Question 1514 Marks
Show that $\text{A}=\begin{bmatrix} 6 & 5 \\ 7 & 6 \end{bmatrix}$ satisfies the equation x2 - 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 I2 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 x2 - 12x + 1 = 0.
Now,
A2 - 12A + I2 = 0
⇒ I2 = 12A - A2
⇒ A-1I2 = A-1 (12A - A2) [Pre-multiplying both sides by A-1]
⇒ A-1 = 12I2 - 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 1524 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 1534 Marks
If $\text{A} = \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},$ find A-1. Using A-1 solve the system of equations:
2x - 3y + 5z = 11
3x + 2y - 4z = -5
x + y - 2z = -3
Answer$\text{Given:}\ \text{Matrix A}=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}$ $\therefore\ \text{|A|}=\begin{vmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{vmatrix}$ $\Rightarrow\ \text{|A|}=2(-4+4)-3(-3)(-6+4)+5(3-2)=0-6+5=-1\neq0$ $\therefore\ \text{A}^{-1}\ \text{exists and A}^{-1}=\frac{1}{\text{|A|}}\text{(adj. A)} \dots\dots(1)$ Now, A11 = 0, A12 = 2, A13 = 1 and A21 = -1, A22 = -9, A23 = -5 and A31 = 2, A32 = 23, A13 = 13 $\therefore\ \text{adj. A}=\begin{bmatrix}0&2&1\\-1&-9&-5\\2&23&13\end{bmatrix}=\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}$ $\therefore$ From eq. (1), $\text{A}^{-1}=\frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}$ Now, Matrix form of given equations is AX = B $ \Rightarrow\ \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$ $\text{Here}\ \text{A}=\begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\ \text{X}=\begin{bmatrix}x\\y\\z\end{bmatrix}\text{and B}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$ Therefore, solution is unique and X = A-1B $\Rightarrow\ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}$ $=\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$ Therefore, x = 1, y = 2 and z = 3
View full question & answer→Question 1544 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 Cij be the co-factors of aij 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 1554 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 1564 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: C3 → C3 - C2, C4 → C4 - C1
$=\begin{vmatrix}1^1&2^2&3^2-2^2&4^2-1^2\\2^2&3^2&4^2-3^2&5^2-2^2\\3^3&4^2&5^2-4^2&6^2-3^2\\4^2&5^2&6^2-5^2&7^2-4^2 \end{vmatrix}$
$=\begin{vmatrix}1^1&2^2&5&15\\2^2&3^2&7&21\\3^3&4^2&9&27\\4^2&5^2&11&33 \end{vmatrix}$
Take 3 common from C4
$=0$
$\because\text{C}_3=\text{C}_4$
View full question & answer→Question 1574 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 1584 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 C3, we get
$\sin3\theta\times(28-21)-\cos2\theta\times(-7-7)-2(3+4)=0$
$\Rightarrow\ 7\sin3\theta+14\cos2\theta-14=0$
$\Rightarrow\ \sin3\theta+2\cos2\theta-2=0$
$\Rightarrow\ (3\sin\theta-4\sin^3\theta)+2(1-2\sin^2\theta)-2=0$
$\Rightarrow\ 4\sin^3\theta-4\sin^2\theta+3\sin\theta=0$
$\Rightarrow\ \sin\theta(4\sin^2\theta-4\sin\theta+3)=0$
$\Rightarrow\ \sin\theta(2\sin\theta+1)(2\sin\theta-3)=0$
$\Rightarrow\ \sin\theta=0\text{ or }\sin\theta=-\frac{1}{2}\text{or }\sin\theta=\frac{3}{2}$
$\Rightarrow\ \theta=\text{n}\pi\text{ or }\theta=\text{m}\pi+(-1)^\text{n}\Big(-\frac{\pi}{6}\Big);\text{ m, n}\in\text{Z}$
$\sin\theta=\frac{-3}{2}\text{ is not possible}$
View full question & answer→Question 1594 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 1604 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 1614 Marks
Show that $\text{A}=\begin{bmatrix} 5 & 3 \\-1 & -2 \end{bmatrix}$ satisfies the equation x2 - 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 I2 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 x2 - 3x - 7 = 0
Now,
A2 - 3A - 7I2 = 0
⇒ A2 - 3A = 7I2
⇒ A-1 (A2 - 3A) = A-1 × 7I2 [Pre-multiplying both sides by A-1]
⇒ A - 3I2 = 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 1624 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 1634 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 1644 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 1654 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 1664 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 1674 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 1684 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 C1 → C1 + C2 + C3]
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\1&\text{x}+\text{a}&\text{x}\\1&\text{x}&\text{x}+\text{a}\end{vmatrix}$
