Question 14 Marks
Find matrix A such that
$\begin{pmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{pmatrix}\text{A} = \begin{pmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{pmatrix}$
Answer$\text{Let} \begin{pmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{pmatrix}\begin{pmatrix} \text{a} & \text{b} \\ \text{c} & \text{d} \end{pmatrix} = \begin{pmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{pmatrix}$
$\Rightarrow \begin{pmatrix} \text{2a - c} & \text{2b - d} \\ \text{a} & \text{b} \\ \text{-3a + 4c} & \text{-3b + 4d} \end{pmatrix} = \begin{pmatrix} -1 & -8 \\ 1 & -2 \\ 9 & 22 \end{pmatrix}$
$\Rightarrow $ 2a – c = –1, 2b – d = –8
a = 1, b = –2
–3a + 4c = 9, –3b + 4d = 22
Solving to get a = 1, b = –2, c = 3, d = 4
$\therefore \text{A} = \begin{pmatrix} 1 & -2 \\ 3 & 4 \\ \end{pmatrix}$
View full question & answer→Question 24 Marks
If A = $\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 &-4 \\ 1 & 1 & -2 \end{bmatrix} $, then find A–1 and hence solve the system of linear equations 2x – 3y + 5z = 11, 3x + 2y – 4z = – 5 and x + y – 2z = – 3.
Answer$\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 &-4 \\ 1 & 1 & -2 \end{bmatrix} \Rightarrow|\text{A}|=2(0)+3(-2)+5(1)=-1\neq0$
A11 = 0, A12 = 2, A13 = 1
A21 = –1, A22 = –9, A23 = –5
A31 = 2, A32 = 23, A33 = 13
$ \Rightarrow\text{A}^{-1}=-1\begin{bmatrix} 0 & 2 & 1 \\ -1 & -9 &-5 \\ 2 & 23 & 13 \end{bmatrix}^\text{T}=-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}$
Given equations can be written as
$\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 &-4 \\ 1 & 1 & -2 \end{bmatrix}\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix}=\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}\ \text{or}\ \text{AX}=\text{B}$
$\Rightarrow\text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix}\begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 &-23 \\ -1 & 5 & -13 \end{bmatrix}\begin{bmatrix}11\\ -5\\ -3\end{bmatrix}=\begin{bmatrix} \text{1} \\ \text{2} \\ \text{3} \end{bmatrix}$
⇒ x = 1, y = 2, z = 3.
View full question & answer→Question 34 Marks
$\text{If A} = \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -8 & 0 \end{bmatrix}, \text{B} = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}, \text{C} = \begin{bmatrix} 2 \\ -2 \\ 3 \end{bmatrix},$ then calculate AC, BC and (A + B) C. Also verify that (A + B) C = AC + BC.
Answer$\text{AC} = \begin{bmatrix} 0 & 6 & 7 \\ -6 & 0 & 8 \\ 7 & -7 & 0 \end{bmatrix} \begin{bmatrix} 2\\ -2\\ 3\end{bmatrix} =\begin{bmatrix} 9\\ 12\\ 30\end{bmatrix} $
$\text{BC} = \begin{bmatrix} 0 & 1& 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 2\\ -2\\ 3\end{bmatrix} =\begin{bmatrix} 1\\ 8\\ -2\end{bmatrix} $
$\text{AC + BC} = \begin{bmatrix} 10\\ 20\\ 28\end{bmatrix} $
$\text{(A+B) C} = \begin{bmatrix} 0 & 7 & 8 \\ -5 & 0 & 10 \\ 8 & -6& 0 \end{bmatrix} \begin{bmatrix} 2\\ -2\\ 3\end{bmatrix} $
$ =\begin{bmatrix} 10\\ 20\\ 28\end{bmatrix} $
$\text{Yes, (A + B) C AC + BC}$
View full question & answer→Question 44 Marks
Using elementary row operations (transformations), find the inverse of the following matrix:
$\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix}$
Answer$\text{A = IA}$
$\begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \text{A}$
Using elementry row trans formations to get
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -2 & 1 \\ -9 & 6 & -2 \\ 5 & -3 & 1 \end{bmatrix}\text{A}$
$\Rightarrow \text{A}^{-1} = \begin{bmatrix} 3 & -2 & 1 \\ -9 & 6 & -2 \\ 5 & -3& 1 \end{bmatrix}$
View full question & answer→Question 54 Marks
Using properties of determinants, prove that
$\begin{vmatrix} \text{a}^2+2\text{a}& 2\text{a}+1 & 1\\[0.3em] 2\text{a}+1 & \text{a}+2 & 1 \\[0.3em] 3 & 3 & 1 \end{vmatrix}=(\text{a}-1)^3$
Answer$\triangle=\begin{vmatrix} \text{a}^2+2\text{a}& 2\text{a}+1 & 1\\[0.3em] 2\text{a}+1 &\text{a}+2 &1 \\[0.3em] 3 & 3 & 1 \end{vmatrix}$
R1 → R1 – R2 and R2 → R2 – R3
$\triangle=\begin{vmatrix} \text{a}^2+1& \text{a}-1 & 0\\[0.3em] 2(\text{a}-1) &\text{a}-1 &0 \\[0.3em] 3 & 3 & 1 \end{vmatrix}$
$\triangle=(\text{a}-1)^2\begin{vmatrix} \text{a}+1& 1 & 0\\[0.3em] 2 &1 &0 \\[0.3em] 3 & 3 & 1 \end{vmatrix}$
Expanding
(a – 1)2.(a – 1) = (a – 1)3.
View full question & answer→Question 64 Marks
Find matrix A such that
$\begin{bmatrix} 2& -1\\[0.3em] 1 & 0 \\[0.3em] -3 & 4 \end{bmatrix}\text{A}=\begin{bmatrix} -1& -8\\[0.3em] 1 & -2 \\[0.3em] 9 & 22 \end{bmatrix}$
AnswerLet $\begin{bmatrix} 2& -1 \\[0.3em] 1 &1 \\[0.3em] -3 & 4 \end{bmatrix}\begin{bmatrix} \text{a}& \text{b}\\[0.3em] \text{c} &\text{d} \\[0.3em] \end{bmatrix}=\begin{bmatrix} -1& -8 \\[0.3em] 1 &-2 \\[0.3em] 9 & 22 \end{bmatrix}$ ⇒$\begin{bmatrix} 2\text{a}-\text{c}& 2\text{b}-\text{d} \\[0.3em] \text{a} &\text{b} \\[0.3em] -3\text{a}+4\text{c} & -3\text{b} +4\text{d} \end{bmatrix}=\begin{bmatrix} -1&-8 \\[0.3em] 1 &-2 \\[0.3em] 9 & 22 \end{bmatrix}$
⇒ 2a – c = –1, 2b – d = –8 a = 1, b = –2 –3a + 4c = 9, –3b + 4d = 22 Solving to get a = 1, b = –2, c = 3, d = 4 $\therefore\ \text{A}=\begin{bmatrix} 1& -2 \\[0.3em] 3&4 \\[0.3em] \end{bmatrix}$ View full question & answer→Question 74 Marks
If $\text{A} = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 &1 & -2 \end{pmatrix}$ find A–1. Hence using A–1 solve the system of equations 2x – 3y + 5z = 11, 3x + 2y – 4z = – 5, x + y – 2z = – 3.
Answer$\text{A} = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 &1 & -2 \end{pmatrix}$⇒ |A| = 2(0) + 3(–2) + 5(1) = –1 $\neq$ 0
A11 = 0, A12 = 2, A13 = 1
A21 = –1, A22 = –9, A23 = –5
A31 = 2, A32 = 23, A33 = 13
$\Rightarrow\text{A}^{-1} =-1 \begin{pmatrix} 0 & 2 & 1 \\ -1 & -9& -5\\ 2&23 &13 \end{pmatrix}=-1 \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9& 23\\ 1 & -5 &13 \end{pmatrix}= \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9& -23\\ -1 & 5 &-13 \end{pmatrix}$
Given equations can be written as
$\begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4\\ 1 & 1 &-2 \end{pmatrix} \begin{pmatrix} \text{x} \\ \text{y}\\\text{z} \end{pmatrix}= \begin{pmatrix} 11 \\ -5\\ -3 \end{pmatrix}\text{ }\text{or}\text{ }\text{AX}=\text{B}$
$\Rightarrow\text{x}=\text{x}^{-1}\text{B}$
$\Rightarrow\begin{pmatrix} \text{x}\\ \text{y}\\ \text{z} \end{pmatrix} = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9& -23\\ -1 &5 &-13 \end{pmatrix} \begin{pmatrix} 11 \\ -5\\ -3 \end{pmatrix}= \begin{pmatrix} 1 \\ 2\\ 3 \end{pmatrix}$
⇒ x = 1, y = 2, z = 3.
View full question & answer→Question 84 Marks
Determine the product and use it to $\begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix}$solve the system of equations x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1.
Answer$\text{Getting}\begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} =\begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} \text{ }\text{ }\text{ }\text{ }\text{ } \dots \text{(i)}$
Given equations can be written as $\begin{pmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{pmatrix} \begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix} = \begin{pmatrix} 4 \\ 9 \\ 1 \end{pmatrix}$
$\Rightarrow \text{AX = B}$
$\text{From (i) } \text{A}^{-1} = \frac{1}{8}\begin{pmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{pmatrix} $
$\therefore \text{ } \text{X = A}^{-1} \text{B} = \frac{1}{8} \begin{pmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{pmatrix} \begin{pmatrix} 4 \\ 9 \\ 1 \end{pmatrix}$
$= \frac{1}{8} \begin{pmatrix} 24 \\ -16 \\ -8 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ -1 \end{pmatrix}$
$\Rightarrow \text{x = 3, y = -2, z = -1}$
View full question & answer→Question 94 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.
