{a}+\text{b}+\text{c}&\text{c}+\text{a}+\text{b}&\text{a}+\text{b}+\text{c}\end{vmatrix}$ [Applying R3 → R3 + R2]
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{a}-\text{b}&\text{b}-\text{c}&\text{c}-\text{a}\\1&1&1\end{vmatrix}$ [Taking (a + b + c) common]
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{b}&\text{c}&\text{a}\\1&1&1\end{vmatrix}$ [Applying R2 → R1 - R2]
$=(\text{a}+\text{b}+\text{c})\begin{vmatrix}\text{a}-\text{b}&\text{b}-\text{c}&\text{c}\\\text{b}-\text{c}&\text{c}-\text{a}&\text{a}\\0&0&1\end{vmatrix}$ [C1 → C1 - C2 and C2 → C2 - C3]
$=(\text{a}+\text{b}+\text{c})\big[-1\{(\text{a}-\text{b})(\text{c}-\text{a})-(\text{b}-\text{c})^2\}\big]$
$=(\text{a}+\text{b}+\text{c})\big[-\{\text{ac}-\text{bc}-\text{a}^2+\text{ab}-\text{b}^2-\text{c}^2+2\text{bc}\}\big]$
$=(\text{a}+\text{b}+\text{c})(\text{a}^2+\text{b}^2+\text{c}^2-\text{ab}-\text{bc}-\text{ca})$
$=\text{a}^3+\text{b}^3+\text{c}^3-3\text{abc}=\text{ R.H.S}$
View full question & answer→Question 1064 Marks
Let $\text{A}=\begin{bmatrix}3 & 2 \\7 & 5 \end{bmatrix}\text{and B}=\begin{bmatrix}6 & 7 \\8 & 9 \end{bmatrix}$. Find (AB)-1.
Answer$\text{A}=\begin{bmatrix}3 & 2\\7 & 5 \end{bmatrix}\therefore\ |\text{A}|=1\neq0\text{ and adj A}=\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$\therefore\ \text{A}^{-1}\frac{\text{adj A}}{|\text{A}|}=\frac{1}{1}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$\text{B}=\begin{bmatrix}6 & 7 \\7 & 9 \end{bmatrix}\therefore\ |\text{B}|=-2\neq0\text{ and adj B}=\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}$
$\therefore\ \text{B}^{-1}=\frac{\text{adj B}}{|\text{B}|}=\frac{1}{(-2)}=\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}$
Now, $(\text{AB})^{-1}=\text{B}^{-1}\text{A}^{-1}$
$(\text{AB})^{-1}=\frac{1}{(-2)}\begin{bmatrix}9 & -7 \\-8 & 6 \end{bmatrix}\begin{bmatrix}5 & -2 \\-7 & 3 \end{bmatrix}$
$(\text{AB})^{-1}=-\frac{1}{2}\begin{bmatrix}94 & -39 \\-82 & 34 \end{bmatrix}$
$\text{(AB)}^{-1}=\begin{bmatrix}-47 & \frac{39}{2} \\41 & -17 \end{bmatrix}$
View full question & answer→Question 1074 Marks
If $\text{A}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix},$ find the value of $\lambda$ so that $\text{A}^2=\lambda\text{A}-2\text{I}.$ Hence, find A-1.
Answer$\text{A}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$\text{A}^2=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$=\begin{bmatrix}1 & -2 \\4 & -4 \end{bmatrix}$
If $\text{A}^2=\lambda\text{A}-2\text{I}$
$\lambda\text{A}=\text{A}^2+2\text{I}$
$=\begin{bmatrix}1 & -2 \\4 & -4 \end{bmatrix}+\begin{bmatrix}2 & 0 \\0 & 2 \end{bmatrix}$
$\lambda\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$\lambda\begin{bmatrix}3\lambda & -2\lambda \\4\lambda & -2\lambda \end{bmatrix}=\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}$
$3\lambda=3$
$\lambda=1$
$\text{A}^2=\text{A}-2\text{I}$
Px multiplying by A-1
A-1 AA = A-1 A - A-1 I
A = I - 2A-1
$2\text{A}^{-1}=\text{I}-\text{A}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-\begin{bmatrix}3 & -2 \\4 & -2 \end{bmatrix}=\begin{bmatrix}-2 & 2 \\-4 & 3 \end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}-2 & 2 \\-4 & 3 \end{bmatrix}$
View full question & answer→Question 1084 Marks
Find the matrix X satisfying the equation:
$\begin{bmatrix}2 & 1 \\5 & 3 \end{bmatrix}\text{X}\begin{bmatrix}5 & 3 \\3 & 2 \end{bmatrix}=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}.$
AnswerLet $\text{A}=\begin{bmatrix}2 & 1 \\5 & 3 \end{bmatrix}$
$\text{B}=\begin{bmatrix}5 & 3 \\3 & 2 \end{bmatrix}$
Then the given equation can be wirtten as
A × B = I
⇒ X = A-1B-1
Now |A| = 6 - 5 = 1
|B| = 10 - 9 = 1
$\text{A}^{-1}=\frac{\text{adj (A)}}{|\text{A}|}=\begin{bmatrix}3 & -1 \\-5 & 2 \end{bmatrix}$
$\text{B}^{-1}=\frac{\text{adj (B)}}{|\text{B}|}=\begin{bmatrix}2 & -3 \\-3 & 5 \end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix} 3 & -1 \\-5 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\-3 & 2 \end{bmatrix}$
$=\begin{bmatrix} 9 & -14 \\-16 & 25 \end{bmatrix}$
View full question & answer→Question 1094 Marks
Prove that:
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
Answer$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}=-2$
$\text{L.H.S}=\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)(\text{a}+4)&\text{a}+4 &1\end{vmatrix}$
Apply R3 → R3 - R2
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)(\text{a}+3)&\text{a}+3&1\$\text{a}+3)2&1&0\end{vmatrix}$
Apply R2 → R2 - R1
$\begin{vmatrix}(\text{a}+1)(\text{a}+2)&\text{a}+2&1\$\text{a}+2)2&1&0\$\text{a}+3)2&1&0\end{vmatrix}$
$=[(2\text{a}+4)(1)-(1)(2\text{a}+6)]$
$=-2$
$=\text{R.H.S}$
View full question & answer→Question 1104 Marks
Solve the following determinant equations:
$\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
AnswerLet $\begin{vmatrix}15-2\text{x}&11-3\text{x}&7-\text{x}\\11&17&14\\10&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}15-2\text{x}-14+2\text{x}&11-3\text{x}&7-\text{x}\\11-28&17&14\\10-26&16&13\end{vmatrix}=0$ [Applying C1 → C1 - 2C3]
$\Rightarrow\begin{vmatrix}1&11-3\text{x}&7-\text{x}\\-17&17&14\\-16&16&13\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}12-3\text{x}&4-2\text{x}&7-\text{x}\\0&3&14\\0&3&13\end{vmatrix}=0$ [Applying C1 → C1 + C2 and C2 → C2 - C3]
$\Rightarrow(12-3\text{x})((3)\times13-(3\times14))=0$
$\Rightarrow(12-3\text{x})(-3)=0$
$\Rightarrow12-3\text{x}=0$
$\Rightarrow3\text{x}=12$
$\Rightarrow\text{x}=4$
View full question & answer→Question 1114 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&43&6\\7&35&4\\3&17&2\end{vmatrix}$
$=\begin{vmatrix}1&1&6\\7&7&4\\3&3&2\end{vmatrix}=0$ [Appliying C2 → C2 - 7C3]
View full question & answer→Question 1124 Marks
Show that x = 2 is a root of the equation $\begin{vmatrix}\text{x}&-6&-1\\2&-3\text{x}&\text{x}-3\\-3&2\text{x}&\text{x}+2\end{vmatrix}=0$ and solve it completely.
