Question 14 Marks
The cost of $4$ pencils, $3$ pens and $2$ erasers is $₹ 60$ . The cost of $2$ pencils, $4$ pens and $6$ erasers is $₹ 90$ . Whereas the cost of $6$ pencils, $2$ pens and $3$ erasers is $₹ 70$ . Find the cost of each item by using matrix inversion method.
AnswerLet the cost of one pencil, one pen and one eraser be $x, y$ and $z$ respectively.
From the given conditions:
$4x + 3y + 2z = 60 ....(1)$
$2x + 4y + 6z = 90$
$i.e. x + 2y + 3z = 45 ....(2)$
$6x + 2y + 3z = 70 ....(3)$
The matrix form of the equations is
$\left[\begin{array}{lll}4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}60 \\ 45 \\ 70\end{array}\right]$
This is in the form of $A X=B$,
where $A=\left[\begin{array}{lll}4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right], B=\left[\begin{array}{l}60 \\ 45 \\ 70\end{array}\right]$
Now, $| A |=\left|\begin{array}{lll}4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3\end{array}\right|$
$=4(6-6)-3(3-18)+2(2+12)$
$=25 \neq 0$
$\therefore A ^{-1}$ exists.
Consider $AX = B$
$\therefore \quad A ^{-1} AX = A ^{-1} B$,
$\therefore \quad$ IX $= A ^{-1} B$
$\therefore \quad X = A ^{-1} B...(4)$
We write : $AA ^{-1}= I$
$\therefore\left[\begin{array}{lll}4 & 3 & 2 \\ 1 & 2 & 3 \\ 6 & 2 & 3\end{array}\right] A ^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Using : $R_1 \leftrightarrow R_2$
$\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 3 & 2 \\ 6 & 2 & 3\end{array}\right] A^{-1}=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $R_2 \rightarrow R_2-4 R_1, R_3 \rightarrow R_3-6 R_1$
$\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & -5 & -10 \\ 0 & -10 & -15\end{array}\right] A^{-1}=\left[\begin{array}{ccc}0 & 1 & 0 \\ 1 & -4 & 0 \\ 0 & -6 & 1\end{array}\right]$
By $R_2 \rightarrow-\frac{1}{5} R_2, R_3 \rightarrow-\frac{1}{5} R_3$
$\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 2 & 3\end{array}\right] A^{-1}=\left[\begin{array}{ccc}0 & 1 & 0 \\ -1 / 5 & 4 / 5 & 0 \\ 0 & 6 / 5 & -1 / 5\end{array}\right]$
By $R_1 \rightarrow R_1-2 R_2, R_3 \rightarrow R_3-2 R_2$
$\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & -1\end{array}\right] A^{-1}=\left[\begin{array}{ccc}2 / 5 & -3 / 5 & 0 \\ -1 / 5 & 4 / 5 & 0 \\ 2 / 5 & -2 / 5 & -1 / 5\end{array}\right]$
By $R_3 \rightarrow-R_3$
$\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ccc}2 / 5 & -3 / 5 & 0 \\ -1 / 5 & 4 / 5 & 0 \\ -2 / 5 & 2 / 5 & 1 / 5\end{array}\right]$
By $R_1 \rightarrow R_1+R_3, R_2 \rightarrow R_2-2 R_3$
$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ccc}0 & -1 / 5 & 1 / 5 \\ 3 / 5 & 0 & -2 / 5 \\ -2 / 5 & 2 / 5 & 1 / 5\end{array}\right]$
$\therefore A^{-1}=\left|\begin{array}{ccc}0 & -1 / 5 & 1 / 5 \\ 3 / 5 & 0 & -2 / 5 \\ -2 / 5 & 2 / 5 & 1 / 5\end{array}\right|$
$\therefore$ From (4), $\quad X=A^{-1} B$
$\therefore\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\left.\begin{array}{ccc}0 & -1 / 5 & 1 / 5 \\ 3 / 5 & 0 & -2 / 5 \\ -2 / 5 & 2 / 5 & 1 / 5\end{array} \right\rvert\,\left[\begin{array}{l}60 \\ 45 \\ 70\end{array}\right]\right.$
$=\left[\left.\begin{array}{c}-9+14 \\ 36-28 \\ -24+18+14\end{array} \right\rvert\,\right.$
$=\left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$
Cost of one pencil $= ₹5$
Cost of one pen $= ₹8$
Cost of eraser $= ₹8$
View full question & answer→Question 24 Marks
Find the inverse of the matrix, $A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$ by using column transformations.