AnswerAccording to question
x + y + z = 12
2x + 3y + 3z = 33
x - 2y + z = 0
The above system of linear equation can be written in matrix form as
AX = B
Where A $ = \begin{bmatrix} 1 & 1 &1 \\[0.3em] 2& 3 & 3 \\[0.3em] 1 &-2 &1 \end{bmatrix},\text{X} =\begin{bmatrix} \text{x} \\[0.3em] \text{y} \\[0.3em] \text{z} \end{bmatrix},\text{B} = \begin{bmatrix} 12 \\[0.3em] 33\\[0.3em] 0 \end{bmatrix}$
$ \text{Now}|\text{A}|= \begin{vmatrix} 1 & 1 &1 \\[0.3em] 2& 3 & 3 \\[0.3em] 1 &-2 &1 \end{vmatrix} =1 (3 + 6) - 1 (2 -3) + 1 (-4 - 3) = 9 + 1 -7 = 3$
$\begin{matrix} \text{A}_{11} = 9,& \text{A}_{12} = 1, &\text{A}_{13} = -7 \\ \text{A}_{21} = -3 , & \text{A}_{22} = 0 , &\text{A}_{23} = 3 \\ \text{A}_{31} = 0, &\text{A}_{32} = -1, &\text{A}_{33} = 1 \end{matrix} $
Adj A $= \begin{bmatrix} 9& 1 &-7 \\ -3&0 &3 \\ 0&-1&1 \end{bmatrix}^\text{T} = \begin{bmatrix} 9& -3 &0 \\ 1&0 &-1 \\ -7&3&1 \end{bmatrix}$
$\therefore\text{A}^{-1} = \frac{1}{3} \begin{bmatrix} 9& -3 &0 \\ 1&0 &-1 \\ -7&3&1 \end{bmatrix} $
$\because\text{AX} = \text{B}\Rightarrow\text{X} = \text{A}^{-1}\text{B}$
$\therefore\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 9 & -3 & 0 \\ 1 & 0 & - 1 \\-7 & 3 &1 \end{bmatrix} .\begin{bmatrix} 12\\ 33 \\0\end{bmatrix}$
$\therefore\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 108 - 99 \\ 12 + 0 + 0 \\-84 + 99 \end{bmatrix}$
$\therefore\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 9 \\ 12 \\15 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\5 \end{bmatrix}\Rightarrow\text{x} = 3,\text{y} = 4,\text{z} = 5$
No. of awards for honesty = 3
No. of awards for helping others = 4
No. of awards for supervising = 5.
The persons, who work in the field of health and hygiene should also be awarded.
View full question & answer→Question 104 Marks
Using matrices, solve the following system of equations:
2x + 3y + 3z = 5, x – 2y + z = – 4, 3x – y – 2z = 3.
AnswerThe given system of equations can be written as
$\text{AX = B, where A = } \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1\\ 3 & -1 & -2 \end{bmatrix},\text{X}=\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix},\text{B}=\begin{bmatrix} \text{5} \\ \text{-4} \\ \text{3} \end{bmatrix}$
$|\text{A}|=\begin{vmatrix}2& 3&3 \\1 &-2&1\\3&-1&-2\end{vmatrix}$
$=2[4+1]-3[-2-3]+3[-1+6]$
$=10+15+15$
$=40$
$|\text{A}|=40$
$\text{A}^{-1}\ \text{exists}$
$\text{For Adj A}$
$\text{A}_{11}=\begin{vmatrix}-2& 1 \\-1 &-2\end{vmatrix}=4+1=5$
$\text{A}_{12}=-\begin{vmatrix}1& 1 \\3 &-2\end{vmatrix}=-(-2-3)=5$
$\text{A}_{13}=\begin{vmatrix}1& -2 \\3 &-1\end{vmatrix}=-1+6=5$
$\text{A}_{21}=-\begin{vmatrix}3& 3 \\-1 &-2\end{vmatrix}=-(-6+3)=3$
$\text{A}_{22}=\begin{vmatrix}2& 3 \\3 &-2\end{vmatrix}=-4-9=-13$
$\text{A}_{23}=-\begin{vmatrix}2& 3 \\3 &-1\end{vmatrix}=-(-2-9)=11$
$\text{A}_{31}=\begin{vmatrix}3& 3 \\-2 &1\end{vmatrix}=3+6=9$
$\text{A}_{32}=-\begin{vmatrix}2& 3 \\1 &1\end{vmatrix}=-(2-3)=1$
$\text{A}_{33}=\begin{vmatrix}2& 3 \\1 &-2\end{vmatrix}=-4-3=-7$
$\text{Adj A}=\begin{bmatrix}5& 3&9 \\5 &-13&1\\5&11&-7\end{bmatrix}$
$\text{A}^{-1}=\frac{\text{Adj A}}{|\text{A}|}$
$\text{A}^{-1}=\frac{1}{40}\begin{bmatrix}5& 3&9 \\5 &-13&1\\5&11&-7\end{bmatrix}$
$\text{x}=\text{A}^{-1}\text{B}$
$\begin{bmatrix}\text{x}\\\text{y}\\\text{z} \end{bmatrix}=\frac{1}{40}\begin{bmatrix}5& 3&9 \\5 &-13&1\\5&11&-7\end{bmatrix}\begin{bmatrix}5\\-4\\3 \end{bmatrix}$
$\text{x}=1,\text{ y}=2,\text{ z}=-1$
View full question & answer→Question 114 Marks
Using matrices, solve the following system of equations:
4x + 3y + 2z = 60.
x + 2y + 3z = 45.
6x + 2y + 3z = 70.
AnswerGiven system of equations can be written as $\begin{pmatrix} 4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3 \end{pmatrix}\begin{pmatrix} \text{x} \\ \text{y}\\ \text{z} \end{pmatrix}=\begin{pmatrix} \text{60} \\ \text{45}\\ \text{70} \end{pmatrix}\text{OR A}\cdot\text{X = B}$
|A | = 4(0) –3 (–15) + 2 (–10) = 45 – 20 = 25 $\neq$ 0 $\therefore$ X = A–1 B Cofactors are $\begin{pmatrix} \text{C}_{11}=0 & \text{C}_{12}=+15 & \text{C}_{13}=-10 \\ \text{C}_{21}=-5 & \text{C}_{22}=0 & \text{C}_{23}=10 \\ \text{C}_{31}=5 & \text{C}_{32}=-10 & \text{C}_{33}=5 \end{pmatrix}$ $\therefore\text{A}^{-1}=\frac{1}{25}\begin{pmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{pmatrix}$
$\therefore\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}=\frac{1}{25}\begin{pmatrix} 0 & -5 & 5 \\ 15 & 0 & -10 \\ -10 & 10 & 5 \end{pmatrix}\begin{pmatrix} \text{60} \\ \text{45} \\ \text{70} \end{pmatrix}=\begin{pmatrix} \text{5} \\ \text{8} \\ \text{8} \end{pmatrix}$
$\therefore$ x = 5, y = 8, z = 8.
View full question & answer→Question 124 Marks
Express the following matrix as the sum of a symmetric and a skew symmetric matrix, and verify your result:
$\begin{pmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{pmatrix}$
Answer$\text{A}=\begin{pmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{pmatrix},\text{then} \text{ A}'=\begin{pmatrix}3&3&-1\\-2&-2&1\\-4&-5&2\end{pmatrix}$
$\text{Wrinting A}=\frac{1}{2}(\text{A+A}')+\frac{1}{2}(\text{A-A}')$
$\frac{1}{2}\text{(A+A}')=\begin{pmatrix}3&1/2&-5/2\\1/2&-2&-2\\-5/2&-2&2\end{pmatrix}$
$\frac{1}{2}\text{(A-A}')=\begin{pmatrix}0&-5/2&-3/2\\5/2&0&-3\\3/2&3&0\end{pmatrix}$
$\text{and}\begin{pmatrix}3&-2&-4\\3&-2&-5\\-1&1&2\end{pmatrix}=\begin{pmatrix}3&1/2&-5/2\\1/2&-2&-2\\-5/2&-2&2\end{pmatrix}+\begin{pmatrix}0&-5/2&-3/2\\5/2&0&-3\\3/2&3&0\end{pmatrix}$
Thus A = B + C
Where B is Symmetric matrix and C is skew symmetric matrix
View full question & answer→Question 134 Marks
Find the inverse of fhe following matrix using elementary operations:
$\text{A}=\begin{pmatrix} 1 & 2& -2\\ -1 & 3 & 0 \\ 0 & -2& 1 \end{pmatrix}$
AnswerWriting $\begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 &0 \\ 0 & -2 &1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 &0 \\ 0 & 0 &1 \end{pmatrix}\text{ A} $ $\text{R}_{2}\rightarrow\text{R}_{2}+\text{R}_{1}\Rightarrow\begin{pmatrix} 1 & 2 & -2 \\ 0 & 5 & -2 \\ 0 & -2 &1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 1& 1 &0 \\ 0 & 0 &1 \end{pmatrix}\text{ A} $ $\text{R}_{2}\rightarrow\text{R}_{2}+\text{2R}_{3}\Rightarrow\begin{pmatrix} 1 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 &2 \\ 0 & 0 &1 \end{pmatrix}\text{ A} $ $\text{R}_{3}\rightarrow\text{R}_{3}+\text{2R}_{2}\Rightarrow\begin{pmatrix} 1 & 2 & -2 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 &2 \\ 2 & 2 &5 \end{pmatrix}\text{ A} $ $\text{R}_{1}\rightarrow\text{R}_{1}+\text{2R}_{3}\Rightarrow\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 5 & 4 & 10 \\ 1 & 1 &2 \\ 2 & 2 &5 \end{pmatrix}\text{ A} $ $\text{R}_{1}\rightarrow\text{R}_{1}-\text{2R}_{2}\Rightarrow\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 &2 \\ 2 & 2 &5 \end{pmatrix}\text{ A} $ $\text{Hence A}^{-1}=\begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 &2 \\ 2 & 2 &5 \end{pmatrix} $.