AnswerLet us show that x = 2 is a root of the given equation:
Putting x = 2 in the L.H.S, we get
$\begin{vmatrix}2&-6&-1\\2&-6&-1\\-3&4&4 \end{vmatrix}=0$
$\because\text{R}_1=\text{R}_2$
Hence, x = 2 is a root of the given equation.
Now, we see if there are any other roots. For this we need to solve the equation:
$\begin{vmatrix}\text{x}&-6&-1\\2&-3\text{x}&\text{x}-3\\-3&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-1&-6&-1\\\text{x}-1&-3\text{x}&\text{x}-3\\\text{x}-1&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\1&-3\text{x}&\text{x}-3\\1&2\text{x}&\text{x}+2\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\0&-3\text{x}+6&\text{x}-3+1\\0&2\text{x}+6&\text{x}+2+1\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)\begin{vmatrix}1&-6&-1\\0&-3(\text{x}-2)&\text{x}-2\\0&2(\text{x}+3)&\text{x}+3\end{vmatrix}=0$
$\Rightarrow(\text{x}-1)(\text{x}-2)(\text{x}+3)\begin{vmatrix}1&-6&-1\\0&-3&1\\0&2&1 \end{vmatrix}=0\\$
$\Rightarrow(\text{x}-1)(\text{x}-2)(\text{x}+3)=0$
$\Rightarrow(\text{x}-1)=0,(\text{x}-2)=0,(\text{x}+3)=0$
$\Rightarrow\text{x}=1,\text{x}=2,\text{x}=-3$
View full question & answer→Question 1134 Marks
Prove that:
$\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}=16(3\text{x}+4)$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{x}+4&\text{x}&\text{x}\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
$=\begin{vmatrix}3\text{x}+4&3\text{x}+4&3\text{x}+4\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
[Applying R1 → R2 + R2 + R3]
$=(3\text{x}+4)\begin{vmatrix}1&1&1\\\text{x}&\text{x}+4&\text{x}\\\text{x}&\text{x}&\text{x}+4\end{vmatrix}$
[Taking out (3x + 4) common from R1]
$=(3\text{x}+4)\begin{vmatrix}1&0&0\\\text{x}&4&0\\\text{x}&0&4\end{vmatrix}$
[Applying C2 → C2 - C1 and C3 → C3 - C1]
$=(3\text{x}+4)(4)^2$ [Expanding along R1]
$=16(3\text{x}+4)$
$=\text{R.H.S}$
View full question & answer→Question 1144 Marks
Find the adjoint of the matrix $\text{A}=\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$ and hence show that A (adj A) = |A|I3.
Answer$\text{A}=\begin{bmatrix} -9 & 8 &-2 \\ -8 & 7 & 2 \\ -5 & 4 & -1 \end{bmatrix}$
Now, to find Adj A
A11 = (-1)1+1 (-3) = -3, A21 = (-1)2+1 (-6) = 6, A31 = (-1)3+1 (6) = 6
A12 = (-1)1+2 (6) = -6, A22 = (-1)2+2 (3) = 3, A32 = (-1)3+2 (6) = -6
A13 = (-1)1+3 (-6) = -6, A23 = (-1)2+3 (6) = -6, A33 = (-1)3+3 (3) = 3
Therefore,
$\text{Adj .A}=\begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}$
$|\text{A}|=\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$
= -1(1 - 4) - 2(-2 - 4) + 2(4 + 2)
= 3 + 12 + 12
= 27
To show: A(adj A) = |A|I3
$\text{L.H.S}=\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}\begin{bmatrix} -3 & 6 & 6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}$
$\text{L.H.S}=\begin{bmatrix} 3+12+12 & -6-6+12 & -6+12-6 \\ -6-6+12 & 12+3+12 & 12-6-6 \\ -6+12-6 & 12-6-6 & 12+12+3 \end{bmatrix}$
$=\begin{bmatrix} 27 & 0 & 0 \\ 0 & 27 & 0 \\ 0 & 0 & 27 \end{bmatrix}$
$=27\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=|\text{A}|\text{I}_3$
$=\text{R.H.S}$
Hence, A(adj A) = |A|I3.