Answer$A=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$
$|A|=\left|\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right|$
$=1(16-9)-3(4-3)+3(3-4)$
$=1 \neq 0$
$A^{-1}$ Exists
consider $A^{-1}A=I$
$A^{-1}\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Applying $C_2 \rightarrow C_2 -3C_1$ and $C_3\rightarrow C_3 - 3C_1,$
$A^{-1}\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -3 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Applying $C_1\rightarrow C_1 - C_2,$
$A^{-1}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}4 & -3 & -3 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Applying $C_3 \rightarrow C_3 - C_1$
$A^{-1}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
View full question & answer→Question 34 Marks
The cost of $2$ books, $6$ notebooks and $3$ pens is $₹ 40$ . The cost of $3$ books, $4$ notebooks and $2$ pens is $₹ 35$ , while the cost of $5$ books, $7$ notebooks and $4$ pens is $₹ 61$ . Using this information and matrix method, find the cost of $1$ book, $1$ notebook and $1$ pen separately.
AnswerLet the cost of $1$ book, $1$ notebook and $1$ pen be $Rs. x, Rs. y$ and $Rs. z$ respectively.
According to the given conditions,
$2x + 6y + 3z = 40$
$3x + 4y + 2z = 35$
$5x + 7y + 4z = 61$
These equations can be written in the matrix form as
$\left[\begin{array}{lll}2 & 6 & 3 \\ 3 & 4 & 2 \\ 5 & 7 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}40 \\ 35 \\ 61\end{array}\right]$
Applying $R _1 \rightarrow R _1- R _2$
$\left[\begin{array}{ccc}-1 & 2 & 1 \\ 3 & 4 & 2 \\ 5 & 7 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}5 \\ 35 \\ 61\end{array}\right]$
Applying $R_1 \rightarrow(-1) R_1$,
$\left[\begin{array}{ccc}1 & -2 & -1 \\ 3 & 4 & 2 \\ 5 & 7 & 4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-5 \\ 35 \\ 61\end{array}\right]$
Applying $R_2 \rightarrow R_2-3 R_1$ and $R_3 \rightarrow R_3-5 R_1$,
$\left[\begin{array}{ccc}1 & -2 & -1 \\ 0 & 10 & 5 \\ 0 & 17 & 9\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-5 \\ 50 \\ 86\end{array}\right]$
Applying $R_2 \rightarrow\left(\frac{1}{10}\right) R_2$,
$\left[\begin{array}{ccc}1 & -2 & -1 \\ 0 & 1 & \frac{1}{2} \\ 0 & 17 & 9\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-5 \\ 5 \\ 86\end{array}\right]$
Applying $R_3 \rightarrow R_3-17 R_2$,
$\left[\begin{array}{ccc}1 & -2 & -1 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}-5 \\ 5 \\ 1\end{array}\right]$
By equality of matrices,
$x − 2y − z = −5 ….(i)$
$y + z/2= 5 ….(ii)$
$z/2 = 1 ….(iii)$
From $(iii), z = 2$
Putting $z = 2$ in $(ii)$, we get
$y + 1 = 5$
$\therefore y = 4$
Putting $y = 4, z = 2$ in $(i)$, we get
$x − 8 − 2 = − 5$
$\therefore x = 5$
Thus, the cost of $1$ book, $1$ notebook and $1$ pen are $5, 4$ and $2$ respectively
View full question & answer→Question 44 Marks
Solve the following system of equations by the method of inversion. $x-y+z=4,2 x+y-3 z=0, x+y+z=2$
Answer$x - y + z = 4,$
$2x + y - 3z = 0,$
$x+y+z=2$
The matrix form of given equations is
$\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
i.e. $AX = B,$ where $A = \left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
$B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
Now $AX = B $
$\therefore(A^{-1}A)X = A-B,$
$\therefore IX = A^{-1}B$
$\therefore X = A^{-1}B ....(1)$
We write: $AA^{-1} = \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\therefore\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
By: $R_2 \rightarrow R_2+2 R_1, R_3 \rightarrow R_3-R_1$
$\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0\end{array}\right] A^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ -2 & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$
By: $R _2 \rightarrow R _2- R _3$
$\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 1 & -5 \\ 0 & 2 & 0\end{array}\right] A^{-1}=\left[\begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & -1 \\ -1 & 0 & 1\end{array}\right]$
By: $R_1 \rightarrow R_1+R_2, R_3 \rightarrow R_3-2 R_2$
$\left[\begin{array}{lll}1 & 0 & -4 \\ 0 & 1 & -5 \\ 0 & 0 & 10\end{array}\right] A^{-1}=\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 1 & -1 \\ 1 & -2 & 3\end{array}\right]$