View full question & answer→Question 144 Marks
Using matrices, solve the following system of equations:
x + y + z = 6
x + 2z = 7
3x + y + z = 12
AnswerWriting as $ \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{pmatrix} \begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}=\begin{pmatrix} \text{6} \\ \text{7} \\ \text{12} \end{pmatrix}\text{or AX = B}$ |A| = 1 (-2) - 1 (-5) + 1 = 4 $\Rightarrow$ X = A-1 B | A11 = -2 | A12 = 5 | A13 = 1 |
| A21 = 0 | A22 = -2 | A23 = 2 |
| A31 = 2 | A32 = -1 | A33 = -1 |
$\text{A}^{-1}=\frac{1}{4} \begin{pmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix}$ $\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}=\frac{1}{4} \begin{pmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix} \cdot\begin{pmatrix} \text{6} \\ \text{7} \\ \text{12} \end{pmatrix}=\begin{pmatrix} \text{3} \\ \text{1} \\ \text{2} \end{pmatrix}$ x = 3, y = 1, z = 2.
View full question & answer→Question 154 Marks
Using matrices, solve the following system of linear equations;
x + 2y - 3z = –4.
2x + 3y + 2z = 2.
3x – 3y – 4z = 11.
AnswerThe given system of equations can be written as AX = B, where
$\text{A}=\begin{pmatrix} \text{1} & \text{2}& \text{-3} \\ \text{2} & \text{3}& \text{2} \\ \text{3} & \text{-3}& \text{-4} \end{pmatrix},\text{ X}=\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}\text{B}=\begin{pmatrix} \text{-4} \\ \text{2} \\ \text{11} \end{pmatrix}$ $\therefore$ X = A−1 B, if A−1 exists
|A| = 1 (– 12 + 6) –2 (– 8 – 6) – 3 (– 6 – 9) = – 6 + 28 + 45 = 67 ≠ 0
$\therefore$ A−1 exists
$\text{A}^{-1}=\frac{1}{|\text{A}|}\text{Adj. A}$ $\text{Adj. A}=\begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix}$ $\Rightarrow\text{ A}^{-1}=\frac{1}{67}\begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix}$ $\therefore\text{ X}=\begin{pmatrix} \text{x} \\ \text{y}\\ \text{z} \end{pmatrix}=\frac{1}{67}\begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix}\begin{pmatrix} \text{-4} \\ \text{2}\\ \text{11} \end{pmatrix}=\begin{pmatrix} \text{3}\\ \text{-2}\\ \text{1} \end{pmatrix}$ ∴ x = 3, y = – 2 z = 1.
View full question & answer→Question 164 Marks
Using matrices, solve the following system of equations:
x + 2y + z = 7.
x + 3z = 11.
3x - 3y = 1.
AnswerFor $\begin{pmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{pmatrix}\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}\begin{pmatrix} \text{7} \\ \text{11} \\ \text{1} \end{pmatrix}$
$\therefore$ A . X = B $\Rightarrow$ X = A-1 . B
|A| = 1(9) - 2(-6) + 1(-3) = 9 + 12 - 3 =18
C11 = 9, C12 = 6 C13 = -3
C21 = -3, C22 = -2 C23 = 7
C31 = 6, C32 = -2 C33 = -2
$\text{A}^{-1}=\frac{1}{18}\begin{pmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{pmatrix}$
$\Rightarrow\begin{pmatrix} \text{x}\\ \text{y} \\ \text{z} \end{pmatrix}=\frac{1}{18}\begin{pmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{pmatrix}\begin{pmatrix} \text{7}\\ \text{11} \\ \text{1} \end{pmatrix}$
$=\frac{1}{18}\begin{pmatrix} 63-33+6 =36\\ 42-22-2=18 \\ -21+77-2=54 \end{pmatrix}=\begin{pmatrix} \text{2}\\ \text{1} \\ \text{3} \end{pmatrix}$
$\Rightarrow$ x = 2, y = 1, z = 3.
View full question & answer→Question 174 Marks
Using matrices, solve the following system of equations:
3x - y + z = 5
2x - 2y + 3z = 7
x + y - z = -1.
AnswerWriting the system of equations as $ \begin{pmatrix} 3 & -1 & 1\\ 2 & -2 & 3\\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}=\begin{pmatrix} \text{5} \\ \text{7} \\ \text{-1} \end{pmatrix}$ i.e. AX = B $\therefore$ X = A-1B |A| = 3 (-1) + 1 (-5) + 1(4) = -3 - 5 + 4 = -4 $\neq$ 0 $\text{A}_{11}=+\begin{vmatrix}-2&3\\1&-1 \end{vmatrix}=(2-3)=-1$
$\text{A}_{12}=-\begin{vmatrix}2&3\\1&-1 \end{vmatrix}=-(-2-3)=5$
$\text{A}_{13}=+\begin{vmatrix}2&-2\\1&1 \end{vmatrix}=(2+2)=4$
$\text{A}_{21}=-\begin{vmatrix}-1&1\\1&-1 \end{vmatrix}=-(1-1)=0$
$\text{A}_{22}=+\begin{vmatrix}3&1\\1&-1 \end{vmatrix}=(-3-1)=-4$
$\text{A}_{23}=-\begin{vmatrix}3&-1\\1&1 \end{vmatrix}=-(3+1)=-4$
$\text{A}_{31}=+\begin{vmatrix}-1&1\\-2&3 \end{vmatrix}=(-3+2)=-1$
$\text{A}_{32}=-\begin{vmatrix}3&1\\2&3 \end{vmatrix}=-(9-2)=-7$
$\text{A}_{33}=+\begin{vmatrix}3&-1\\2&-2 \end{vmatrix}=(-6+2)=-4$
$\text{Adj A}=+\begin{vmatrix}-1&0&-1\\5&-4&-7\\4&-4&-4 \end{vmatrix}$
a11 = -1, a12 = 5, a13 = 4, a21 = 0, a22 = -4, a23 = -4 a31 = -1, a32 = -7, a33 = -4, $\therefore\text{ }\text{ } \text{ A}^{-1}=-\frac{1}{4}\begin{pmatrix} -1 & 0 & -1\\ 5 & -4 & -7\\ 4 & -4 & -4 \end{pmatrix}$ $\therefore\text{ }\text{ } \begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}=-\frac{1}{4}\begin{pmatrix} -1 & 0 & -1\\ 5 & -4 & -7\\ 4 & -4 & -4 \end{pmatrix} =\begin{pmatrix} \text{5} \\ \text{7} \\ \text{-1} \end{pmatrix}\begin{pmatrix} \text{1} \\ \text{-1} \\ \text{1} \end{pmatrix}$ $\therefore$ x = 1, y = -1, x = 1. View full question & answer→Question 184 Marks
Find matrix X so that $\text{X} \begin{pmatrix} 1 & 2 & 3 \\ \\ 4 & 5 & 6 \end{pmatrix} = \begin{pmatrix} -7 & -8 & -9 \\ \\ 4 & 5 & 6 \end{pmatrix}.$
AnswerGiven: $\text{X} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 4 & 6 \\ \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{bmatrix}$
Since $\text{X} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 4 & 6 \\ \end{bmatrix}$ is a $2\times 3$ matrix and the product is also a $2\times 3$ matrix so, X will be a $2\times 2$ matrix.
Let $\text{X} = \begin{bmatrix} x & y \\ a & b \\ \end{bmatrix}$ Then the given equation becomes,
$\begin{bmatrix} x & y \\ a & b \\ \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{bmatrix}$
$\Rightarrow \begin{bmatrix} x + 4y & 2x + 5y & 3x + 6y \\ a + 4b & 2a + 5b & 3a + 6b \\ \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{bmatrix}$
$\Rightarrow x + 4y = -7, 2x + 5y = -8, 3x + 6y = -9\\ \text{ }\text{ }\text{ }\text{ }\text{ }a + 4b = 2, 2a + 5b = 4, 3a + 6b = 6\\ \Rightarrow x = 1, y = -2, a = 2, b = 0$
Thus, X will be $\text{X} = \begin{bmatrix} 1 & -2 \\ 2 & 0 \\ \end{bmatrix}$
View full question & answer→Question 194 Marks
If $\text{A} = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{bmatrix} $ If A =, find A–1 and hence solve the system of equations x – 2y = 10, 2x + y + 3z = 8 and – 2y + z = 7.
Answer|A| = 11 $\neq$ 0, A–1 will exist
A11 = 7, A21 = 2, A31 = –6
A12 = –2, A22 = 1, A32 = –3
A13 = –4, A23 = 2, A33 = 5
$\text{A}^{-1} =\frac{1}{11} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} $
Given system of equations can be written as AX = B, where
$\text{A}=\begin{bmatrix} 1 & -2 & 0 \\ 2 & 1& 3 \\ 0 & -2 & 1 \end{bmatrix},\text{X}=\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix} ,\text{B}= \begin{bmatrix} 10 \\ 8\\ 7 \end{bmatrix}$
$\text{X}=\text{A}^{-1}\text{B}= \frac{1}{11} \begin{bmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{bmatrix} \begin{bmatrix} 10 \\8 \\ 7 \end{bmatrix}$
$ = \begin{bmatrix} 4\\ -3 \\ 1 \end{bmatrix}$
$\therefore\ \text{x = 4, y = -3, z = 1}$
View full question & answer→Question 204 Marks
If $\text{A} = \begin{bmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix}$ find A–1 and hence solve the system of equations x + 2y + 5z = 10, x – y – z = – 2 and 2x + 3y – z = – 11.
