Question 15 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 25 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$\text{A}=\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 &-4 \\ 1 & 1 & -2 \end{bmatrix} \Rightarrow|\text{A}|=2(0)+3(-2)+5(1)=-1\neq0$
$A_{11} = 0, A_{12}= 2, A_{13} = 1$
$A_{21} = –1, A_{22}= –9, A_{23} = –5$
$A_{31}= 2, A_{32}= 23, A_{33} = 13$
$ \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 35 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 45 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 55 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$\triangle=\begin{vmatrix} \text{a}^2+2\text{a}& 2\text{a}+1 & 1\$0.3em] 2\text{a}+1 &\text{a}+2 &1 \$0.3em] 3 & 3 & 1 \end{vmatrix}$
$R_1 → R_1 – R_2$ and $R_2 → R_2– R_3$
$\triangle=\begin{vmatrix} \text{a}^2+1& \text{a}-1 & 0\$0.3em] 2(\text{a}-1) &\text{a}-1 &0 \$0.3em] 3 & 3 & 1 \end{vmatrix}$
$\triangle=(\text{a}-1)^2\begin{vmatrix} \text{a}+1& 1 & 0\$0.3em] 2 &1 &0 \$0.3em] 3 & 3 & 1 \end{vmatrix}$
Expanding
$(a – 1)^2.(a – 1) = (a – 1)^3$.
View full question & answer→Question 65 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 75 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
$A_{11} = 0, A_{12} = 2, A_{13}= 1$
$A_{21} = –1, A_{22} = –9, A_{23} = –5$
$A_{31} = 2, A_{32} = 23, A_{33} = 13$
$\Rightarrow\text{A}^{-1} =-1 \begin{pmatrix} 0 & 2 & 1 \\ -1 & -9& -5\\ 2&23 &13 \end{pmatrix}=-1 \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9& 23\\ 1 & -5 &13 \end{pmatrix}= \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9& -23\\ -1 & 5 &-13 \end{pmatrix}$
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 85 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 95 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 105 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 115 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 125 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 135 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 145 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}$
$|A| = 1 (-2) - 1 (-5) + 1 = 4 \Rightarrow X = A^{-1} B$
| $A_{11} = -2$ |
$A_{12} = 5$ |
$A_{13} = 1$ |
| $A_{21} = 0$ |
$A_{22} = -2$ |
$A_{23} = 2$ |
| $A_{31} = 2$ |
$A_{32} = -1$ |
$A_{33} = -1$ |
$\text{A}^{-1}=\frac{1}{4} \begin{pmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix}$
$\begin{pmatrix} \text{x} \\ \text{y} \\ \text{z} \end{pmatrix}=\frac{1}{4} \begin{pmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{pmatrix} \cdot\begin{pmatrix} \text{6} \\ \text{7} \\ \text{12} \end{pmatrix}=\begin{pmatrix} \text{3} \\ \text{1} \\ \text{2} \end{pmatrix}$
$x = 3, y = 1, z = 2.$ View full question & answer→Question 155 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 165 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
$|A| = 1(9) - 2(-6) + 1(-3) = 9 + 12 - 3 =18$
$C_{11} = 9, C_{12} = 6\ C_{13} = -3$
$C_{21} = -3, C_{22} = -2\ C_{23} = 7$
$C_{31} = 6, C_{32} = -2\ C_{33} = -2$
$\text{A}^{-1}=\frac{1}{18}\begin{pmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{pmatrix}$
$\Rightarrow\begin{pmatrix} \text{x}\\ \text{y} \\ \text{z} \end{pmatrix}=\frac{1}{18}\begin{pmatrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{pmatrix}\begin{pmatrix} \text{7}\\ \text{11} \\ \text{1} \end{pmatrix}$
$=\frac{1}{18}\begin{pmatrix} 63-33+6 =36\\ 42-22-2=18 \\ -21+77-2=54 \end{pmatrix}=\begin{pmatrix} \text{2}\\ \text{1} \\ \text{3} \end{pmatrix}$
$\Rightarrow$ x = 2, y = 1, z = 3.