By $R_3 \rightarrow \frac{1}{10} R_3$
$\left[\begin{array}{ccc}1 & 0 & -4 \\ 0 & 1 & -5 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 1 & -1 \\ \frac{1}{10} & -\frac{2}{10} & \frac{3}{10}\end{array}\right]$
By $R_1 \rightarrow R_1+4 R_3, R_2 \rightarrow R_2+5 R_3$
$\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A ^{-1}=\left[\begin{array}{ccc}\frac{4}{10} & \frac{2}{10} & \frac{2}{10} \\ -\frac{5}{10} & 0 & \frac{5}{10} \\ \frac{1}{10} & -\frac{2}{10} & \frac{3}{10}\end{array}\right]$
$A^{-1}=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$\therefore$ From $(1)$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
$=\frac{1}{10}\left[\begin{array}{c}20 \\ -10 \\ 10\end{array}\right]=\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]$
$\therefore x = 2, y = -1, z = 1$
View full question & answer→Question 54 Marks
If three numbers are added, their sum is $2.$ If two times the second number is subtracted from the sum of first and third numbers we get $8$ and if three times the first number is added to the sum of second and third numbers we get $4.$ Find the numbers using matrices.
AnswerLet the three numbers $\text{x , y , z.}$
From given condition, we have
$x + y + z = 2 .......(1)$
$x + z - 2y = 8$
$x - 2y + z = 8 ......(2)$
And
$3x + y + z = 4 .....(3)$
Given all equation can be written in matrix form as ,
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & -2 & 1 \\ 3 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 8 \\ 4\end{array}\right]$
Consider , $AX = B$
On multiplying $A^{-1}$ both sides , we get
$X = A^{-1} . B ......(4)$
Now
$| A |=\left|\begin{array}{ccc}1 & 1 & 1 \\ 0 & -3 & 0 \\ 0 & -2 & -2\end{array}\right|\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}2 \\ 6 \\ -2\end{array}\right]$
$\left|\begin{array}{ccc}x+ & y+ & z \\ 0- & 3 y & +0 \\ 0 & -2 & -2\end{array}\right|=\left[\begin{array}{c}2 \\ 6 \\ -2\end{array}\right]$
By equality of matrices,
$x + y + z = 2 ……(1)$
$-3y = 6 ……(2)$
$– 2y – 2z = -2 ……..(3)$
From $(2), y = -2$
Substituting $y = -2$ in $(3),$ we get,
$-2(-2) – 2z$
$= -2$
$\therefore -2z = -6$
$\therefore z = 3$
Substituting $y = -2, z = 3$ in $(1),$ we get,
$x – 2 + 3$
$= 2$
$\therefore x = 1$
Hence, the required numbers are $1, -2$ and $3.$
View full question & answer→Question 64 Marks
Find the inverse of the matrix,
$A=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$ using elementary row transformations.
Answer$A=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$
= 1[3] - 2[-1] - 2[2]
$=3+2-4=1 \neq 0 \Rightarrow A^{-1}$ exist
we know
$A A^{-1}=I$
$\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$R_2 \rightarrow R_2+R_1$
$\left[\begin{array}{ccc}1 & 2 & -2 \\ 0 & 5 & -2 \\ 0 & -2 & 1\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$R_2 \rightarrow R_2+2 R_3$
$\left[\begin{array}{ccc}1 & 2 & -2 \\ 0 & 1 & 0 \\ 0 & -2 & 1\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 2 \\ 0 & 0 & 1\end{array}\right]$
$R_1 \rightarrow R_1-2 R_2$ and $R_3 \rightarrow R_3+2 R_2$
$\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ccc}-1 & -2 & -4 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$
$R_1 \rightarrow R_1+2 R_3$
$\therefore A^{-1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right]$
View full question & answer→Question 74 Marks
Solve the following equations by method of reduction:
$x-y+z=4,2 x+y-3 z=0, x+y+z=2 \text.$
Answer$x-y + z = 4,$
$2x + y - 3z = 0,$
$x + y + z = 2$
$\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
$R_2-2 R_1$ and $R_3-R_1$
$\left[\begin{array}{ccc}1 & -1 & 1 \\ 0 & 3 & -5 \\ 0 & 2 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}4 \\ -8 \\ -2\end{array}\right]$
$x-y+z=4 \quad \ldots .(1)$
$3 y-5 z=-8 \ldots(2)$
$2 y=-2...(3)$
$y=-1$
By equation $(2)$
$-3-5 z=-8$
$-5 z=-5$
$z=1$
By equation $(1)$
$x+1+1=4$
$x=2$
Ans : $x=2, y=-1, z=1$
View full question & answer→MCQ 84 Marks
The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is $₹ 60$. The cost of 2 dozen pencils, 4 dozen pen and 6 dozen erasers is ₹ 90 whereas the cost of 6 dozen pencils, 2 dozen pen and 3 dozen erasers is ₹ 70 . Find the cost of each item per dozen by using matrices.