Answer|A| = 27 $\ne$ 0, A–1 exist
A11 = 4, A21 = 17, A13 = 3
A12 = –1, A22 = –11, A32 = 6
A13 = 5, A23 = 1, A33 = –3
$\therefore\ \text{A}^{-1} = \frac{1}{27}\begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix}$
Given system of equations can be written as AX = B where
$\text{A}=\begin{bmatrix} 1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1 \end{bmatrix},\text{X}=\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix},\text {B}= \begin{bmatrix} 10 \\ -2 \\ -11 \end{bmatrix}$
Now, AX = B ⇒ X = A–1B $=\frac{1}{27}\begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix}\begin{bmatrix} 10 \\ -2 \\ -11 \end{bmatrix}= \begin{bmatrix} -1 \\ -2 \\ 3 \end{bmatrix}$
$\therefore\ $ x = –1, y = –2, z = 3
View full question & answer→Question 214 Marks
A coaching institute of English (subject) conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, it has 20 poor and 5 rich children and total monthly collection is ₹ 9,000, whereas in batch II, it has 5 poor and 25 rich children and total monthly collection is ₹ 26,000. Using matrix method, find monthly fees paid by each child of two types. What values the coaching institute is inculcating in the society?
AnswerLet each poor child pay ₹ x per month and each rich child pay ₹ y per month
$\therefore \text{20x + 5y = 9000}$
$\text{5x + 25y = 26000}$
In matrix form,
$\begin{bmatrix} 20 & 5 \\ 5 & 25 \\ \end{bmatrix}\begin{bmatrix} \text{x} \\ \text{y} \\ \end{bmatrix} = \begin{bmatrix} 9000 \\ 26000 \\ \end{bmatrix}$
$\text{AX = B} \Rightarrow \text{X} = \text{A}^{-1} \text{B}$
$\text{A}^{-1} = \frac{1}{475} \begin{bmatrix} 25 & -5 \\ -5 & 20 \\ \end{bmatrix}$
$\begin{bmatrix} \text{x} \\ \text{y} \\ \end{bmatrix} = \frac{1}{475} \begin{bmatrix} 25 & -5 \\ -5 & 20 \\ \end{bmatrix}\begin{bmatrix} 9000 \\ 26000 \\ \end{bmatrix} = \begin{bmatrix} 200 \\ 1000 \\ \end{bmatrix}$
$\Rightarrow \text{x} = 200, \text{y} = 1000$
View full question & answer→Question 224 Marks
If $\text{A} = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \text{and B = } \text{A} = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}$ are two square matrices, find AB and hence solve the system of linear
equations x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7.
AnswerGetting $\text{AB} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = \text{6I}$
Given system of equations can be written as
$\begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix} = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$
$\text{i.e., AX = C} \Rightarrow \text{X} = \text{A}^{-1} \text{C} = \frac{1}{6} . \text{BC} \text{ }\text{ }\text{ }\text{ }\text{ }\bigg( \because \text{AB = 6I} \Rightarrow \text{A}^{-1} = \frac{1}{6} \text{B}\bigg)$
$= \frac{1}{6} \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}$
$ = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}$
$\Rightarrow \text{x = 2, y = -1, z = 4}$
View full question & answer→Question 234 Marks
$\text{If A}=\begin{bmatrix} 2 & 3 & 1 \\ 1 & 2 & 2 \\ -3 & 1 & -1 \end{bmatrix}$, find A–1 and hence solve the system of equations 2x + y – 3z = 13,
3x + 2y + z = 4, x + 2y – z = 8.
Answer|A| = –16 Co-factors are C11 = –4, C21 = 4, C31 = 4 C12 = –5, C22 = 1, C32 = –3 C13 = 7, C23 = –11, C33 = 1 $\text{A}^{-1}=\frac{-1}{16}\begin{bmatrix} -4 & 4& 4 \\ -5 & 1 & -3\\ 7 & -11 & 1 \end{bmatrix}$
Given equations can be written as A/X = C ⇒ X =( A-1 )/C $\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix}=\frac{-1}{16}\begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix}\begin{bmatrix}13 \\ 4 \\ 8 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix}$
⇒ x = 1, y = 2, z = –3
View full question & answer→Question 244 Marks
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
AnswerLet the income be 3x, 4x and expenditures, 5y, 7y $\therefore\text{3x - 5y = 15000}$ $\therefore\text{4x - 7y = 15000}$ $ \begin{bmatrix} 3 & -5 \\ 4 & -7 \\ \end{bmatrix} \begin{bmatrix} \text{x} \\ \text{y} \\ \end{bmatrix} = \begin{bmatrix} 15000 \\ 15000 \\ \end{bmatrix} $ $ \begin{bmatrix} \text{x} \\ \text{y} \\ \end{bmatrix} = -1 \begin{bmatrix} -7 & 5 \\ -4 & 3 \\ \end{bmatrix} \begin{bmatrix} 15000 \\ 15000 \\ \end{bmatrix} $ $\Rightarrow \text{x} = 30000, \text{y} = 15000$ $\therefore$ Incomes are ₹ 90,000 and ₹ 1,20,000 respectively, “Expenditure must be less than income”
View full question & answer→Question 254 Marks
$\text{If A} = \begin{bmatrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{bmatrix}, \text{find (A')}^{-1}. $
Answer$\text{A'} = \begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix} $
$|\text{A'|} = 1(-9) -2 (-5) = -9 + 10 = 1 \neq 0$
$\text{Adj A'} = \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} $
$\therefore \text{(A')}^{-1} = \begin{bmatrix} -9 & -8 & -2 \\ 8 & 7 & 2 \\ -5 & -4 & -1 \end{bmatrix} $
View full question & answer→Question 264 Marks
Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to its 3, 2 and 1 students with a total award money of ₹ 1,000. School Q wants to spend ₹ 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize one each value is ₹ 600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.
AnswerHere $ \begin{matrix} 3\text{x }+ & 2\text{y } +& \text{z } = 1000 \\ 4\text{x }+ & \text{y } +& 3\text{z } = 1500 \\ \text{x } +& \text{y } +& \text{z } = 600 \end{matrix}$
$\therefore \begin{pmatrix} 3 & 2 & 1 \\ 4 & 1& 3 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} \text{x}\\ \text{y} \\ \text{z} \end{pmatrix} = \begin{pmatrix} 1000 \\ 1500 \\ 600 \end{pmatrix}\text{ or }\text{A. X} = \text{B}$
|A | = 3 (– 2) – 2 (1) + 1 (3) = – 5 $\neq$ 0 $\therefore$ X = A–1 B
Co-factors are
$\begin{matrix} \text{A}_{11} = -2, & \text{A}_{12} = -1, & \text{A}_{13} = 3 \\ \text{A}_{21} = -1, & \text{A}_{22} = 2, & \text{A}_{23} = -1 \\ \text{A}_{31} = 5, & \text{A}_{32}= -5, & \text{A}_{33} = -5 \end{matrix}$
$\therefore\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix} = -\frac{1}{5}\begin{pmatrix} -2 & - 1 & 5 \\ - 1 & 2 & -2 \\ 3 & -1 & -5 \end{pmatrix}\begin{pmatrix} 1000\\ 1500 \\ 600 \end{pmatrix}$
$\therefore$ x = 100, y = 200, z = 300
i.e.Rs. 100 for discipline,Rs 200 for politeness&Rs. 300 for punctuality
One more value like sincerity, truthfulness etc.
View full question & answer→Question 274 Marks
$\text{If A}= \begin{pmatrix} 2 & 3 & 10 \\ 4& -6 & 5 \\ 6& 9& -20 \end{pmatrix},\text{ find A}^{-1}.\text{ Using A}^{-1}\text{ Solve the system of equation }$
$\frac{2}{\text{x}}+\frac{3}{\text{y}}+\frac{10}{\text{z}}=2;\frac{4}{\text{x}}-\frac{6}{\text{y}}+\frac{5}{\text{z}}=5;\frac{6}{x}+\frac{9}{\text{y}}-\frac{20}{\text{z}}=-4$
AnswerHere |A| = 1200
Co-factors are
C11 = 75, C21 = 150, = C31 = 75
C12 = 110, C22 = –100, C32 = 30
C13 = 72, C23 = 0, C33 = –24
$\text{A}^{-1} =\frac{1}{1200}\begin{bmatrix} 75 & 150& 75\\[0.3em] 110& -100 &30 \\[0.3em] 72 &0 & -24 \end{bmatrix}$
Given equation in matrix from is:
$\begin{bmatrix} 2 & 3& 10\\[0.3em] 4& -6 &5 \\[0.3em] 6 &9 & -20 \end{bmatrix}\begin{bmatrix} \frac{1}{\text{x}}\\[0.3em] \frac{1}{\text{y}} \\[0.3em] \frac{1}{\text{z}} \end{bmatrix}=\begin{bmatrix} 2\\[0.3em] 5 \\[0.3em] -4\end{bmatrix}$
⇒ A X = B
⇒ X = A–1B
$\Rightarrow\begin{bmatrix} \frac{1}{\text{x}}\\[0.3em] \frac{1}{\text{y}} \\[0.3em] \frac{1}{\text{z}} \end{bmatrix}=\begin{bmatrix} \frac{1}{2}\\[0.3em] \frac{-1}{3} \\[0.3em] \frac{1}{5} \end{bmatrix}$
⇒ x = 2, y = –3, z = 5
View full question & answer→Question 284 Marks
$\text{Let A} = \begin{pmatrix} 2 & -1 \\ 3 & 4 \\ \end{pmatrix}, = \text{B} = \begin{pmatrix} 5 & 2 \\ 7 & 4 \\ \end{pmatrix}, \text{C} = \begin{pmatrix} 2 & 5 \\ 3 & 8 \\ \end{pmatrix},$find a matrix D such that CD – AB = O.