View full question & answer→Question 1154 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=\begin{vmatrix}0&\text{a}-\text{b}&\text{a}^2-\text{bc}-\text{b}^2+\text{ac}\\0&\text{b}-\text{c}&\text{b}^2-\text{ac}-\text{c}^2+\text{ab}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$ [applying R1 → R1 - R2, R2 → R2 - R3]
$=\begin{vmatrix}0&\text{a}-\text{b}&(\text{a}-\text{b})(\text{a}+\text{b})+\text{c}(\text{a}-\text{b})\\0&\text{b}-\text{c}&(\text{b}-\text{c})(\text{b}+\text{c})+\text{a}(\text{b}-\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$=(\text{a}-\text{b})(\text{b}-\text{c})\begin{vmatrix}0&1&(\text{a}+\text{b}+\text{c})\\0&1&(\text{a}+\text{b}+\text{c})\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 1164 Marks
If $\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}=0,$ find the value of $\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}},$ $\text{p}\neq\text{a},\text{q}\neq\text{b},\text{r}\neq\text{c}.$
Answer$\text{L.H.S}=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}$
$=\begin{vmatrix}\text{p}&\text{q}&\text{c}\\0&\text{q}-\text{b}&\text{c}-\text{r}\\\text{a}&\text{b}&\text{r} \end{vmatrix}$ [Applying R2 → R2 - R3]
$=\text{p}[\text{r}(\text{q}-\text{b})-\text{b}(\text{c}-\text{r})]+\text{a}[\text{b}(\text{c}-\text{r})-\text{c}(\text{q}-\text{b})]$
$=\text{pr}(\text{q}-\text{b})+\text{pb}(\text{r}-\text{c})-\text{ab}(\text{r}-\text{c})-\text{ac}(\text{q}-\text{b})$
$=(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})$
Since, $\triangle=0$
$\therefore(\text{pr}-\text{ac})(\text{q}-\text{b})+\text{b}(\text{p}-\text{a})(\text{r}-\text{c})=0$
$\Rightarrow\frac{\text{pr}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{pr}-\text{ar}+\text{ar}-\text{ac}}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}(\text{p}-\text{a})+\text{a}(\text{r}-\text{c})}{(\text{p}-\text{a})(\text{r}-\text{c})}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{r}}{\text{r}-\text{c}}+\frac{\text{a}}{\text{p}-\text{a}}+\frac{\text{b}}{\text{q}-\text{b}}=0$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}\\=\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}-\frac{\text{a}}{\text{p}-\text{a}}-\frac{\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=\frac{\text{p}-\text{a}}{\text{p}-\text{a}}+\frac{\text{q}-\text{b}}{\text{q}-\text{b}}$
$\Rightarrow\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}}=2$
View full question & answer→Question 1174 Marks
Prove that:
$\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}=4\text{abc}$
Answer$\text{L.H.S}=\begin{vmatrix}\frac{\text{a}^2+\text{b}^2}{\text{c}}&\text{c}&\text{c}\\\text{a}&\frac{\text{b}^2+\text{c}^2}{\text{a}}&\text{a}\\\text{b}&\text{b}&\frac{\text{c}^2+\text{a}^2}{\text{b}}\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2&\text{c}^2\\\text{a}^2&\text{b}^2+\text{c}^2&\text{a}^2\\\text{b}^2&\text{b}^2&\text{c}^2+\text{a}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}\text{a}^2+\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2&\text{c}^2-\text{a}^2-\text{b}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}\begin{vmatrix}0&-2\text{b}^2&-2\text{a}^2\\\text{a}^2&\text{b}^2+\text{c}^2-\text{a}^2&0\\\text{b}^2&0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}$
$=\frac{1}{\text{abc}}(-\text{a}^2)\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\0&\text{c}^2+\text{a}^2-\text{b}^2\end{vmatrix}+\text{b}^2\begin{vmatrix}-2\text{b}^2&-2\text{a}^2\\\text{b}^2+\text{c}^2-\text{a}^2&0\end{vmatrix}$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2(\text{c}^2+\text{a}^2-\text{b}^2)\}+\text{b}^2\{0+2\text{a}^2(\text{b}^2+\text{c}^2-\text{a}^2)\}\big]$
$=\frac{1}{\text{abc}}\big[-\text{a}^2\{-2\text{b}^2\text{c}^2-2\text{b}^2\text{a}^2+2\text{b}^4\}+\text{b}^2\{2\text{a}^2\text{b}^2+2\text{a}^2\text{c}^2-2\text{a}^4\}\big]$
$=\frac{1}{\text{abc}}\big[2\text{a}^2\text{b}^2\text{c}^2+2\text{a}^4\text{b}^2-2\text{a}^4\text{b}^4+2\text{a}^2\text{b}^4+2\text{a}^2\text{b}^2\text{c}^2-2\text{a}^4\text{b}^2\big]$
$=\frac{1}{\text{abc}}4\text{a}^2\text{b}^2\text{c}^2$
$=4\text{abc}$
$=\text{R.H.S}$
View full question & answer→Question 1184 Marks
Solve the following systems of linear equations by cramer's rule:
x - 2y = 4,
-3x + 5y = -7
AnswerGiven, x - 2y = 4
-3x + 5y = -7
Using the properties of determinants, we get
$\text{D}=\begin{vmatrix}1&-2\\-3&5 \end{vmatrix}=5-6=-1\neq0$
$\text{D}_1=\begin{vmatrix}4&-2\\-7&5 \end{vmatrix}=20-14=6$
$\text{D}_2=\begin{vmatrix}1&4\\-3&-7 \end{vmatrix}=-7+12=5$
Using cramer's Rule, we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{6}{-1}=-6$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{5}{-1}=-5$
$\therefore\text{x}=-6$ and $\text{y}=-5$
View full question & answer→Question 1194 Marks
If $\text{A}=\begin{vmatrix}2&5\\2&1\end{vmatrix}$ and $\text{B}=\begin{vmatrix}4&-3\\2&5\end{vmatrix},$ verify that |AB| = |A| |B|.