View full question & answer→Question 175 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
$\therefore X = A^{-1}B |A| = 3 (-1) + 1 (-5) + 1(4) = -3 - 5 + 4 = -4 \neq$ 0$\text{A}_{11}=+\begin{vmatrix}-2&3\\1&-1 \end{vmatrix}=(2-3)=-1$
$\text{A}_{12}=-\begin{vmatrix}2&3\\1&-1 \end{vmatrix}=-(-2-3)=5$
$\text{A}_{13}=+\begin{vmatrix}2&-2\\1&1 \end{vmatrix}=(2+2)=4$
$\text{A}_{21}=-\begin{vmatrix}-1&1\\1&-1 \end{vmatrix}=-(1-1)=0$
$\text{A}_{22}=+\begin{vmatrix}3&1\\1&-1 \end{vmatrix}=(-3-1)=-4$
$\text{A}_{23}=-\begin{vmatrix}3&-1\\1&1 \end{vmatrix}=-(3+1)=-4$
$\text{A}_{31}=+\begin{vmatrix}-1&1\\-2&3 \end{vmatrix}=(-3+2)=-1$
$\text{A}_{32}=-\begin{vmatrix}3&1\\2&3 \end{vmatrix}=-(9-2)=-7$
$\text{A}_{33}=+\begin{vmatrix}3&-1\\2&-2 \end{vmatrix}=(-6+2)=-4$
$\text{Adj A}=+\begin{vmatrix}-1&0&-1\\5&-4&-7\\4&-4&-4 \end{vmatrix}$
$a_{11} = -1, a_{12} = 5, a_{13} = 4, a_{21} = 0, a_{22} = -4, a_{23} = -4 a_{31} = -1, a_{32} = -7, a_{33} = -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 185 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 195 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
$A_{11} = 7, A_{21} = 2, A_{31} = –6$
$A_{12} = –2, A_{22}= 1, A_{32} = –3$
$A_{13} = –4, A_{23} = 2, A_{33} = 5$
$\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 205 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
$A_{11} = 4, A_{21} = 17, A_{13} = 3$
$A_{12}= –1, A_{22} = –11, A_{32} = 6$
$A_{13} = 5, A_{23} = 1, A_{33} = –3$
$\therefore\ \text{A}^{-1} = \frac{1}{27}\begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix}$
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, $AX = B ⇒ X = A^{–1}B =\frac{1}{27}\begin{bmatrix} 4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3 \end{bmatrix}\begin{bmatrix} 10 \\ -2 \\ -11 \end{bmatrix}= \begin{bmatrix} -1 \\ -2 \\ 3 \end{bmatrix}$
$\therefore\ $ x = –1, y = –2, z = 3
View full question & answer→Question 215 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 225 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 235 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 $C_{11} = –4, C_{21} = 4, C_{31} = 4 C_{12} = –5, C_{22} = 1, C_{32}= –3 C_{13} = 7, C_{23} = –11, C_{33} = 1$
$\text{A}^{-1}=\frac{-1}{16}\begin{bmatrix} -4 & 4& 4 \\ -5 & 1 & -3\\ 7 & -11 & 1 \end{bmatrix}$
Given equations can be written as $A/X = C \Rightarrow 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 245 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 255 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 265 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 275 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
$C_{11} = 75, C_{21} = 150, = C_{31} = 75$
$C_{12} = 110, C_{22} = –100, C_{32} = 30$
$C_{13} = 72, C_{23} = 0, C_{33}= –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}$
$\Rightarrow A X = B$
$\Rightarrow X = A^{–1}B$
$\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 285 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 295 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 305 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 315 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 325 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 335 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^{–1}B$
$a_{11} = 7, a_{12} = -19, a_{13}= -11$
$a_{21} = 1, a_{22} = -1, a_{23} = -1$
$a_{31} = -3, a_{32} = 11, a_{33} = 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 345 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 355 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$.