Answer$\text{Let D} = \begin{bmatrix} \text{x} & \text{y} \\ \text{z} & \text{w}\\ \end{bmatrix}$
$\text{CD} = \text{AB} \Rightarrow\begin{bmatrix} \text{2x + 5z} & \text{2y + 5w} \\ \text{3x + 8z} & \text{3y + 8w}\\ \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 43 & 22\\ \end{bmatrix}$
$\text{2x + 5z = 3, 3x + 8z = 43; 2y + 5w = 0, 3y + 8w = 22.}$
$\text{Solving, we get x = –191, y = –110, z = 77, w = 44}$
$\therefore \text{D} = \begin{bmatrix} -191 & -110\\ 77 & 44\\ \end{bmatrix}$
View full question & answer→Question 294 Marks
Use product $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$ to solve the system of equations x + 3z = 9, –x + 2y – 2z = 4, 2x – 3y + 4z = –3.
Answer$\text{A} = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}.\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\text{AB = I} \Rightarrow \text{A}^{-1} = \text{B} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} $
Given equations in matrix form are:
$\begin{bmatrix} 1 & 0 & 3 \\ -1 & 2 & -2 \\ 2 & -3 & 4 \end{bmatrix}\begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix} = \begin{bmatrix} 9 \\ 4 \\ -3 \end{bmatrix}$
$\text{A}'\text{X = C}$
$\Rightarrow \text{X = (A}')^{-1} \text{ C = (A})' \text{C}$
$\Rightarrow \begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix}= \begin{bmatrix} -2 & 9 & 6\\ 0 & 2 & 1 \\ 1 & -3 & -2 \end{bmatrix} . \begin{bmatrix} 9 \\ 4 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}$
$\Rightarrow \text{x = 0, y = 5, z = 3}$
View full question & answer→Question 304 Marks
Using elementary transformations, find the inverse of the matrix $\text{A} = \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ and use it to solve the following system of linear equations:
$\text{8x + 4y +3z = 19}$
$\text{2x + y + z = 5} $
$\text{x + 2y + 2z = 7}$
Answer$\text{Writing} \begin{bmatrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A} $
$\text{R}_{1}\leftrightarrow \text{R}_{3} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 1 \\ 8 & 4 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\text{A} $
$ \begin{matrix} \text{R}_{1}\rightarrow & \text{R}_{1} - & 2\text{R} _{2} \\ \text{R}_{3}\rightarrow& \text{R}_{3} - & 4\text{R}_{2} \\ \end{matrix} $ $ \begin{bmatrix} -3 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -2 & 1 \\ 0 & 1 & 0 \\ 1 & -4 & 0 \end{bmatrix}\text{A} $
$ \begin{matrix} \text{R}_{1}\rightarrow & -\frac{1}{3}\text{R}_{1} \\ \text{R}_{3}\rightarrow & -\text{R}_{3}\\ \end{matrix} $ $ \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 0 & 1 & 0 \\ -1 & 4 & 0 \end{bmatrix}\text{A} $
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 0 & -\frac{1}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2} - \text{R}_{3} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix}\text{A} $
$\therefore\text{A}^{-1} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix} $
$\text{AX = B}\Rightarrow\text{X = A}^{-1}\text{B}$
$\therefore \begin{bmatrix} \text{x} \\ \text{y}\\ \text{z} \end{bmatrix} = \begin{bmatrix} 0 & \frac{2}{3} & -\frac{1}{3} \\ 1 & -\frac{13}{3} & \frac{2}{3} \\ -1 & 4 & 0 \end{bmatrix} \begin{bmatrix} 19 \\ 5 \\ 7 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} $
$\therefore \text{x = 1, y = 2, z = 1}$
View full question & answer→Question 314 Marks
$\text{If A} = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \text{find A}^{2} - \text{5 A + 4I} $ and hence find a matrix X such that $\text{A}^{2} - \text{5A + 4I + X = 0}$
Answer$\text{Getting A}^{2} = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{bmatrix} $
$\text{A}^{2} - \text{5A + 4I} = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} + \begin{bmatrix} -10 & 0 & -5 \\ -10 & -5 & -15 \\ -5 & 5 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 4 \end{bmatrix} $
$= \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix} $
$\therefore \text{X} = \begin{bmatrix} 1 & 1 & 3 \\ 1 &3 & 10\\ 5 & -4 & -2 \end{bmatrix} $
View full question & answer→Question 324 Marks
Two schools P and Q want to award their selected students on the values of Discipline, Politeness and Punctuality. The school P wants to award ₹ x each,₹ y each and ₹ z each for the three respective values to its 3, 2 and 1 students with a total award money of ₹ 1,000. School Q wants to spend ₹ 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize one each value is ₹ 600, using matrices, find the award money for each value.
Apart from the above three values, suggest one more value for awards.
AnswerHere $ \begin{matrix} 3\text{x }+ & 2\text{y } +& \text{z } = 1000 \\ 4\text{x }+ & \text{y } +& 3\text{z } = 1500 \\ \text{x } +& \text{y } +& \text{z } = 600 \end{matrix}$
$\therefore \begin{pmatrix} 3 & 2 & 1 \\ 4 & 1& 3 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} \text{x}\\ \text{y} \\ \text{z} \end{pmatrix} = \begin{pmatrix} 1000 \\ 1500 \\ 600 \end{pmatrix}\text{ or }\text{A. X} = \text{B}$
|A | = 3 (– 2) – 2 (1) + 1 (3) = – 5 $\neq$ 0 $\therefore$ X = A–1 B
Co-factors are
$\begin{matrix} \text{A}_{11} = -2, & \text{A}_{12} = -1, & \text{A}_{13} = 3 \\ \text{A}_{21} = -1, & \text{A}_{22} = 2, & \text{A}_{23} = -1 \\ \text{A}_{31} = 5, & \text{A}_{32}= -5, & \text{A}_{33} = -5 \end{matrix}$
$\therefore\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix} = -\frac{1}{5}\begin{pmatrix} -2 & - 1 & 5 \\ - 1 & 2 & -2 \\ 3 & -1 & -5 \end{pmatrix}\begin{pmatrix} 1000\\ 1500 \\ 600 \end{pmatrix}$
$\therefore$ x = 100, y = 200, z = 300
i.e.Rs. 100 for discipline,Rs 200 for politeness&Rs. 300 for punctuality
One more value like sincerity, truthfulness etc.
View full question & answer→Question 334 Marks
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of
6,000. Three times the award money for Hard work added to that given for honesty amounts to
11,000. The award money given for Honesty and Hardwork together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
AnswerLet x, y and z be the awarded money for honesty, Regularity and hard work.
From question
x + y +z = 6000 - - - -(i)
x + 3z =11000 - - - - (ii)
x +z = 2y $\Rightarrow$x – 2y +z = 0 - - - - - - (iii)
The above system of three equations may be written in matrix form as
AX = B,
where $\text{A} = \begin{bmatrix} 1 & 1 & 1 \\[0.3em] 1 & 0 & 3 \\[0.3em] 1 & -2 & 1 \end{bmatrix},\text{X} = \begin{bmatrix} \text{x} \\[0.3em] \text{y} \\[0.3em] \text{z} \end{bmatrix},\text{B} = \begin{bmatrix} 6000 \\[0.3em] 11000 \\[0.3em] 0 \end{bmatrix}$
Now $|\text{A}| = \begin{bmatrix} 1 & 1 & 1 \\[0.3em] 1 & 0 & 3 \\[0.3em] 1 & -2 & 1 \end{bmatrix} = 1 ( 0 + 6) - 1 (1 - 3) + 1 (- 2 - 0)$
= 6 + 2 – 2 = 6 $\neq$0
Hence A–1 exist
If Aij is co-factor of aij then
$\text{A}_{11} =(-1)^{1+1}\begin{bmatrix} 0 & 3\\[0.3em] -2 &1 \\[0.3em] \end{bmatrix} = 0 + 6 = 6 $
$\text{A}_{12} =(-1)^{1+2}\begin{bmatrix} 1 & 3 \\[0.3em] 1 &1 \\[0.3em] \end{bmatrix} = -(1-3) = 2; \text{A}_{13} = (-1)^{1+3}\begin{bmatrix} 1 & 0 \\[0.3em] 1 &-2 \\[0.3em] \end{bmatrix} = (-2-0) = -2 $
$\text{A}_{21} =(-1)^{2+1}\begin{bmatrix} 1 & 1 \\[0.3em] -2 &1 \\[0.3em] \end{bmatrix} = -(1+2) = -3; \text{A}_{22} = (-1)^{2+2}\begin{bmatrix} 1 & 1 \\[0.3em] 1 &1 \\[0.3em] \end{bmatrix} = 0$
$\text{A}_{23} =(-1)^{2+3}\begin{bmatrix} 1 & 1 \\[0.3em] 1 &-2 \\[0.3em] \end{bmatrix} = (-2-1) = 3; \text{A}_{31} = (-1)^{3+1}\begin{bmatrix} 1 & 1 \\[0.3em] 0 &3 \\[0.3em] \end{bmatrix} =3-0=3$
$\text{A}_{32} =(-1)^{3+2}\begin{bmatrix} 1 & 1 \\[0.3em] 1 &3\\[0.3em] \end{bmatrix} = -(3-1) = -2; \text{A}_{33} = (-1)^{3+3}\begin{bmatrix} 1 & 1 \\[0.3em] 1 &0 \\[0.3em] \end{bmatrix} =0-1=-1$
$adj\text{A} = \begin{bmatrix} 6 & 2 &-2 \\[0.3em] -3 &0 & 3 \\[0.3em] 3 & -2 & -1 \end{bmatrix}^{\text{T}} = \begin{bmatrix} 6 & -3 &3 \\[0.3em] 2 &0 & -2 \\[0.3em] -2 & 3 & -1 \end{bmatrix}$
$\therefore \text{A}^{-1} = .\frac{1}{|\text{A}|}.adj\text{A}=\frac{1}{6} = \begin{bmatrix} 6 & -3 &3 \\[0.3em] 2 &0 & -2 \\[0.3em] -2 & 3 & -1 \end{bmatrix}$
$\because \text{ AX} = \text{B}\Rightarrow\text{X} = \text{A}^{-1}\text{B}$
$\Rightarrow \begin{bmatrix} \text{x} \\[0.3em] \text{y} \\[0.3em] \text{z} \end{bmatrix} =\frac{1}{6}\begin{bmatrix} 6 & -3 & 3 \\[0.3em] 2 & 0 & -2 \\[0.3em] -2 & 3 & -1 \end{bmatrix}\begin{bmatrix} 6000 \\[0.3em] 11000 \\[0.3em] 0 \end{bmatrix}$
$\frac{1}{6}\begin{bmatrix} 36000 - 33000 + 0 \\[0.3em] 12000 + 0 + 0 \\[0.3em] -12000 + 33000 + 0 \end{bmatrix} = \frac{1}{6}\begin{bmatrix} 3000 \\[0.3em] 12000 \\[0.3em] 21000 \end{bmatrix}$
$\Rightarrow\begin{bmatrix} \text{x} \\[0.3em] \text{y} \\[0.3em] \text{z} \end{bmatrix} =\begin{bmatrix} 500 \\[0.3em] 2000 \\[0.3em] 3500 \end{bmatrix}$
$\Rightarrow$ x =500, y = 2000, z = 3500
Except above three values, school must include discipline for award as discipline has great importance in student’s life.