AnswerLet $\text{A}=\begin{vmatrix}2&5\\2&1\end{vmatrix}$
$\Rightarrow|\text{A}|=2-10=-8$
$\text{B}=\begin{vmatrix}4&-3\\2&5\end{vmatrix}$
$\Rightarrow|\text{B}|=20+6=26$
Now $\text{AB}=\begin{vmatrix}2&5\\2&1\end{vmatrix}\begin{vmatrix}4&-3\\2&5\end{vmatrix}$
$=\begin{vmatrix}2\times4+5\times2&2\times(-3)+5\times5\\2\times4+1\times2&2\times(-3)+1\times5\end{vmatrix}$
$=\begin{vmatrix}8+10&-6+25\\8+2&-6+5\end{vmatrix}$
$=\begin{vmatrix}18&19\\10&-1\end{vmatrix}$
$\Rightarrow|\text{AB}|=18\times(-1)-(10)(19)$
$=-18-190=-208$
Now $|\text{AB}|=|\text{A}|\times|\text{B}|$
$-208=(-8)\times(26)$
$-208=-208$
Hence verified.
View full question & answer→Question 1204 Marks
Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
$\Rightarrow\triangle=\begin{vmatrix}0&-3&2\\0&-1&2\\0&5&2 \end{vmatrix}$ [Applying C1 → C1 + 2C2]
$\Rightarrow\triangle=0$
View full question & answer→Question 1214 Marks
Given $\text{A}=\begin{bmatrix}5 & 0 & 4 \\2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix},\text{B}^{-1}=\begin{bmatrix}1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}.$ Compute (AB)-1.
AnswerWe have,
$\text{A}=\begin{bmatrix}5 & 0 & 4 \\2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix}$
$\text{B}^{-1}=\begin{bmatrix}1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}$
We have (AB)-1 = B-1A-1
For matrix A,
$\text{C}_{11}=\begin{vmatrix}3 & 2 \\2 & 1 \end{vmatrix}=-1,\ \text{C}_{12}=-\begin{vmatrix}2 & 2 \\1 & 1 \end{vmatrix}=0$
$\text{and C}_{13}=\begin{vmatrix}2 & 3 \\1 & 2 \end{vmatrix}=1$
$\text{C}_{21}=-\begin{vmatrix}0 & 4 \\2 & 1 \end{vmatrix}=8,\ \text{C}_{22}=\begin{vmatrix}5 & 4 \\1 & 1 \end{vmatrix}=1$
$\text{and C}_{23}=-\begin{vmatrix}5 & 0 \\1 & 2 \end{vmatrix}=-10$
$\text{C}_{31}=\begin{vmatrix}0 & 4 \\3 & 1 \end{vmatrix}=-12,\ \text{C}_{32}=-\begin{vmatrix}5 & 4 \\2 & 2 \end{vmatrix}=-2$
$\text{and C}_{33}=\begin{vmatrix}5 & 0 \\2 & 3 \end{vmatrix}=15$
Now, $\text{adj (A)}=\begin{bmatrix}-1 & 0 & 1 \\8 & 1 & -2 \\ -12 & -2 & 15 \end{bmatrix}^\text{T}$
$=\begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix}$
and |A| = -1
$\therefore\text{A}^{-1}=-\begin{bmatrix} -1 & 8 & -12 \\ 0 & 1 & -2 \\ 1 & -10 & 15 \end{bmatrix}$
$=\begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix}$
So, $\text{B}^{-1}\text{A}^{-1}=\begin{bmatrix} 1 & 3 & 3 \\1 & 4 & 3 \\ 1 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & -8 & 12 \\ 0 & -1 & 2 \\ -1 & 10 & -15 \end{bmatrix}$
$=\begin{bmatrix} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{bmatrix}$
View full question & answer→Question 1224 Marks
If $\text{A}=\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix},$ show that A2 = A-1.
AnswerWe have $\text{A}=\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
Now,
$\text{A}^2=\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
$=\begin{bmatrix}1-2+0 & -2+2+0 & 0+2+0 \\ 1-1+0 & -2+1+1 & 0+1+0 \\ 0-1+0 & 0+1+0 & 0+1+0 \end{bmatrix}$
$=\begin{bmatrix}-1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{bmatrix}$
And $\text{A}^2\times\text{A}=\begin{bmatrix}-1 & 0 & 2 \\ 0 & 0 & 1 \\ -1 & 1 & 1\end{bmatrix}\begin{bmatrix}-1 & 2 & 0 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$
$=\begin{bmatrix}1+0+0 & -2+0+2 & 0 \\ 0+0+0 & 0+0+1 & 0 \\ 0 & -2+1+1 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}=\text{I}_3\ \big[\text{Identity matrix of order }3\big]$
$\Rightarrow\ \text{A}^2\times\text{A}=\text{I}_3$
$\Rightarrow\ \text{A}^2=\text{A}^{-1}$
View full question & answer→Question 1234 Marks
Find the real values of $\lambda$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions:
$2\lambda\text{x}-2\text{y}+3\text{z}=0,$
$\text{x}+\lambda\text{y}+2\text{z}=0,$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
AnswerThe given system of equations can be written as,
$2\lambda\text{x}-2\text{y}+3\text{z}=0$
$\text{x}+\lambda\text{y}+2\text{z}=0$
$2\text{x}+0\text{y}+\lambda\text{z}=0$
The given system of equations will have non-trivial solutions if D = 0
$\Rightarrow\begin{vmatrix}2\lambda&-2&3\\1&\lambda&2\\2&0&\lambda\end{vmatrix}=0$
$\Rightarrow 2\lambda(\lambda^2)+2(\lambda-4)+3(-2\lambda)=0$
$\Rightarrow 2\lambda^3 - 4\lambda - 8 = 0$
$\Rightarrow \lambda = 2$
So, the given system of equations will have non-trivial solutions if $\lambda = 2$
Now, we shall find solutions for $\lambda = 2$
Replacing z by k in the first two equations, we get
$2\lambda\text{x}-2\text{y}=-3\text{k}$
$\text{x}+\lambda\text{y}=-2\text{k}$
$\text{x}=\frac{\begin{vmatrix}-3\text{k}&-2\\-2\text{k}&\lambda\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-3\text{k}\lambda-4\text{k}}{2\lambda^2+2} $