- A
$\frac{1}{9}\left[\begin{array}{ccc}1 & 4 & -2 \\ -2 & -5 & 4 \\ 1 & -2 & 1\end{array}\right]$
- B
$\left[\begin{array}{ccc}1 & -2 & 1 \\ 4 & -5 & -2 \\ -2 & 4 & 1\end{array}\right]$
- C
$\left[\begin{array}{ccc}1 & 4 & -2 \\ -2 & -5 & 4 \\ 1 & -2 & 1\end{array}\right]$
- D
$\left[\begin{array}{ccc}-1 & -4 & 2 \\ 2 & 5 & -4 \\ 1 & -2 & 1\end{array}\right]$
View full question & answer→Question 94 Marks
The sum of three members is $6$ . When second number is subtracted from thrice the sum of first and we get number $10.$ Four times the sum of third number is subtracted from five times the sum of first and second number, the result is $3$ . Using above information, find these three numbers by matrix method.
AnswerGiven that the sum of three numbers, $x, y$ and $z$ is $6$.
From the given statement, we have,
$x + y + z = 6$
$3 ( x + z ) - y = 10$
$5 ( x + y ) - 4z = 3$
Thus, the system of equations are :
$x + y + z = 6 (i)$
$3x - y + 3z = 10 (ii)$
$5x + 5y - 4z = 3 (iii)$
Let us write the above equations in the matrix form as:
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 3 & -1 & 3 \\ 5 & 5 & -4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}6 \\ 10 \\ 3\end{array}\right]$
$AX = B$
$A =\left[\begin{array}{ccc}1 & 1 & 1 \\ 3 & -1 & 3 \\ 5 & 5 & -4\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right], B =\left[\begin{array}{c}6 \\ 10 \\ 3\end{array}\right]$
Applying $R_{3\rightarrow } R_3- 5R_1$ , we have
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 3 & -1 & 3 \\ 0 & 0 & -9\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}6 \\ 10 \\ -27\end{array}\right]$
Applying $R_{2 \rightarrow } R_{2 }- 3R_1$ , we get
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & -4 & 0 \\ 0 & 0 & -9\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}6 \\ -8 \\ -27\end{array}\right]$
Thus,
$x + y + z = 6 [$From$(i)]$
$- 4y = - 8 (iv)$
$- 9z = - 27 (v)$
From equation $(iv),$ we get
$y = 2$
From equation $(v),$ we get
$z = 3$
Putting the value of $y$ and $z$ in equation
$(i),$ we get $x = 6 - 2 - 3 = 1$
Hence, numbers are $1, 2,$ and $3.$
View full question & answer→Question 104 Marks
Express the following equations in the matrix form and solve them by method of reduction: $2 x-y+z=1, x+2 y+3 z=8,3 x+y-4 z=1$.