View full question & answer→Question 344 Marks
Using matrices, solve the following system of linear equations:
x – y + 2z = 7.
3x + 4y - 5z = –5.
2x – y + 3z = 12.
AnswerGiven equations can be written as $\begin{pmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{pmatrix}\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}=\begin{pmatrix} \text{7} \\ \text{-5} \\ \text{12} \end{pmatrix} \text{or AX = B}$
| A | = 1(7) + 1(19) + 2 (–11) = 4 $\neq$ 0 $\therefore$ X = A–1B
a11 = 7, a12 = -19, a13 = -11
a21 = 1, a22 = -1, a23 = -1
a31 = -3, a32 = 11, a33 = 7
$\Rightarrow$ $\text{A}^{-1}=\frac{1}{4}\begin{pmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{pmatrix}$
$\therefore$ $\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}$ = $\frac{1}{4}\begin{pmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{pmatrix}\begin{pmatrix} \text{7} \\ \text{-5} \\ \text{12} \end{pmatrix}=\begin{pmatrix} \text{2} \\ \text{1} \\ \text{3} \end{pmatrix}$
$\therefore$
x = 2, y = 1, z = 3. View full question & answer→Question 354 Marks
Using elementary operations, find the inverse of the following matrix: $\begin{pmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}$
Answerlet A =$\begin{pmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}\therefore\text{ Writing} \begin{pmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}=\text{A} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $
$\text{c}_{1}\leftrightarrow\text{c}_{2}\Rightarrow\begin{pmatrix} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & 3 & 1 \end{pmatrix}=\text{A} $ $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $
$\DeclareMathOperator*{\median}{{\text{c}_{2}\rightarrow\text{c}_{2}+\text{c}_{1}}}\median,\ {\text{c}_{3}\rightarrow{\text{c}_{3}-\text{2c}_{1}}}\begin{pmatrix} 1 & 0 & 0 \\ 2 & 3 & 1 \\ 1 & 4 & -1 \end{pmatrix}=\text{ A} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}$
$\DeclareMathOperator*{\median}{{\text{c}_{1}\rightarrow\text{c}_{1}+\text{2c}_{3}}}\median,\ {\text{c}_{2}\rightarrow{\text{c}_{2}-\text{2c}_{3}}}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ -1 & 2 & -1 \end{pmatrix}=\text{ A} \begin{pmatrix} 0 & 1 & 0 \\ -3 & -3 & -2 \\ 2 & 2 & 1 \end{pmatrix}$
$\text{c}_{3}\rightarrow\text{c}_{3}+\text{c}_{2}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 2 & 1 \end{pmatrix}=\text{ A} \begin{pmatrix} 0 & 1 & 1 \\ -3 & -3 & -5 \\ 2 & 2 & 3 \end{pmatrix}$
$\DeclareMathOperator*{\median}{{\text{c}_{1}\rightarrow\text{c}_{1}+\text{c}_{3}}}\median,\ {\text{c}_{2}\rightarrow{\text{c}_{2}-\text{2c}_{3}}}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}=\text{ A} \begin{pmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{pmatrix}$
$\Rightarrow$ $\text{A}^{-1}=$ $\begin{pmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{pmatrix}$.
View full question & answer→Question 364 Marks
Using matrix method, solve the following system of equations:
$\frac{2}{\text{x}}+\frac{3}{\text{y}}+\frac{10}{\text{z}}=4,\frac{4}{\text{x}}-\frac{6}{\text{x}}+\frac{5}{\text{z}}=1,\frac{6}{\text{x}}+\frac{9}{\text{y}}-\frac{20}{\text{z}}=2; \text{ x,y,z,}\neq0$
AnswerWriting the given system of equations as $\begin{pmatrix} 2 & 3 & 10 \\ 4 & -6 & 5\\ 6 & 9 & -20 \end{pmatrix}\begin{pmatrix} \frac{1}{\text{x}} \\ \frac{1}{\text{y}} \\ \frac{1}{\text{z}} \end{pmatrix}=\begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix} \text{or } \text{A}\cdot\text{X}=B$ |A| = 2(120-45)-3(-80-30)+10(36+36) = 1200, $\therefore$ X = A-1 B. C11 = 75, C12 = 110, C13 = 75
cofactors are C21 = 150, C22 = -100, C23 = 0 C31 = 75, C32 = 30, C33 = -24
A-1 $=\frac{1}{1200}\begin{pmatrix} 75 & 150 & 75 \\ 110 & -100 & 30\\ 72 & 0 & -24 \end{pmatrix}$
$\therefore$ $\begin{pmatrix} \frac{1}{\text{x}} \\ \frac{1}{\text{y}} \\ \frac{1}{\text{z}} \end{pmatrix}$$=\frac{1}{1200}\begin{pmatrix} 75 & 150 & 75 \\ 110 & -100 & 30\\ 72 & 0 & -24 \end{pmatrix}$$\begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}$ $=\begin{pmatrix} \frac{1}{\text{2}} \\ \frac{1}{\text{3}} \\ \frac{1}{\text{5}} \end{pmatrix}$
$\therefore$ x = 2, y = 3, z = 5.
View full question & answer→Question 374 Marks
Using elementary transformations, find the inverse of the matrix
$\begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -1\\ 2 & 1 & 0 \end{pmatrix}.$
AnswerGiven matrix A can be written as
$\begin{pmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\text{A}$
$\text{Applying R}_{2}\rightarrow\text{R}_{2}+\text{3R}_{1}\text{ R}_{3}\rightarrow{\text{R}_{3}-\text{2R}_{1}}$ $\begin{pmatrix} 1 & 3 & -2 \\ 0 & 9 & -7 \\ 0 & -5 & 4 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2}+\text{2R}_{3}\Rightarrow\begin{pmatrix} 1 & 3 & -2 \\ 0 & -1& 1 \\ 0 & -5 & 4 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 2 \\ -2 & 0 & 1 \end{pmatrix}\text{A}$
$\text{R}_{3}\rightarrow\text{R}_{3}-\text{5R}_{2}\Rightarrow\begin{pmatrix} 1 & 3 & -2 \\ 0 & -1& 1 \\ 0 & 0 & -1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 2 \\ 3 & -5 & -9 \end{pmatrix}\text{A}$
$\text{R}_{2}\rightarrow\text{R}_{2}+\text{R}_{3},\text{R}_{3}\rightarrow\begin{pmatrix} 1 & 3 & -2 \\ 0 & -1& 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 2 & -4 & -7 \\ -3 & 5 & 9 \end{pmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1}+\text{2R}_{3},\text{R}_{2}\rightarrow\text{-R}_{3}\begin{pmatrix} 1 & 3 & 0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} -5 & 10 & 18 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{pmatrix}\text{A}$
$\text{R}_{1}\rightarrow\text{R}_{1}-\text{3R}_{2}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{pmatrix}\text{A}$
$\therefore\text{A}^{-1}\begin{pmatrix} 1 & -2 & -3 \\ -2 & 4& 7 \\ -3 & 5 & 9 \end{pmatrix}.$
View full question & answer→Question 384 Marks
Using elementary row operations, find the inverse of the following matrix: $\begin{pmatrix} 2 & 5 \\ 1 & 3 \\ \end{pmatrix}$.
AnswerHere A = $\begin{pmatrix} 2 & 5 \\ 1 & 3 \\ \end{pmatrix}$
Writing A= IA $\Rightarrow$ $\begin{vmatrix} 2 & 5 \\ 1 & 3 \\ \end{vmatrix}=\begin{vmatrix} 1 & 0 \\ 0 & 1 \\ \end{vmatrix}\text{A}$
Applying R1 $\rightarrow$ R1 - R2, we get
$\begin{vmatrix} 1 & 2 \\ 1 & 3 \\ \end{vmatrix}=\begin{vmatrix} 1 & -1 \\ 0 & 1 \\ \end{vmatrix}\text{A}$
Applying R2 $\rightarrow$ R2 - R1, we get $\begin{bmatrix} 1 & 2 \\ 1 & 3 \\ \end{bmatrix}$= $\begin{bmatrix} 1 & -1 \\ -1 & 2 \\ \end{bmatrix}\text{A}$
Applying R1 $\rightarrow$ R1 - 2R2, we get $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$= $\begin{bmatrix} 3 & -5 \\ -1 & 2 \\ \end{bmatrix}\text{A}$
$\Rightarrow\text{A}^{-1}=\begin{bmatrix} 3 & -5 \\ -1 & 2 \\ \end{bmatrix}$.