$=\frac{-3\text{k}(2)-4\text{k}}{2(2)^2+2}=\frac{-6\text{k}-4\text{k}}{10}=-\text{k}$
$\text{y}=\frac{\begin{vmatrix}2\lambda&-3\text{k}\\1&-2\text{k}\end{vmatrix}}{\begin{vmatrix}2\lambda&-2\\1&\lambda\end{vmatrix}}=\frac{-4\text{k}\lambda+3\text{k}}{2\lambda^2+2}$
$=\frac{-4\text{k}(2)+3\text{k}}{2(2)^2+2}=\frac{-5\text{k}}{10}=\frac{-\text{k}}{2} $
Substituting these value of x and y in the third equation, we get
$\text{L.H.S}= 2(-\text{k})+0\Big(-\frac{\text{k}}{2}\Big)+2\text{k}$
$=0=\text{R.H.S}$
Thus,
$\lambda=2, \text{x}=-\text{k},\text{y}=-\frac{\text{k}}{2}$ and $\text{z}=\text{k}\ [\text{k}\in\text{R}]$
View full question & answer→Question 1244 Marks
Solve the following systems of linear equations by cramer's rule:
2x - y = -2,
3x + 4y = 3
AnswerLet $\text{D}=\begin{vmatrix}2&-1\\3&4\end{vmatrix}=11$
$\text{D}_1=\begin{vmatrix}-2&-1\\3&4\end{vmatrix}=-5$
$\text{D}_2=\begin{vmatrix}2&-2\\3&3\end{vmatrix}=12$
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-5}{11}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{12}{11}$
View full question & answer→Question 1254 Marks
Evaluate the following determinant:
$\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
Answer$\triangle=\begin{vmatrix}6&-3&2\\2&-1&2\\-10&5&2 \end{vmatrix}$
$=6(-2-10)-(-3)(4+20)+(10-10)$
$=-72+72+0$
$=0$
View full question & answer→Question 1264 Marks
Prove that:
$\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}=1+\text{a}^2+\text{b}^2+\text{c}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}^2+1&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2+1&\text{bc}\\\text{ca}&\text{cb}&\text{c}^2+1 \end{vmatrix}$
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\\text{a}&\text{b}+\frac{1}{\text{b}}&\text{c}\\\text{a}&\text{b}&\text{c}+\frac{1}{\text{c}} \end{vmatrix}$
[Taking out a, b and c common from R1, R2 and R3]
$=(\text{abc})\begin{vmatrix}\text{a}+\frac{1}{\text{a}}&\text{b}&\text{c}\\-\frac{1}{\text{a}}&\frac{1}{\text{b}}&0\\-\frac{1}{\text{a}}&0&\frac{1}{\text{c}} \end{vmatrix}$
[Applying R2 → R2 - R1 and R3 → R3 - R1]
$=(\text{abc})\Big(\frac{1}{\text{abc}}\Big)\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
[Applying C1 → aC1, C2 → bC2 and C3 → cC3]
$=\begin{vmatrix}\text{a}^2+1&\text{b}^2&\text{c}^2\\-1&1&0\\-1&0&1\end{vmatrix}$
$=(-1)\begin{vmatrix}\text{b}^2&\text{c}^2\\1&0\end{vmatrix}+(1)\begin{vmatrix}\text{a}^2+1&\text{b}^2\\-1&1\end{vmatrix}$
$=(-1)(-\text{c}^2)+(\text{a}^2+1+\text{b}^2)$
$=(\text{a}^2+1+\text{b}^2+\text{c}^2)$
$=(\text{a}^2+\text{b}^2+\text{c}^2+1)$
$=\text{R.H.S}$
View full question & answer→Question 1274 Marks
Solve the following system of equations by matrix method:
5x + 7y + 2 = 0
4x + 6y + 3 = 0
AnswerThe given system of equations can be written in matrix from as follws: $\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ $\text{AX = B}$ Here,
$\text{A}=\begin{bmatrix}5&7\\ 4&6\end{bmatrix},\text{X=}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}$ And$\text{ B}=\begin{bmatrix}-2\\ -3\end{bmatrix}$ Now, $|\text{A}|=\begin{bmatrix}5&7\\ 4&6\end{bmatrix}\\ $ $=30-28$ $=2\neq0$ The given system has a unique solution given by $\text{X}=\text{A}^{-1 }\text{B.}$ Let Cij be the cofactors of the elements aij in A = [aij]. Then, $\text{C}_{11}=(-1)^{1+1}(6)=6,\text{C}_{12}=(-1)^{1+2}(4)=-4$ $\text{C}_{21}=(-1)^{2+1}(7)=-7,\text{C}_{22}=(-1)^{2+2}(5)$ $\text{adj}\ \text{A}=\begin{bmatrix}6&-4\\ -7&5\end{bmatrix}=\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{A}^{-1}=\frac{1}{\text{|A|}}\text{ adj}\text{ A}$ $\Rightarrow \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}$ $\text{X}=\text{A}^{-1}\text{B}$ $=\frac{1}{2}\begin{bmatrix}6&-7\\ -4&5\end{bmatrix}\begin{bmatrix}-2\\-3\end{bmatrix} $ $=\frac{1}{2}\begin{bmatrix}-12+21\\ 8-15\end{bmatrix} $ $\Rightarrow\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{9}{2}\\ \frac{-7}{2}\end{bmatrix}$ $\therefore\text{X}=\frac{9}{2}$ And $\text{ y }=\frac{-7}{2}$ View full question & answer→Question 1284 Marks
Evaluate the following:
$\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
AnswerLet $\triangle=\begin{vmatrix}\text{a}+\text{x}&\text{y}&\text{z}\\\text{x}&\text{a}+\text{y}&\text{z}\\\text{x}&\text{y}&\text{a}+\text{z}\end{vmatrix}$
Applying R1 → R1 - R2 and R3 → R3 - R2
$\triangle=\begin{vmatrix}\text{a}&-\text{a}&0\\\text{x}&\text{a}+\text{y}&\text{z}\\0&-\text{a}&\text{a}\end{vmatrix}$
Applying C2 → C2 + C1
$\triangle=\begin{vmatrix}\text{a}&0&0\\\text{x}&\text{a}+\text{y}&\text{z}\\0&-\text{a}&\text{a}\end{vmatrix}$
$\triangle=\text{a}[\text{a}(\text{a}+\text{x}+\text{y})+\text{az}]+0+0$