$C_{11} = 75, C_{12}= 110, C_{13} = 75$
cofactors are $C_{21} = 150, C_{22} = -100, C_{23} = 0$
$C_{31} = 75, C_{32} = 30, C_{33} = -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 365 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 375 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 $R_1 \rightarrow R_1- R_2$, we get
$\begin{vmatrix} 1 & 2 \\ 1 & 3 \\ \end{vmatrix}=\begin{vmatrix} 1 & -1 \\ 0 & 1 \\ \end{vmatrix}\text{A}$
Applying $R_2 \rightarrow R_2- R_1$, we get $\begin{bmatrix} 1 & 2 \\ 1 & 3 \\ \end{bmatrix}$= $\begin{bmatrix} 1 & -1 \\ -1 & 2 \\ \end{bmatrix}\text{A}$
Applying $R_1 \rightarrow R_1- 2R_2$, 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 385 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 395 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 405 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 415 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 425 Marks
Using elementary row transformations, find the inverse of the matrix $\text{A}=\begin{bmatrix}1 & 2&3 \\2 & 5&7\\-2&-4&-5 \end{bmatrix}.$
Answer$A^{-1} = IA$ (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 435 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 445 Marks
Find the values of x, y, z if the matrix A = $\begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}\\\text{x}&-\text{y}&\text{z}\end{bmatrix}$satisfies the equation A’A = I.
AnswerGiven: $\text{A} = \begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}\\\text{x}&-\text{y}&\text{z}\end{bmatrix}$ $\Rightarrow\text{A}'= \begin{bmatrix}0&\text{x}&\text{x}\\2\text{y}&\text{y}&-\text{y}\\\text{z}&-\text{z}&\text{z}\end{bmatrix}$
Now $\text{A}'\text{A} =\text{ I}\Rightarrow\begin{bmatrix}0&2\text{y}&\text{z}\\\text{x}&\text{y}&-\text{z}&\\\text{x}&-\text{y}&\text{z}\end{bmatrix} \begin{bmatrix}0&\text{x}&\text{x}\\2\text{y}&\text{y}&-\text{y}\\\text{z}&-\text{z}&\text{z}\end{bmatrix} = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}0+\text{x}^{2}+\text{x}^{2}&0+\text{xy}-\text{xy}&0-\text{xz}+\text{xz}\\0+\text{xy}-\text{xy}&4\text{y}^{\text{y}}+\text{y}^{2}+\text{y}^{2}&2\text{yz}-\text{yz}-\text{yz}\\0-\text{zx}+\text{zx}&2\text{yz}-\text{yz}-\text{yz}&\text{z}^{2}+\text{z}^{2}+\text{z}^{2}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\ \begin{bmatrix}2\text{x}^{2}&0&0\\0&6\text{y}^{2}&0\\0&0&3\text{z}^{2}\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
Equating corresponding entries, we have
$2\text{x}^{2}=1\ \Rightarrow \ \text{x}^{2}=\frac{1}{2} \ \Rightarrow\ \text{x}=\pm\frac{1}{\sqrt{2}}$
And $6\text{y}^{2}=1\ \Rightarrow \ \text{y}^{2}=\frac{1}{6} \ \Rightarrow\ \text{y}=\pm\frac{1}{\sqrt{6}}$
And $3\text{z}^{2}=1\ \Rightarrow \ \text{z}^{2}=\frac{1}{3} \ \Rightarrow\ \text{z}=\pm\frac{1}{\sqrt{3}}$
$\therefore \ \text{x}=\pm \frac{1}{\sqrt{2}},\text{y}=\pm\frac{1}{\sqrt{6}},\text{z}=\pm\frac{1}{\sqrt{3}}$
View full question & answer→Question 455 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 465 Marks
If $\text{A}=\begin{bmatrix}2&3\\-1&0\end{bmatrix},$ show that $A^2 - 2A + 3I_2 = 0$.