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&\text{x}&\text{x}+\text{a}\end{vmatrix}$ [Applying R2 → R2 - R1]
$=(3\text{x}+\text{a})\begin{vmatrix}1&\text{x}&\text{x}\\0&\text{a}&0\\1&0&\text{a}\end{vmatrix}$ [Applying R3 → R3 - R1]
$=(3\text{x}+\text{a})=(\text{a}^2-0)=0$
$\text{x}=\frac{-\text{a}}{3}$
View full question & answer→Question 1694 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 R2 → R2 - R1]
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\3&\text{x}+1&\text{x}+2\end{vmatrix}$
$=\text{p}\begin{vmatrix}1&1&\text{x}\\1&1&1\\2&\text{x}&2\end{vmatrix}$
$=\text{p}\begin{vmatrix}0&1&\text{x}\\0&1&1\\2-\text{x}&\text{x}&2\end{vmatrix}$ [Applying C1 → C2 - C1]
$=\text{p}\left\{(2-\text{x})\times\begin{vmatrix}1&\text{x}\\1&1 \end{vmatrix}\right\}$ [Expanding along C1]
$=\text{p}(2-\text{x})(1-\text{x})=0$
$\text{x}=1,2$
View full question & answer→Question 1704 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 Cij be the co-factors of the elemennts aij in A = [aij]. 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 1714 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 1724 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 1734 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 1744 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-1B.
Let cij be the co-factors of the elements aij 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 1754 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
A11 = 5, A12 = -7, A21 = -2, A22 = 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
B11 = 1, B12 = 2, B21 = -1, B22 = -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
⇒ (A-1 A) X (BB-1) = A-1CB-1
⇒ (I) X (I) = A-1CB-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 1764 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 1774 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 1784 Marks
Show that $\text{A}=\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$ sastisfies the equation A2 + 4A - 42I = 0. Hence find A-1.
Answer$\text{A}=\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$
Now A2 + 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,
⇒ A2 + 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, A2 + 4A - 42I = 0
⇒ A-1AA + 4A-1A - 42A-1I = 0
⇒ IA + 4I - 42A-1 = 0
⇒ 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 1794 Marks
Solve the following systems of linear equations by cramer's rule:
x + y + z + 1 = 0,
ax + by + cz + d = 0,
a2x + b2y + x2z + d2 = 0
AnswerThese equations can be written as
x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + x2z + d2 = 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 C2 → C1 - C2, C3 → C2 - C3]
Taking (b - a) and (c - a) common from C1 and C2, 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 1804 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 1814 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 R1
$=6(14)-1(8)-3(-5)$
$=84-8+15=91$
Also, $\text{D}_1=\begin{vmatrix}5&1&-3\\5&-3&-2\\8&1&4\end{vmatrix}$
Expanding along R1
$=5(14)-1(36)-3(-19)$
$=70-36+57=91$
Again $\text{D}_2=\begin{vmatrix}6&5&-3\\1&5&-2\\2&8&4\end{vmatrix}$
Expanding along R1
$=6(36)-5(8)-3(-2)$
$=216-40+6=182$
Also $\text{D}_3=\begin{vmatrix}6&1&5\\1&3&5\\2&1&8\end{vmatrix}$
Expanding along R1
$=6(19)-1(-2)+5(-5)$
$=114+2-25=91$
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 1824 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 Cij be the co-factors of aij 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 1834 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 1844 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.
A11 = (-1)1+1 (-1) = -1; A12 = (-1)1+2 (-1) = 1; A13 = (-1)1+3 (1) = 1
A21 = (-1)2+1 (-1) = 1; A22 = (-1)2+2 (-1) = -1; A23 = (-1)2+3 (-1) = 1
A31 = (-1)3+1 (1) = 1; A32 = (-1)3+2 (-1) = 1; A33 = (-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 1854 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 1864 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 Cij be the co-factors of the elements aij in A [aij]. 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}\text{ and }\text{z}=\frac{-3}{-1}$
$\therefore$ x = 1, y = 2 and z = 3
View full question & answer→Question 1874 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 1884 Marks
Prove that |A adj A| = |A|n.