$\triangle=\text{a}^2(\text{a}+\text{x}+\text{y}+\text{z})$
View full question & answer→Question 1294 Marks
Prove that:
$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z}).$
Answer$\begin{vmatrix}\text{z}&\text{x}&\text{y}\\\text{z}^2&\text{x}^2&\text{y}^2\\\text{z}^4&\text{x}^4&\text{y}^4 \end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}=\begin{vmatrix}\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^2\\\text{x}&\text{y}&\text{z}\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$\text{L.H.S}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{x}^2&\text{y}^2&\text{z}^2\\\text{x}^4&\text{y}^4&\text{z}^4 \end{vmatrix}$
$=\text{xyz}\begin{vmatrix}1&1&1\\\text{x}&\text{y}&\text{z}\\\text{x}^3&\text{y}^3&\text{z}^3\end{vmatrix}$
$=\text{xyz}\begin{vmatrix}0&1&0\\\text{x}-\text{y}&\text{y}&\text{z}-\text{y}\\\text{x}^3-\text{y}^3&\text{y}^3&\text{z}^3-\text{y}^3\end{vmatrix}$
$=\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})\begin{vmatrix}0&1&0\\1&\text{y}&1\\\text{x}^2+\text{y}^2+\text{xy}&\text{y}^3&\text{z}^2+\text{y}^2+\text{zy}\end{vmatrix}$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[\text{z}^2+\text{y}^2+\text{zy}-\text{x}^2-\text{y}^2-\text{xy}]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})[(\text{z}-\text{x})(\text{z}+\text{x})+\text{y}(\text{z}-\text{x})]$
$=-\text{xyz}(\text{x}-\text{y})(\text{z}-\text{y})(\text{z}-\text{x})[\text{z}+\text{x}+\text{y}]$
$=\text{xyz}(\text{x}-\text{y})(\text{y}-\text{z})(\text{z}-\text{x})(\text{x}+\text{y}+\text{z})$
$=\text{R.H.S}$
View full question & answer→Question 1304 Marks
Solve the following systems of linear equations by cramer's rule:
9x + 5y = 10,
3x - 2y = 8
AnswerGiven, 9x + 5y = 10
3y - 2x = 8
Rearranging the second equation, the two equations can be written as
9x + 5y = 10
-2x + 3y = 8
Now,
$\text{D}=\begin{vmatrix}9&5\\-2&3\end{vmatrix}=27+10=37$
$\text{D}_1=\begin{vmatrix}10&5\\8&3\end{vmatrix}=30-40=-10$
$\text{D}_2=\begin{vmatrix}9&10\\-2&8\end{vmatrix}=72+20=-92$
Using Cramer's rule we get
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{-10}{37}$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{92}{37}$
$\therefore\text{x}=\frac{-10}{37}$ and $\text{y}=\frac{92}{37}$
View full question & answer→Question 1314 Marks
For the matrix $\text{A} = \begin{bmatrix}3&2\\1&1\end{bmatrix},$ find the numbers a and b such that A2 + aA + bI = 0.
Answer$\text{Given:}\ \text{A}=\begin{bmatrix}3&2\\1&1\end{bmatrix}$ $\therefore\ \text{A}^2=\text{A.A}=\begin{bmatrix}3&2\\1&1\end{bmatrix}\begin{bmatrix}3&2\\1&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}9+2&6+2\\3+1&2+1\end{bmatrix}=\begin{bmatrix}11&8\\4&3\end{bmatrix}$
$\therefore\ \text{A}^2+\text{aA}+\text{bI}=0$ $\Rightarrow\ \begin{bmatrix}11&8\\4&3\end{bmatrix}+a\begin{bmatrix}3&2\\1&1\end{bmatrix}+b\begin{bmatrix}1&0\\0&1\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}11&8\\4&3\end{bmatrix}+\begin{bmatrix}3a&2a\\a&a\end{bmatrix}+\begin{bmatrix}b&0\\0&b\end{bmatrix}=0$
$\Rightarrow\ \begin{bmatrix}11+3a+b&8+2a+0\\4+a+0&3+a+b\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\therefore$ We have 11 + 3a + b = 0 .....(1)
8 + 2a + 0 = 0 ⇒ 2a = -8 ⇒ a = -4
Here a = -4 satisfies 4 + a + 0 = 0 also, therefore a = -4
Putting a = -4 in eq. (1), 11 - 12 + b = 0 ⇒ b - 1 = 0 ⇒ b = 1
Here also b = 1 satisfies 3 + a + b = 0, therefore b = 1
Therefore, a = -4 and b = 1
View full question & answer→Question 1324 Marks
Write the minors and cofactors of element of the first column of the following matrices and hence evaluate the determinant in case:
$\text{A}=\begin{vmatrix}\text{a}&\text{h}&\text{g}\\\text{h}&\text{b}&\text{f}\\\text{g}&\text{f}&\text{c} \end{vmatrix}$
Answer$\text{M}_{11}=\begin{vmatrix}\text{b}&\text{f}\\\text{f}&\text{c} \end{vmatrix}=\text{bc}-\text{f}^2$
$\text{M}_{21}=\begin{vmatrix}\text{h}&\text{g}\\\text{f}&\text{c} \end{vmatrix}=\text{hc}-\text{fg}$
$\text{M}_{31}=\begin{vmatrix}\text{h}&\text{g}\\\text{b}&\text{f} \end{vmatrix}=\text{hf}-\text{gb}$
$\text{C}_{11}=(-1)^{1+1}\text{M}_{11}=\text{bc}-\text{f}^2$
$\text{C}_{21}=(-1)^{2+1}\text{M}_{11}=-(\text{hc}-\text{fg})=\text{fg}-\text{hc}$
$\text{C}_{31}=(-1)^{3+1}\text{M}_{31}=\text{hf}-\text{gb}$
$\text{D}=\text{a}(\text{bc}-\text{f}^2)-\text{h}(\text{hc}-\text{fg})+\text{g}(\text{fh}-\text{bg})$
$=\text{abc}-\text{af}^2-\text{h}^2\text{c}+\text{fgh}+\text{fgh}-\text{bg}^2$
$=\text{abc}+2\text{hfg}-\text{af}^2-\text{bg}^2-\text{ch}^2$
View full question & answer→Question 1334 Marks
$\text{If A} = \begin{bmatrix}3&1\\-1&2\end{bmatrix},$ show that A2 - 5A + 7I = 0. Hence find A-1.