View full question & answer→Question 394 Marks
Using matrices, solve the following system of equations:
$\text{2x - 3y + 5z} = 11$
$\text{3x + 2y - 4z} = -5$
$\text{x + y - 2z} = -3$
AnswerWriting as $\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{i. e A X = B} $
$|\text{A}| = 2(0) + 3(-2) + 5(1) = -1\ \therefore\ \text{X = A}^{-1} \text{B}$
$\text{A}_{11}=\begin{vmatrix}2 & -4 \\1 & -2 \end{vmatrix}=-4+4=0$
$\text{A}_{12}=-\begin{vmatrix}3 & -4 \\1 & -2 \end{vmatrix}=-(-6+4)=2$
$\text{A}_{13}=\begin{vmatrix}3 & 2 \\1 & 1 \end{vmatrix}=3-2=1$
$\text{A}_{21}=-\begin{vmatrix}-3 & 5 \\1 & -2 \end{vmatrix}=-(6-5)=-1$
$\text{A}_{22}=\begin{vmatrix}2 & 5 \\1 & -2 \end{vmatrix}=-4-5=-9$
$\text{A}_{23}=-\begin{vmatrix}2 & -3 \\1 & 1 \end{vmatrix}=-(2+3)=-5$
$\text{A}_{31}=\begin{vmatrix}-3 & 5 \\2 & -4 \end{vmatrix}=12-10=2$
$\text{A}_{32}=-\begin{vmatrix}2 &5\\3 & -4 \end{vmatrix}=-(-8-15)=23$
$\text{A}_{33}=\begin{vmatrix}2 & -3 \\3 & 2 \end{vmatrix}=4+9=13$
$\text{A}^{-1} = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} $
$\begin{bmatrix} \text{x} & \\ \text{y} & \\ \text{z} & \end{bmatrix} = - \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} \begin{bmatrix} \text{11} & \\ \text{-5} & \\ \text{-3} & \end{bmatrix} = \begin{bmatrix} \text{1} & \\ \text{2} & \\ \text{3} & \end{bmatrix}$
$\text{x = 1, y = 2, z = 3}$
View full question & answer→Question 404 Marks
$\text{Let A} = \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix}. $ Express A as sum of two matrices such that one is symmetric and the other is skew symmetric.
Answer$\text{A}= \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix}, \text{A}' = \begin{bmatrix} 3 & 4 &0 \\ 2 & 1 & 6 \\ 5 & 3 & 7 \end{bmatrix} $ $\therefore \frac{A + A'}{2} = \frac{1}{2} \begin{bmatrix} 6 & 6 & 5 \\ 6& 2 & 9 \\ 5 & 9 & 14 \end{bmatrix} = \begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{bmatrix} \Rightarrow \text{Symmetric} $
$\frac{A- A'}{2} = \frac{1}{2} \begin{bmatrix} 0 & -2 & 5 \\ 2 & 0 & -3 \\ -5 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & \frac{5}{2} \\ 1 & 0 & \frac{-3}{2} \\ \frac{-5}{2} & \frac{3}{2} & 0 \end{bmatrix} \Rightarrow \text{Skew symmetric} $
$\therefore \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} = \begin{bmatrix} 3 & 3 & \frac{5}{2} \\ 3 & 1 & \frac{9}{2} \\ \frac{5}{2} & \frac{9}{2} & 7 \end{bmatrix} + \begin{bmatrix} 0 & -1 & \frac{5}{2}] \\ 1 & 0 & \frac{-3}{2} \\ \frac{-5}{2} & \frac{3}{2} & 0 \end{bmatrix} $
View full question & answer→Question 414 Marks
Using matrices, solve the following system of equations: $x + 2y - 3z = 6$
$3x + 2y - 2z =3$
$2x - y + z = 2$
AnswerThe given system of equations can be written as: $\text{A X = B ,where A} = \begin{bmatrix} 1 & 2 & -3 \\ 3 & 2 & -2 \\ 2 & -1 & 1 \end{bmatrix}_{,} \text{X} = \begin{bmatrix} x \\ y \\z \end{bmatrix}_{,} \text{B}= \begin{bmatrix} 6 \\ 3 \\2 \end{bmatrix}$
$\therefore \text{X} = \text{A}^{-1} \text{B}$
$| \text{A}| = 7 \neq 0$
For finding cofactors:
$\begin{matrix} \text{C}_{11} = 0 & \text{C}_{12} = -7 & \text{C}_{13} = -7 \\ \text{C}_{21} = 1 & \text{C}_{22} = 7 & \text{C}_{23} = 5 \\ \text{C}_{31} = 2 & \text{C}_{32} = -7 & \text{C}_{33} = -4 \end{matrix} $
$\Rightarrow \text{A}^{-1} = \frac{1}{7} \begin{bmatrix} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{bmatrix} $
$\therefore \text{X} = \begin{bmatrix} \text{x} \\ \text{y} \\ \text{z} \end{bmatrix}= \begin{bmatrix} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{bmatrix} \begin{bmatrix} \text{6} \\ \text{3} \\ \text{2} \end{bmatrix} = \begin{bmatrix} \text{1} \\ \text{-5} \\ \text{-5} \end{bmatrix}$
$\therefore\text{ x = 1, y = -5, z = -5}$
View full question & answer→Question 424 Marks
Using matrices, solve the following system of equations: $\text{x + y + z} = 3, \text{x} - 2{\text{y}} + 3{z}=2\ \text{and} \ 2 \text{x - y + z} = 2 $
AnswerThe system of equations can be written as: AX = B, where
$\text{A}= \begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & 3 \\ 2 & -1 & 1 \end{bmatrix} , X = \begin{bmatrix} x\\ y\\ z\end{bmatrix} , B = \begin{bmatrix} 3\\ 2\\ 2\end{bmatrix} $ OR
$X = A^{-1} B$
$|A| = 1 (- 2 + 3) -1 (1 - 6) + 1(-1+4) = 1+5+3 = 9$
$\text{A}_{11}=+\begin{vmatrix}-2&3\\-1 & 1 \end{vmatrix}=-2+3=1$
$\text{A}_{12}=-\begin{vmatrix}1&3\\2 & 1 \end{vmatrix}=-(1-6)=5$
$\text{A}_{13}=+\begin{vmatrix}1&-2\\2 & -1 \end{vmatrix}=-1+4=3$
$\text{A}_{21}=-\begin{vmatrix}1&1\\-1 & 1 \end{vmatrix}=-(1+1)=-2$
$\text{A}_{22}=+\begin{vmatrix}1&1\\2 &1 \end{vmatrix}=1-2=-1$
$\text{A}_{23}=-\begin{vmatrix}1&1\\2 & -1 \end{vmatrix}=-(-1-2)=3$
$\text{A}_{31}=+\begin{vmatrix}1&1\\-2 &3 \end{vmatrix}=3+2=5$
$\text{A}_{32}=-\begin{vmatrix}1&1\\1 & 3 \end{vmatrix}=-(3-1)=-2$
$\text{A}_{33}=+\begin{vmatrix}1&1\\1 & -2 \end{vmatrix}=-2-1=-3$
$Adj. A = \begin{bmatrix} 1 & -2 & 5 \\ 5 & -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} $
$\therefore A^-1 = \frac{1}{9} \begin{bmatrix} 1 & -2 & 5 \\ 5 & -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} $
$\therefore \begin{bmatrix} x \\ y \\z \end{bmatrix} =\frac{1}{9} \begin{bmatrix} 1 &-2 & 5 \\ 5& -1 & -2 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} 3 \\ 2 \\2 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 \\ 9 \\9 \end{bmatrix}$
$\begin{bmatrix} 1 \\ 1 \\1 \end{bmatrix} \Rightarrow x = y = z = 1$
View full question & answer→Question 434 Marks
Using elementary row transformations, find the inverse of the matrix $\text{A}=\begin{bmatrix}1 & 2&3 \\2 & 5&7\\-2&-4&-5 \end{bmatrix}.$
AnswerA-1 = IA (Inverse of matrix)
$\text{A}^{-1}=\begin{bmatrix}1&0&0 \\0 &1&0\\0&0&1 \end{bmatrix}\begin{bmatrix}1&2&3\\2&5&7\\-2&-4&-5 \end{bmatrix}$
$\text{R}_2\rightarrow\text{R}_2-2\text{R}_1$
$\text{R}_3\rightarrow\text{R}_3-2\text{R}_1$
$\text{A}^{-1}=\begin{bmatrix}1&0&0 \\-2 &1&0\\2&0&1 \end{bmatrix}\begin{bmatrix}1&2&3\\0&1&1\\0&0&1 \end{bmatrix}$
$\text{R}_1\rightarrow\text{R}_1-3\text{R}_3$
$\text{R}_2\rightarrow\text{R}_2-\text{R}_3$
$\text{A}^{-1}=\begin{bmatrix}-5&0&-3 \\-4 &1&-1\\2&0&1 \end{bmatrix}\begin{bmatrix}1&2&0\\0&1&0\\0&0&1 \end{bmatrix}$
$\text{R}_1\rightarrow\text{R}_1-2\text{R}_2$
$\text{A}^{-1}=\begin{bmatrix}3&-2&-1 \\-4 &1&-1\\2&0&1 \end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}3&-2&-1\\-4&1&-1\\2&0&1 \end{bmatrix}$
View full question & answer→Question 444 Marks
if $\text{A}=\begin{bmatrix}2 & -3&5 \\3 & 2&-4\\1&1&-2 \end{bmatrix},$ find A-1. Use it to solve the system of equations
2x - 3y + 5z = 11
3x + 2y – 4z = -5
x + y - 2z = -3.