AnswerLet A = [aij] be a square matrix of order n × n.
If Cij is a cofactor of aij in A, then adj A = [Cij]T = [Cij]. Also, it is a matrix of order n × n.
Because A and ahj A are matrices of order n × n, A × (adj A) exists and is of order n × 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 × (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 1894 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 Cij be the co-factors of the elements aij in A [aij]. 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 1904 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→Question 1914 Marks
Show that the following system of linear equations is consistent and also find solutions:
5x +3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
AnswerThis can be written as: $\begin{bmatrix}5&3&7\\ 3&26&2\\ 7&2&10\end{bmatrix}\begin{bmatrix}\text{X}\\ \text{y}\\ \text{z}\end{bmatrix}=\begin{bmatrix}4\\ 9\\ 5\end{bmatrix}$ Or $\text{A}\text{X}=\text{B}$ $\text{|A|}=5{(256)}-3{(16)}+7{(6-182)}$ $=0$ So, A is singular. Thus, the given system is either inconsistent or it is consistent with infinitely many 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 $\text{C}_{11}=256\\ \text{C}_{21}=-16\\ \text{C}_{31}=-176$ $\text{C}_{12}=-16\\ \text{C}_{22}=1\\ \text{C}_{32}=11$ $\text{C}_{13}=-176\\ \text{C}_{23}=11\\ \text{C}_{33}=121$ $\text{adj A}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}^\text{T}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}$ $\text{adj}\text{A}\times\text{B}=\begin{bmatrix}256&-16&-176\\ -16&1&11\\ -176&11&121\end{bmatrix}\begin{bmatrix}4\\ 9\\ 5\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$ Thus, AX = B has infinite many solutions. Now, let z = k Then, 5x + 3y = 4 - 7k
3x + 26y = 9 - 2k Which can be written as: $\begin{bmatrix}5&3\\ 3&26\end{bmatrix}\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}4-7\text{k}\\ 9-2\text{k}\end{bmatrix}$ Or $\text{AX = B}$ $\text{|A|}=2$ $\text{adj A}=\begin{bmatrix}26&-3\\ -3&5\end{bmatrix}$ Now, $\text{x}=\text{A}^{-1}\text{B}=\frac{1}{\text{|A|}}\times\text{adj A}\times\text{B}$ $=\frac{1}{121}\begin{bmatrix}26&-3\\ -3&5\end{bmatrix}\begin{bmatrix}4-7\text{k}\\ 9-2\text{k}\end{bmatrix}$ $=\frac{1}{121}\begin{bmatrix}77-176\text{k}\\ 11\text{k}+33\end{bmatrix}$ $\begin{bmatrix}\text{x}\\ \text{y}\end{bmatrix}=\begin{bmatrix}\frac{7-16\text{k}}{11}\\ \frac{\text{k}+3}{11}\end{bmatrix}$ There values of x, y, z satisfies the third eq. Hence, $\text{x}=\frac{7-16\text{k}}{11},\text{y}=\frac{\text{k}+3}{11},\text{z}=\text{k}$ View full question & answer→Question 1924 Marks
The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.
AnswerFrom the given data, we get
The following three equations:
x + y + z = 12
2x + 3y + 3z = 33
x - 2y + z = 0
This system of equations can be written in the matrix form as:
$\begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}12\\33\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}^1\begin{bmatrix}12\\33\\0\end{bmatrix}\ \dots(1)$
$\text{A}=\begin{bmatrix}1&1&1\\2&3&3\\1&-2&1\end{bmatrix}$
$|\text{A}|=1(9)-1(-1)+1(-7)=3$
$\text{cof}\cdot\text{A}=\begin{bmatrix}9&1&-7\\-3&0&3\\0&-1&1\end{bmatrix}$
$\text{adj }\text{A}=[\text{cof }\text{A}]^\text{T}=\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{adj }\text{A}}{|\text{A}|}=\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{3}\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}\begin{bmatrix}12\\33\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}9&-3&0\\1&0&-1\\-7&3&1\end{bmatrix}\begin{bmatrix}4\\11\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}36-33+0\\4+0+0\\-28+33+0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3\\4\\5\end{bmatrix}$
An award for organising different festivals in the colony can be included by the management.