AnswerThe matrix form of given equations is
$\left[\begin{array}{ccc}2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 8 \\ 1\end{array}\right]$
$R_1 \leftrightarrow R_2$
$\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & -4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}8 \\ 1 \\ 1\end{array}\right]$
$R_2 \rightarrow R_2+R_1$
$\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 1 & 4 \\ 3 & 1 & -4\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}8 \\ 9 \\ 1\end{array}\right]$
$R_3 \rightarrow R_3-R_2$
$\left[\begin{array}{ccc}1 & 2 & 3 \\ 3 & 1 & 4 \\ 0 & 0 & -8\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}8 \\ 9 \\ -8\end{array}\right]$
$R_2 \rightarrow R_2-3 R_1$
$\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & -5 & -5 \\ 0 & 0 & -8\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}8 \\ -15 \\ -8\end{array}\right]$
$\left[\begin{array}{c}x+2 y+3 z \\ -5 y-5 z \\ -8 z\end{array}\right]=\left[\begin{array}{c}8 \\ -15 \\ -8\end{array}\right]$
therefore
$x + 2y + 3z = 8 .........(1)$
$-5y -5z = -15 ..... (2)$
$-8z = -8 ......(3)$
From $(3),$
$z = 1$
From $(2),$
$-5y - 5(1) = -15 ... ($because $z = 1)$
$-5y =-10$
$y = 2$
From $(1),$
$x + 2(2)+ 3(1) = 8 ... ($because $z = 1$ and $y = 2)$
$x = 8 -7$
$x = 1$
Thus, $x = 1, y = 2, z = 1$
View full question & answer→Question 114 Marks
The sum of three numbers is $9$ . If we multiply third number by $3$ and add to the second number, we get $16$ . By adding the first and the third number and then subtracting twice the second number from this sum, we get $6$ . Use this information and find the system of linear equations. Hence, find the three numbers using matrices.
AnswerLet the three numbers be $\text{x, y, z}$
From the first condition
$x + y + z = 9$
From the second condition
$y + 3z = 16$
From the third condition
$x – 2y + 2 = 6$
$x + y + z = 9$
$y + 3z = 16$
$x – 2y + z = 6$
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}9 \\ 16 \\ 6\end{array}\right]$
$R_3-R_1$
$\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 3 \\ 0 & -3 & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}9 \\ 16 \\ -3\end{array}\right]$
$x+y+z=9$
$y=3 z=16$
$-3 y=-3 \Rightarrow y=1$
$1+3 z=16$
$z=5$
$x+1+5=9$
$x=3$
$\therefore x=3, y=1, z=5$
View full question & answer→Question 124 Marks
If $A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$ verify that $A(\operatorname{adj} A)=|A| I$.
Answer$A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$
$|A|=\left|\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right|$
$=1(0)+1(9+2)+2(0)$
$=0+11+0$
$\therefore|A|=11$
$A_{11}=(-1)^{1+1} M_{11}=1\left|\begin{array}{cc}0 & -2 \\ 0 & 3\end{array}\right|=1(0-0)=1 \times 0=0$
$A_{12}=(-1)^{1+2} M_{12}=-1\left|\begin{array}{cc}3 & -2 \\ 1 & 3\end{array}\right|=-1(9+2)=-11$
$A_{13}=(-1)^{1+3} M_{13}=1\left|\begin{array}{ll}3 & 0 \\ 1 & 0\end{array}\right|=1(0-0)=1 \times 0=0$
$A_{21}=(-1)^{2+1} M_{21}=-1\left|\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right|=-1(-3-0)=1 \times 0=3$
$A_{22}=(-1)^{2+2} M_{22}=1\left|\begin{array}{ll}1 & 2 \\ 1 & 3\end{array}\right|=1(3-2)=1$
$A_{23}=(-1)^{2+3} M_{23}=-1\left|\begin{array}{cc}1 & -1 \\ 1 & 0\end{array}\right|=-1(0+1)=-1$
$A_{31}=(-1)^{3+1} M_{31}=1\left|\begin{array}{cc}-1 & 2 \\ 0 & -2\end{array}\right|=1(2-0)=1 \times 2=2$
$A_{32}=(-1)^{3+2} M_{32}=-1\left|\begin{array}{cc}1 & 2 \\ 3 & -2\end{array}\right|=-1(-2-6)=8$
$A_{33}=(-1)^{3+3} M_{33}=1\left|\begin{array}{cc}1 & -1 \\ 3 & 0\end{array}\right|=1(0+3)=1 \times 3=3$
Hence, matrix of the co-factors is
$\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]=\left[\begin{array}{ccc}0 & -11 & 0 \\ 3 & 1 & -1 \\ 2 & 8 & 3\end{array}\right]=\left[A_{i j}\right]_{3 \times 3}$
Now, $\operatorname{adj} A=\left[A_{i j}\right]_{3 \times 3}^T=\left[\begin{array}{ccc}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]$
$A(\operatorname{adj} A)=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\left[\begin{array}{ccc}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]$
$=\left[\begin{array}{ccc}0+11+0 & 3-1-2 & 2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 & 3+0-3 & 2+0+9\end{array}\right]$
$=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . .(1)$
$|A| . I=11\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . .(2)$
From equations (1) and (2), we get A(adj A) = |A|. I
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