Answer$\text{Given:}\ \text{A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ $\therefore\ \text{A}^2=\text{A.A}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}9-1&3+2\\-3-2&-1+4\end{bmatrix}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$
$\text{L.H.S.}=\text{A}^2-5\text{A}+7\text{I}=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-5\begin{bmatrix}3&1\\-1&2\end{bmatrix}+7\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}8-15&5-5\\-5+5&3-10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}-7&0\\0&-7\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}$
$=\begin{bmatrix}-7+7&0+0\\0+0&-7+7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0=\text{R.H.S.}\ \Rightarrow\ \text{A}^2-5\text{A}+7\text{I}=0$
To find: A-1, multiplying eq. (1) by A-1.
$\Rightarrow\ \text{A}^2\text{A}^{-1}-5\text{A.A}^{-1}+7\text{I}_2\text{A}^{-1}=0.\text{A}^{-1}\ \Rightarrow\ \text{A}-5\text{I}_2+7\text{A}^{-1}=0$
$\Rightarrow\ \ 7\text{A}^{-1}=-\text{A}+5\text{I}_2=\begin{bmatrix}3&1\\-1&2\end{bmatrix}+5\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3&1\\-1&2\end{bmatrix}+\begin{bmatrix}5&0\\0&5\end{bmatrix}=\begin{bmatrix}2&-1\\1&3\end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{7}\begin{bmatrix}2&-1\\1&3\end{bmatrix}$
View full question & answer→Question 1344 Marks
Prove that:
$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
Answer$\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$\text{L.H.S}=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\1&\text{b}^2+\text{ca}&\text{b}^3\\1&\text{c}^2+\text{ab}&\text{c}^3 \end{vmatrix}$
Apply: R2 → R2 - R1 and R3 → R3 - R1
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&\text{b}^2+\text{ca}-\text{a}^2-\text{bc}&\text{b}^3-\text{a}^3\\0&\text{c}^2+\text{abb}+\text{ca}-\text{a}^2-\text{bc}&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}^2-\text{a}^2)-\text{c}(\text{b}-\text{a})&\text{b}^3-\text{a}^3\\0&(\text{c}^2-\text{a}^2)-\text{b}(\text{c}-\text{a})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}-\text{a})(\text{b}+\text{a}-\text{c})&\text{b}^3-\text{a}^3\\0&(\text{c}-\text{a})(\text{c}+\text{a}-\text{b})&\text{c}^3-\text{a}^3 \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\begin{vmatrix}1&\text{a}^2+\text{bc}&\text{a}^3\\0&(\text{b}+\text{a}-\text{c})&\text{b}^2+\text{a}^2+\text{ab}\\0&(\text{c}+\text{a}-\text{b})&\text{c}^2+\text{a}^2+\text{ac} \end{vmatrix}$
$=(\text{b}-\text{a})(\text{c}-\text{a})\big[(\text{b}+\text{a}-\text{c})(\text{c}^2+\text{a}^2+\text{ac})\\-(\text{b}^2+\text{a}^2+\text{ab})(\text{c}^2+\text{a}^2+\text{ac})\big]$
$=-(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{b})(\text{a}^2+\text{b}^2+\text{c}^2)$
$=\text{R.H.S}$
View full question & answer→Question 1354 Marks
For the matrix $\text{A}=\begin{bmatrix}1 & -1 & 1 \\2 & 3 & 0 \\ 18 & 2 & 10 \end{bmatrix},$ show that A (adjoint A) = 0.
Answer$\text{A}=\begin{bmatrix}1 & -1 & 1 \\2 & 3 & 0 \\ 18 & 2 & 10 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{vmatrix}3 & 0 \\2 & 10 \end{vmatrix}= 30\ \text{C}_{12}=-\begin{vmatrix}2 & 0 \\18 & 10 \end{vmatrix}=-20$
$\text{C}_{13}=\begin{vmatrix}2 & 3 \\18 & 2 \end{vmatrix}=-50$
$\text{C}_{21}=\begin{vmatrix}-1 & 1 \\2 & 10 \end{vmatrix}=12\ \text{C}_{22}=-\begin{vmatrix}1 & 1 \\18 & 10 \end{vmatrix}=-8$
$ \text{C}_{23}=\begin{vmatrix}1 & -1 \\2 & 3 \end{vmatrix}=5$
$\text{C}_{31}=\begin{vmatrix}-1 & 1 \\3 & 0 \end{vmatrix}=-3\ \text{C}_{32}=-\begin{vmatrix}1 & 1 \\2 & 0 \end{vmatrix}=2$
$\text{C}_{33}=\begin{vmatrix}1 & -1 \\2 & 3 \end{vmatrix}=5$
$\text{adj A}=\begin{bmatrix}30 & -20 & -50\\12 & -8 & -20 \\ -3 & 2 & 5 \end{bmatrix}^\text{T}$
$=\begin{bmatrix}30 & 12 & -3\\-20 & -8 & 2 \\ -50 & -20 & 5 \end{bmatrix}$
$\therefore\ \text{A(adj A)}=\begin{bmatrix}-1 & -1 & 1\\2 & -8 & -2 \\ 18 & 2 & 10 \end{bmatrix}$
$=\begin{bmatrix}30 & 12 & -3\\-20 & -8 & 2 \\ -50 & -20 & 5 \end{bmatrix}$
$=\begin{bmatrix}30+20-50 & 12+18-20 & -3-2+5 \\60-60-0 & 24-24-0&-6+6+0 \\ 540-40-500 & 216-16-200 & -54+4+50 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 & 0 \\0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
View full question & answer→Question 1364 Marks
Solve the following systems of homogeneous linear equations by matrix method:
3x + y - 2z = 0
x + y + z = 0
x - 2y + z = 0
AnswerThe given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
or, $\text{AX}=\text{O}$
where, $\text{A}=\begin{bmatrix}3&1&-2\\1&1&1\\1&-2&1\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}3&1&-2\\1&1&1\\1&-2&1\end{vmatrix}$
$=3(1+2)-1(1-1)-2(-2-1)$
$=9-0+6$
$=15\neq0$
So, the given system has only trivial solution, which is given below:
x = y = z = 0
View full question & answer→Question 1374 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 R1 → R1 + R2 + R3, 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 C2 → C2 - C1 and C3 → C3 - C1, we have:
$\left(3x+a\right)\begin{vmatrix}1&0&0\\x&a&0\\x&0&a\end{vmatrix}=0$
Expanding along R1, we have:
(3x + a)[1 × a2] = 0
⇒ a2 (3x + a) = 0
But a $\neq$ 0.