View full question & answer→Question 1934 Marks
If $\text{A}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix},$ find A-1 and prove that A2 - 4A - 5I = 0.
Answer$\text{A}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$\Rightarrow\ |\text{A}|=\text{A}=\begin{vmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{vmatrix}$
$=1(1-4)-2(2-4)+2(4-2)$
$=-3+4+4=5$
Since, $|\text{A}|\neq0$
Hence, A is invertible.
Now, $\text{A}^2=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 1+4+4 \end{bmatrix}$
$=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$
Now, $\text{A}^2-4\text{A}-5\text{I}=\begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$
$=-4\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix}$
$=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}=0$
$\Rightarrow\ \text{A}^2=4\text{A}-5\text{I}=0\ \text{[Proved]}$
Again, A2 - 4A - 5I = 0
⇒ A-1 (A2 - 4A - 5I) = A-10 [Pre-multiplying with A-1]
⇒ A-1A2 - 4A-1A - 5A-1 = 0
⇒ A - 4I = 5A-1
$\Rightarrow\ 5\text{A}^{-1}=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}-4\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix} 1-4 & 2-0 & 2-0 \\ 2-0 & 1-4 & 2-0 \\ 2-0 & 2-0 & 1-4 \end{bmatrix}=\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$
$\Rightarrow\ \text{A}^{-1}=\frac{1}{5}\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$
View full question & answer→Question 1944 Marks
Solve the following systems of homogeneous linear equations by matrix method:
x + y - z = 0
x - 2y + z = 0
3x + 6y - 5z = 0
Answerx + y - z = 0 .....(1)
x - 2y + z = 0 .....(2)
3x + 6y - 5z = 0 .....(3)
The given system of homogeneous equations can be written in matrix form as follows:
$\begin{bmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
$\text{AX}=\text{O}$
Here,
$\text{A}=\begin{bmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}\text{and }\text{O}=\begin{bmatrix}0\\0\\0\end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}1&1&-1\\1&-2&1\\3&6&-5\end{vmatrix}$
$=1(10-6)-1(-5-3)-1(6+6)$
$=4+8-12$
$=0$
So, the given system of homogeneous equations has non-trivial solution.
Substituting z = k in eq. (1) & eq. (2), we get
x + y = k and x - 2y = -k
$\Rightarrow\begin{bmatrix}1&1\\1&-2\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}\text{k}\\-\text{k}\end{bmatrix}$
$\text{AX}=\text{B}$
Here,
$\text{A}=\begin{bmatrix}1&1\\1&-2\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}\text{and }\text{B}=\begin{bmatrix}\text{k}\\-\text{k}\end{bmatrix}$
$|\text{A}|=\begin{vmatrix}1&1\\1&-2\end{vmatrix}=-3$
So, A-1 exists.
$\text{adj }\text{A}=\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{adj }\text{A}$
$\Rightarrow\text{A}^{-1}=\frac{1}{-3}\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-2&-1\\-1&1\end{bmatrix}\begin{bmatrix}\text{k}\\-\text{k}\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\frac{1}{-3}\begin{bmatrix}-2\text{k}+\text{k}\\-\text{k}-\text{k}\end{bmatrix}$
Thus, $\text{x}=\frac{\text{k}}{3},\text{y}=\frac{2\text{k}}{3}$ and $\text{z}=\text{k}$ (where k is any real number) satisfy the given system of equations.
View full question & answer→Question 1954 Marks
$\text{If A} = \begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix} $
Verify that A3 - 6A2 + 9A + 4I = 0 and hence find A-1.