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\-1&0\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&3\\-1&0\end{bmatrix}\begin{bmatrix}2&3\\-1&0\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4-3&6+0\\-2+0&-3+0\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}$
$ \text{A}^2-2\text{A}+3\text{I}_2$
$\Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}-2\begin{bmatrix}2&3\\-1&0\end{bmatrix}+3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1&6\\-2&-3\end{bmatrix}-\begin{bmatrix}4&6\\-2&0\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}$
$ \Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}1-4+3&6-6+0\\-2+2+0&-3+0+3\end{bmatrix}$
$\Rightarrow\text{A}^2-2\text{A}+3\text{I}_2=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 475 Marks
Express the matrix $\begin{bmatrix}2&3&1\\1&-1&2\\4&1&2\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
AnswerWe have, $\text{A}=\begin{bmatrix}2&3&1\\1&-1&2\\4&1&2\end{bmatrix}$
$\therefore\ \text{A}'=\begin{bmatrix}2&1&4\\3&-1&1\\1&2&2\end{bmatrix}$
Now, $\Rightarrow\ \frac{\text{A}+\text{A}'}{2}=\frac{1}{2}\begin{bmatrix}4&4&5\\4&-2&3\\5&3&4\end{bmatrix}$
$=\begin{bmatrix}2&2&\frac{5}{2}\\2&-1&\frac{3}{2}\\\frac{5}{2}&\frac{3}{2}&2\end{bmatrix}$
And $\frac{\text{A}-\text{A}'}{2}=\frac{1}{2}\begin{bmatrix}0&2&-3\\-2&0&1\\3&-1&0\end{bmatrix}$
$=\begin{bmatrix}0&1&\frac{-3}{2}\\-1&0&\frac{1}{2}\\\frac{3}{2}&\frac{-1}{2}&0\end{bmatrix}$
$\Rightarrow\ \text{A}=\frac{\text{A}+\text{A}'}{2}+\frac{\text{A}-\text{A}'}{2}$
$=\begin{bmatrix}2&2&\frac{5}{2}\\2&-1&\frac{3}{2}\\\frac{5}{2}&\frac{3}{2}&2\end{bmatrix}+\begin{bmatrix}0&1&\frac{-3}{2}\\-1&0&\frac{1}{2}\\\frac{3}{2}&\frac{-1}{2}&0\end{bmatrix}$
Which is the required expression.
View full question & answer→Question 485 Marks
If $\text{A}=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix},$ show that $A^2 - 7A + 10I_3 = 0$.
AnswerGiven,
$\text{A}=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}$
$ \text{A}^2-7\text{A}+10\text{I}_3$
$=\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}-7\begin{bmatrix}3&2&0\\1&4&0\\0&0&5\end{bmatrix}+10\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$ =\begin{bmatrix}9+2+0&6+8+0&0+0+0\\3+4+0&2+16+0&0+0+0\\0+0+0&0+0+0&0+0+25\end{bmatrix}-\begin{bmatrix}21&14&0\\7&28&0\\0&0&35\end{bmatrix}+\begin{bmatrix}10&0&0\\0&10&0\\0&0&10\end{bmatrix} $
$=\begin{bmatrix}11&14&0\\7&18&0\\0&0&25\end{bmatrix}-\begin{bmatrix}21&14&0\\7&28&0\\0&0&35\end{bmatrix}+\begin{bmatrix}10&0&0\\0&10&0\\0&0&10\end{bmatrix}$
$=\begin{bmatrix}11-21+10&14-14+0&0-0+0\\7-7+0&18-28+10&0-0+0\\0-0+0&0-0+0&25-35+10\end{bmatrix}$
$ =\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}=0$
Hence,
$ \text{A}^2-7\text{A}+10\text{I}_3=0$
View full question & answer→Question 495 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 $A^2 - A + 2I = O$.
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 505 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.
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