Answer$|\text{A}|=\begin{bmatrix}2 & -3&5 \\3 & 2&-4\\1&1&-2 \end{bmatrix}=2(-4+4)+3(-6+4)+5(3-2)$
$\therefore|\text{A}|=0-6+5=-1\neq0$
$\text{Now A}_{11}=0;\text{A}_{12}=2;\text{A}_{13}=1$
$\text{A}_{21}=-1;\text{A}_{22}=-9;\text{A}_{23}=-5$
$\text{A}_{31}=2;\text{A}_{32}=23;\text{A}_{33}=13$
$\text{A}_\text{ij}=\begin{bmatrix}0 & 2&1 \\-1 & -9&-5\\2&23&13 \end{bmatrix}$
$\therefore\ \text{Adj A = [A}_\text{ij}]^\text{T}=\begin{bmatrix}0 & 2&1 \\-1 & -9&-5\\2&23&13 \end{bmatrix}^\text{T}=\begin{bmatrix}0 & -1&2 \\2 & -9&23\\1&-5&13 \end{bmatrix}$
$ \text{A}^{-1}=\frac{\text{AdjA}}{|\text{A}|}=-\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}$
$\text{Now}\begin{bmatrix}2 &-3&5 \\3 & 2&-4\\1&1&-2 \end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\\\text{z} \end{bmatrix}=\begin{bmatrix}11\\-5\\-3 \end{bmatrix}$
$\Rightarrow\text{AX = B}$
$\Rightarrow\text{X = A}^{-1}\text{ B}=\begin{bmatrix}0 &1&-2 \\-2 &9&-23\\-1&5&-13 \end{bmatrix}\begin{bmatrix}11\\-5\\-3 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\\\text{z} \end{bmatrix}\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39 \end{bmatrix}=\begin{bmatrix}1\\2\\3 \end{bmatrix}$
$\therefore\ \text{x}=1;\text{y}=2\text{ and z}=3$
View full question & answer→Question 454 Marks
A manufacturer produces three types of bolts, x, y and z which he sells in two markets. Annual sales (in ₹) are indicated below:
| Markets | Products |
| X | Y | Z |
| I | 10000 | 2000 | 18000 |
| II | 6000 | 20000 | 8000 |
If unit sales prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00 respectively, then answer the following questions using the concept of matrices.- Find the total revenue collected from the Market-I.
- ₹ 44000
- ₹ 48000
- ₹ 46000
- ₹ 53000
- Find the total revenue collected from the Market-II.
- ₹ 51000
- ₹ 53000
- ₹ 46000
- ₹ 49000
- If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively, then find the gross profit from both the markets.
- ₹ 53000
- ₹ 46000
- ₹ 34000
- ₹ 32000
- If matrix $\text{A}=[\text{a}_\text{ij}]_{2\times2},$ where $\text{a}_\text{ij}=1,$ if $\text{i}\neq\text{j}$ and $\text{a}_\text{ij}=0,$ if $\text{i}=\text{j}$ then A2 is equal to:
- I
- A
- OR
- None of these
- If A and B are matrices of same order, then (AB' - BA') is a.
- Skew-synunetric matrix.
- Null matrix.
- Symmetric matrix.
- Unit matrix.
AnswerLet A be the 2 × 3 matrix representing the annual sales of products in two markets. $\begin{matrix}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{X}&\ \ \ \ \ \ \ \text{Y}&\ \ \ \ \ \ \ \text{Z}\end{matrix}\\\therefore\ \ \text{A}=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{matrix}\text{Market I}\\\text{Market II}\end{matrix}$ Let B be the column matrix representing the sale price of each unit of products x, y, z. $\therefore\ \text{B}=\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}$ Now, revenue = sale price × number of items sold $=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2.5\\1.5\\1\end{bmatrix}$ $=\begin{bmatrix}25000+3000+18000\\15000+30000+8000\end{bmatrix}=\begin{bmatrix}46000\\53000\end{bmatrix}$ Therefore, the revenue collected from Market I = ₹ 46000 and the revenue collected from Market II = ₹ 53000.
- (c) ₹ 46000
- (b) ₹ 53000
- (d) ₹ 32000
Solution:
Let C be the column matrix representing cost price of each unit of products x, y, z.
Then, $\text{C}=\begin{bmatrix}2\\1\\0.5\end{bmatrix}$
$\therefore$ Total cost in each market is given by
$\text{AC}=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}\begin{bmatrix}2\\1\\0.5\end{bmatrix}$
$=\begin{bmatrix}20000+2000+9000\\12000+20000+4000\end{bmatrix}=\begin{bmatrix}31000\\36000\end{bmatrix}$
Now, Profit matrix = Revenue matrix - Cost matrix
$=\begin{bmatrix}46000\\53000\end{bmatrix}-\begin{bmatrix}31000\\36000\end{bmatrix}=\begin{bmatrix}15000\\17000\end{bmatrix}$
Therefore, the gross profit from both the markets = ₹ 15000 + ₹ 17000 = ₹ 32000
- (a) I
Solution:
We have, $\text{A}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$
$\therefore\ \ \text{A}^2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}=\text{I}$
- (a) Skew-synunetric matrix.
Solution:
We have, (AB' - BA')' = (B')' A' - (A')' B' = BA' - AB'= -(AB' - BA')
Thus, AB' - BA' is a skew-symmetric matrix.
View full question & answer→Question 464 Marks
If $\text{A}=\begin{bmatrix}1&0\\-1&7\end{bmatrix},$ find k such that A2 - 8A + kI = 0.
AnswerGiven: $\text{A}=\begin{bmatrix}1&0\\-1&7\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\-1&7\end{bmatrix}\begin{bmatrix}1&0\\-1&7\end{bmatrix}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}1-0&0+0\\-1-7&0+49\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0\\-8&49\end{bmatrix}$
$ \text{A}^2-8\text{A}+\text{kI}=0$
$ \Rightarrow\begin{bmatrix}1&0\\-8&49\end{bmatrix}-8\begin{bmatrix}1&0\\-1&7\end{bmatrix}+\text{k}\begin{bmatrix}1&0\\0&1\end{bmatrix}=0$
$ \Rightarrow\begin{bmatrix}1&0\\-8&49\end{bmatrix}-\begin{bmatrix}8&0\\-8&56\end{bmatrix}+\begin{bmatrix}\text{k}&0\\0&\text{k}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}1-8+\text{k}&0-0+0\\-8+8+0&49-56+\text{k}\end{bmatrix}=0$
$\Rightarrow\begin{bmatrix}-7+\text{k}&0\\0&-7+\text{k}\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore-7+\text{k}=0$
$\Rightarrow\text{k}=7$
View full question & answer→Question 474 Marks
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
AnswerGiven: $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
Now,
$\text{P}\text{Q}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{y}\text{b}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{zc}\end{bmatrix}$
$ =\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(4)$
Also,
$\text{Q}\text{P}=\begin{bmatrix}\text{a}&0&0\\0&\text{a}&0\\0&0&\text{a}\end{bmatrix}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$=\begin{bmatrix}\text{ax}+0+0&0+0+0&0+0+0\\0+0+0&0+\text{by}+0&0+0+0\\0+0+0&0+0+0&0+0+\text{cz}\end{bmatrix}$
$=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}\ \dots(5)$
From (4) and (5), we get
$\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
View full question & answer→Question 484 Marks
If $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then prove that A2 - A + 2I = O.
AnswerGiven: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}+2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1-3&-2+2\\4-4&-4+2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2+2&0+0\\0+0&-2+2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=0$
Hence proved.
View full question & answer→Question 494 Marks
Show that A′A and AA′ are both symmetric matrices for any matrix A.
AnswerLet P = A'A
$\therefore$ P' = (AA')'
= A'(A')' $[\because$ (AB')' = BA'$]$
= A'A = P
So, A’A is symmetric matrix for any matrix A.
Similarly, let Q = AA’
$\therefore$ Q' = (AA')' = (A')'(A)'
= A(A')' = Q
So, AA’ is symmetric matrix for any matrix A.
View full question & answer→Question 504 Marks
If $\text{A}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix},$ show that AB = BA = O3 × 3
AnswerHere,
$\text{AB}=\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0+\text{abc}-\text{abc}&0+\text{b}^2\text{c}-\text{b}^2\text{c}&0+\text{bc}^2-\text{bc}^2\\-\text{a}^2\text{c}+0+\text{a}^2\text{c}&-\text{abc}+0+\text{abc}&-\text{ac}^2+0+\text{ac}^2\\\text{a}^2\text{b}-\text{a}^2\text{b}+0&\text{ab}^2-\text{ab}^2+0&\text{abc}-\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{AB}=\text{O}_{3\times3}\ \dots(1)$
$\text{BA}=\begin{bmatrix}\text{a}^2&\text{ab}&\text{ac}\\\text{ab}&\text{b}^2&\text{bc}\\\text{ac}&\text{bc}&\text{c}^2\end{bmatrix}\begin{bmatrix}0&\text{c}&-\text{b}\\-\text{c}&0&\text{a}\\\text{b}&-\text{a}&0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0-\text{abc}+\text{abc}&\text{a}^2\text{c}+0-\text{a}^2\text{c}&-\text{a}^2\text{b}+\text{a}^2\text{b}+0\\0-\text{b}^2\text{c}+\text{b}^2\text{c}&\text{abc}+0-\text{abc}&-\text{ab}^2+\text{ab}^2+0\\0-\text{bc}^2+\text{bc}^2&\text{ac}^2+0-\text{ac}^2&-\text{abc}+\text{abc}+0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\Rightarrow\text{BA}=\text{O}_{3\times3}\ \dots(2)$
$ \Rightarrow\text{AB}=\text{BA}=0_{3\times3}$ [From eqs. (1) and (2)]
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