Answer$\text{Given:}\ \text{A}=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$ $\therefore\ \text{A}^2=\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
$\Rightarrow\ \text{A}^{2}=\begin{bmatrix}4+1+1&-2-2-1&2+1+2\\-2-2-1&1+4+1&-1-2-2\\2+1+2&-1-2-2&1+1+4\end{bmatrix}=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}$
$\text{Now}\ \text{A}^3=\text{A}^2\text{A}= \begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}$
$=\begin{bmatrix}12+5+5&-6-10-5&6+5+10\\-10-6-5&5+12+5&-5-6-10\\10+5+6&-5-10-6&5+5+12\end{bmatrix}=\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}$
$\text{L.H.S.= A}^3-6\text{A}^2+9\text{A}-4\text{I}$
$=\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}-6\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}+9\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}-4\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}22&-21&21\\-21&22&-21\\21&-21&22\end{bmatrix}-\begin{bmatrix}36&-30&30\\-30&36&-30\\30&-30&36\end{bmatrix}+\begin{bmatrix}18&-9&9\\-9&18&-9\\9&-9&18\end{bmatrix}-\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}$
$=\begin{bmatrix}22-36&-21+30&21-30\\-21+30&22-36&-21+30\\21-30&-21+30&22-36\end{bmatrix}+\begin{bmatrix}18-4&-9-0&9-0\\-9-0&18-4&-9-0\\9-0&-9-0&18-4\end{bmatrix}$
$=\begin{bmatrix}-14&9&-9\\9&-14&9\\-9&9&-14\end{bmatrix}+\begin{bmatrix}14&-9&9\\-9&14&-9\\9&-9&14\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
$=\text{R.H.S.}$
Now, to find A-1, multiplying A3 - 6A2 + 9A - 4I by A-1
⇒ A3A-1 - 6A2A-1 + 9AA-1 - 4I.A-1 = 0.A-1
⇒ A2 - 6A + 9I - 4A-1 = 0 ⇒ 4A-1 = A2 - 6A + 9I
$\Rightarrow\ \ 4\text{A}^{-1}=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}-6\begin{bmatrix}2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}+9\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\ \ 4\text{A}^{-1}=\begin{bmatrix}6&-5&5\\-5&6&-5\\5&-5&6\end{bmatrix}-\begin{bmatrix}12&-6&6\\-6&12&-6\\6&-6&12\end{bmatrix}+\begin{bmatrix}9&0&0\\0&9&0\\0&0&9\end{bmatrix}$
$\Rightarrow\ \ 4\text{A}^{-1}=\begin{bmatrix}6-12+9&-5+6+0&5-6+0\\-5+6+0&6-12+9&-5+6+0\\5-6+0&-5+6+0&6-12+9\end{bmatrix}=\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$
$\Rightarrow \ \text{A}^{-1}=\frac{1}{4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}$
View full question & answer→Question 1964 Marks
Solve the following systems of linear equations by cramer's rule:
5x - 7y + z = 11,
6x - 8y - z = 15,
3x + 2y - 6z = 7
Answer$\text{D}=\begin{vmatrix}5&-7&1\\6&-8&-1\\3&2&-6 \end{vmatrix}$
$=5(48+2)+7(-36+3)+1(12+24)$
$=5(50)+7(-33)+1(36)=55$
$\text{D}_1=\begin{vmatrix}11&-7&1\\15&-8&-1\\7&2&-6 \end{vmatrix}$
$=11(48+2)+7(-90+7)+1(30+36)$
$=11(50)+7(-83)+1(86)=55$
$\text{D}_2=\begin{vmatrix}5&11&1\\6&15&-1\\3&7&-6 \end{vmatrix}$
$=5(-90+7)-11(-36+3)+1(42-45)$
$=5(-83)-11(-33)+1(-3)=-55$
$\text{D}_3=\begin{vmatrix}5&-7&11\\6&-8&15\\3&2&7 \end{vmatrix}$
$=5(-56-30)+7(42-45)+11(12+24)$
$=5(-86)+7(-3)+11(36)=-55$
Now,
$\text{x}=\frac{\text{D}_1}{\text{D}}=\frac{55}{55}=1$
$\text{y}=\frac{\text{D}_2}{\text{D}}=\frac{-55}{55}=-1$
$\text{z}=\frac{\text{D}_3}{\text{D}}=\frac{-55}{55}=-1$
$\therefore\text{x}=1,\text{ y}=-1$ and $\text{z}=-1$
View full question & answer→Question 1974 Marks
Using matrix method, solve the system of equation 3x + 2y - 2z = 3, x + 2y + 3z = 6 and 2x - y + z = 2.