Therefore, we have:
3x + a = 0
$\Rightarrow x =-\frac{a}{3}$
View full question & answer→Question 1384 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 R1 → aR1, R2 → bR2 and R3 → cR3]
$=\text{abc}\begin{vmatrix}1&\text{a}^3&1\\1&\text{b}^3&1\\1&\text{c}^3&1 \end{vmatrix}$
$\Rightarrow\triangle=0$
View full question & answer→Question 1394 Marks
$\text{Let A} = \begin{bmatrix}3&7\\2&5\end{bmatrix}\text{and B} = \begin{bmatrix}6&8\\7&9\end{bmatrix}.$Verify that (AB)-1 = B-1A-1.
Answer$\text{Given}:\ \text{Matrix A}=\begin{bmatrix}3&7\\2&5\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}3&7\\2&5\end{vmatrix}=15-14=1\neq0$
$\therefore\ \text{A}^{-1}=\frac{1}{\text{|A|}}\text{adj. A}=\frac{1}{1}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\begin{bmatrix}5&-7\\-2&3\end{bmatrix}$
$\text{Matrix B}=\begin{bmatrix}6&8\\7&9\end{bmatrix}$ $\therefore\ \text{|B|}=\begin{vmatrix}6&8\\7&9\end{vmatrix}=54-56=-2\neq0$
$\therefore\ \text{B}^{-1}=\frac{1}{\text{|B|}}\text{adj. B}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}$
$\text{Now}\ \text{AB}=\begin{bmatrix}3&7\\2&5\end{bmatrix}\begin{bmatrix}6&8\\7&9\end{bmatrix}=\begin{bmatrix}18+49&24+63\\12+35&16+45\end{bmatrix}=\begin{bmatrix}67&87\\47&61\end{bmatrix}$
$\therefore\ \text{|AB|}=\begin{vmatrix}67&87\\47&61\end{vmatrix}=67(61)-87(47)=4087-4089=-2\neq0$
$\text{Now}\ \text{L.H.S.}=\left(\text{AB}\right)^{-1}=\frac{1}{\text{|AB|}}\text{adj. (AB)}=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(1)$
$\text{R.H.S.}=\text{B}^{-1}\text{A}^{-1}=\frac{1}{-2}\begin{bmatrix}9&-8\\-7&6\end{bmatrix}\begin{bmatrix}5&-7\\-2&3\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}45+16&-63-24\\-35-12&49+18\end{bmatrix}$
$=\frac{1}{-2}\begin{bmatrix}61&-87\\-47&67\end{bmatrix} \dots\dots(2)$
From eq. (1) and (2), we get L.H.S. = R.H.S. ⇒ (AB)-1 = B-1A-1
View full question & answer→Question 1404 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 1414 Marks
If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x1, y1), (x2, y2) and (x3, y3), 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 (x1, y1), (x2, y2) and (x3, y3), 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 1424 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 1434 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 R2 → R2 - 2R1
$\begin{bmatrix} 1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ \frac{-2}{3} & 1 \end{bmatrix}\text{A}$
Applying R2 → 3R2
$\begin{bmatrix}1 & \frac{10}{3} \\ 0 & \frac{1}{3} \end{bmatrix}=\begin{bmatrix}\frac{1}{3} & 0 \\ -2 & 3 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\text{R}_1-\frac{10}{3}\text{ R}_3$
$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\text{A}$
I = BA
Hence, B is the inverse of A.
View full question & answer→Question 1444 Marks
For the matrix $\text{A}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}.$ Show that A-3 - 6A2 + 5A + 11I3 = 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, A3 - 6A2 + 5A + 11I = 0
⇒ (AAA)A-1 - 6(AA)A-1 + 5AA-1 + 11IA-1 = 0 [Post-multiplying by A-1 as $|\text{A}|\neq0$]
⇒ AA(AA-1) - 6A(AA-1) + 5(AA-1) = -11(IA-1)
⇒ A2 - 6A + 5I = -11A-1
$\Rightarrow\ \text{A}^{-1}=-\frac{1}{11}(\text{A}^2-6\text{A}+5\text{I})\ .....(\text{i})$
Now,
A2 - 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 1454 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-1A-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.
$\therefore$ B11 = 3, B12 = 1, B13 = 2 and B21 = 2, B22 = 1, B23 = 2 and B31 = 6, B32 = 2, B33 = 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 = $\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}$
⇒ $(\text{AB})^{-1}=\begin{bmatrix}9-30+30&-3+12-12&3-10+12\\3-15+10&-1+6-4&1-5+4\\6-30+25&-2+12-10&2-10+10\end{bmatrix}=\begin{bmatrix}9&-3&5\\-2&3&1\\1&0&2\end{bmatrix}$
View full question & answer→Question 1464 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 1474 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 1484 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-1B.
Let Cij be the co-factors of the elements aij in A = [aij]. 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 1494 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 R1 → R1 + R2 + R3]
$=\frac{1}{\text{xyz}}\begin{vmatrix}0&0&0\\-\text{xy}&0&\text{zy}\\-\text{yx}&-\text{zx}&0\end{vmatrix}=0$ [Applying R1 → R1 - 2R2]
View full question & answer→Question 1504 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 Cij be the co-factors of aij 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→