AnswerGiven system of equation is:
3x + 2y - 2z = 3
x + 2y + 3z = 6
and 2x - y + z = 2
or AX = B
i.e., $\begin{bmatrix}3&2&-2\\1&2&3\\2&-1&1\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\begin{bmatrix}3\\6\\2\end{bmatrix}$
$\therefore$ X = A-1B
For A-1,
Cofactors are
$\text{A}_{11}=5,\text{A}_{12}=5,\text{A}_{13}=-5,$
$\text{A}_{21}=0,\text{A}_{22}=7,\text{A}_{23}=7,$
$\text{A}_{31}=10,\text{A}_{32}=11\text{ and A}_{33}=4$
$\therefore\ \ \text{adj A}=\begin{bmatrix}5&5&-5\\0&7&7\\10&-11&4\end{bmatrix}^\text{T}=\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4 \end{bmatrix}$
$|\text{A}|=3(5)+2(5)+(-2)(-5)=35$
$\therefore\ \ \text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\frac{1}{35}\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4\end{bmatrix}$
Now, X = A-1B
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z}\end{bmatrix}=\frac{1}{35}\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4\end{bmatrix}\begin{bmatrix}3\\6\\2\end{bmatrix}=\frac{1}{35}$ $\begin{bmatrix}15+20\\15+45-22\\-15+42+8\end{bmatrix}=\frac{1}{35}\begin{bmatrix}35\\35\\35\end{bmatrix}=\begin{bmatrix}1\\1\\1\end{bmatrix}$
$\therefore$ x = 1, y = 1 and z = 1
View full question & answer→Question 1984 Marks
If a, b and c are real numbers, and
$\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}=0,$
Show that either a + b + c = 0 or a = b = c.
Answer$\triangle=\begin{vmatrix}b+c&c+a&a+b\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
Applying R1 → R1 + R2 + R3, we have:
$\triangle=\begin{vmatrix}2(a+b+c)&2(a+b+c)&2(a+b+c)\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&1&1\\c+a&a+b&b+c\\a+b&b+c&c+a\end{vmatrix}$
$=2(a+b+c)\begin{vmatrix}1&0&0\\c+a&a-b&b-a\\a+b&c-a&c-b\end{vmatrix}\ \text{C}_1\rightarrow\text{C}_2-\text{C}_1\ \text{and}\ \text{C}_3\rightarrow\text{C}_3-\text{C}_1$
Expanding along R1, we have:
$\triangle$ = 2(a + b + c) (1) [(b - c) (c - b) - (b - a) (c - a)]
= 2(a + b + c) [-b2 - c2 + 2bc - bc + ba + ac - a2]
= 2(a + b + c) [ab + bc + ca - a2 - b2 - c2]
It is given that $\triangle$ = 0.
(a + b + c) [ab + bc + ca - a2 - b2 - c2] = 0
⇒ Either a + b + c = 0, or ab + bc + ca - a2 - b2 - c2 = 0.
Now,
ab + bc + ca - a2 - b2 - c2 = 0
⇒ -2ab - 2bc - 2ca + 2a2 + 2b2 + 2c2 = 0
⇒ (a - b)2 + (b - c)2 + (c - a)2 = 0
⇒ (a -b)2 = (b - c)2 = (c - a)2 = 0 [(a - b)2, (b - c)2, (c - a)2 are non-negative]
⇒ (a - b) = (b - c) = (c - a) = 0
⇒ a = b = c
Hence, if $\triangle$ = 0, then either a + b + c = 0 or a = b = c.
View full question & answer→Question 1994 Marks
Let A $=\begin{bmatrix}1&-2&1\\-2&3&1\\1&1&5\end{bmatrix}.$ Verify that
- [adj. A]-1 = adj.(A-1)
- (A-1)-1 = A
AnswerA = $\begin{bmatrix}1&-2&1\\-2&3&1\\1&1&5\end{bmatrix}$ $\therefore|\text{A}| = 1(15 - 1) + 2(-10-1) + 1(-2-3)=14-22-5=13$ Now, A11 = 14, A12 = 11, A13 = -5 A21 = 11, A22 = 4, A23 = -3 A31 = -5, A32 = -3, A33 = -1 $\therefore\text{adj. A} = \begin{bmatrix}14&11&-5\\11&4&-3\\-5&-3&-1\end{bmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj. A})$ = $-\frac{1}{13}\begin{bmatrix}14&11&-5\\11&4&-3\\-5&-3&-1\end{bmatrix}=\frac{1}{13}\begin{bmatrix}-14&-11&5\\-11&-4&3\\5&3&1\end{bmatrix}$ - |adj. A| = 14(- 4 - 9) - 11(- 11 - 15) - 5(- 33 + 20)
= 14(-13) -11(-26) -5(-13)
= -182 + 286 + 65 = 169
We have,
adj. (adj. A) = $\begin{bmatrix}-13&26&-13\\26&-39&-13\\-13&-13&-65\end{bmatrix}$
$\therefore\ [\text{adj. A}]^{-1}=\frac{1}{|\text{adj. A}|}(\text{adj. (adj. A))}$
= $\frac{1}{169}\begin{bmatrix}-13&26&-13\\26&-39&-13\\-13&-13&-65\end{bmatrix}$
= $\frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}$
Now, A-1 = $\frac{1}{13}\begin{bmatrix}-14&-11&5\\-11&-4&3\\5&3&1\end{bmatrix}=\begin{bmatrix}-\frac{14}{13}&-\frac{11}{13}&\frac{5}{13}\\-\frac{11}{13}&-\frac{4}{13}&\frac{3}{13}\\\frac{5}{13}&\frac{3}{13}&\frac{1}{13}\end{bmatrix}$
$\therefore\ \text{adj.}(\text{A}^{-1})=\begin{bmatrix}-\frac{4}{169}-\frac{9}{169}&-\bigg(-\frac{11}{169}-\frac{15}{169}\bigg)&-\frac{33}{169}+\frac{20}{169}\\-\bigg(-\frac{11}{169}-\frac{15}{169}\bigg)&-\frac{14}{169}-\frac{25}{169}&-\bigg(-\frac{42}{169}+\frac{55}{169}\bigg)\\-\frac{33}{169}+\frac{20}{169}&-\bigg(-\frac{42}{169}+\frac{55}{169}\bigg)&\frac{56}{169}-\frac{121}{169}\end{bmatrix}$
$=\frac{1}{169}\begin{bmatrix}-13&26&-13\\26&-39&-13\\-13&-13&-65\end{bmatrix}=\frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}$
Hence, [adj. A]-1 = adj. (A-1).
- We have shown that:
A-1 = $\frac{1}{13}\begin{bmatrix}-14&-11&5\\-11&-4&3\\5&3&1\end{bmatrix}$
And, adj. A-1 = $\frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}$
Now, $\bigg|\text{A}^{-1}\bigg|=\bigg(\frac{1}{13}\bigg)^3[-14\times(-13)+11\times(-26)+5\times(-13)]=\bigg(\frac{1}{13}\bigg)^3\times(-169)=-\frac{1}{13}$
$\therefore\ (\text{A}^{-1})^{-1}=\frac{\text{adj. A}^{-1}}{|\text{A}^{-1}|}=\frac{1}{\bigg(-\frac{1}{13}\bigg)}\times\frac{1}{13}\begin{bmatrix}-1&2&-1\\2&-3&-1\\-1&-1&-5\end{bmatrix}=\begin{bmatrix}1&-2&1\\-2&3&1\\1&1&5\end{bmatrix}=\text{A}$
$\therefore\ (\text{A}^{-1})^{-1}=\text{A}$
View full question & answer→Question 2004 Marks
Prove that:
$\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}=9(\text{a}+\text{b})\text{b}^2$
AnswerLet $\text{L.H.S}=\begin{vmatrix}\text{a}&\text{a}+\text{b}&\text{a}+2\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$
$=\begin{vmatrix}3\text{a}+3\text{b}&3\text{a}+3\text{b}&3\text{a}+3\text{b} \\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Applying R1 → R2 + R2 + R3]
$=3(\text{a}+\text{b})\begin{vmatrix}1&1&1\\\text{a}+2\text{b}&\text{a}&\text{a}+\text{b}\\\text{a}+\text{b}&\text{a}+2\text{b}&\text{a} \end{vmatrix}$ [Taking out 3(a + b) common from R1]
$=3(\text{a}+\text{b})\begin{vmatrix}0&0&1\\2\text{b}&-\text{b}&\text{a}+\text{b}\\-\text{b}&2\text{b}&\text{a} \end{vmatrix}$ [Applying C1 → C1 - C2 and C2 → C2 - C3]
$=3(\text{a}+\text{b})\text{b}^2\begin{vmatrix}0&0&1\\2&-1&\text{a}+\text{b}\\-1&2&\text{a} \end{vmatrix}$ [Taking out b common from C1 and C2]
$=3(\text{a}+\text{b})\text{b}^2\times3$
$=9(\text{a}+\text{b})\text{b}^2$
$=\text{R.H.S}$
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