Advertisements
Advertisements
प्रश्न
The prices of three commodities P, Q and R are Rs x, y and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R. Cpurchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using matrix method.
Advertisements
उत्तर
According to the question,
\[3x + 5y - 4z = 6000\]
\[2x - 3y + z = 5000\]
\[ - x + 4y + 6z = 13000\]
The given system of equations can be written in matrix form as follows:
\[ \begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[AX = B\]
Here,
\[A = \begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix} X = \begin{bmatrix}x \\ y \\ z\end{bmatrix} B = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
Now,
\[\left| A \right|=3 \left( - 18 - 4 \right) - 5\left( 12 + 1 \right) - 4\left( 8 - 3 \right)\]
\[ = - 66 - 65 - 20\]
\[ = - 151 \neq 0\]
\[\text{ So, }A^{- 1}\text{ exists .} \]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A=\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 3 & 1 \\ 4 & 6\end{vmatrix} = - 22, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ - 1 & 6\end{vmatrix} = - 13, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 3 \\ - 1 & 4\end{vmatrix} = 5\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}5 & - 4 \\ 4 & 6\end{vmatrix} = - 46, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 4 \\ - 1 & 6\end{vmatrix} = 14, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 5 \\ - 1 & 4\end{vmatrix} = - 17\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}5 & - 4 \\ - 3 & 1\end{vmatrix} = - 7, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 4 \\ 2 & 1\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 5 \\ 2 & - 3\end{vmatrix} = - 19\]
\[adj A = \begin{bmatrix}- 22 & - 13 & 5 \\ - 46 & 14 & - 17 \\ - 7 & - 11 & - 19\end{bmatrix}^T \]
\[ = \begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}\]
\[ \Rightarrow X = \frac{1}{- 151}\begin{bmatrix}- 453000 \\ - 151000 \\ - 302000\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3000 \\ 1000 \\ 2000\end{bmatrix}\]
\[ \therefore x = 3000, y = 1000\text{ and }z = 2000\]
APPEARS IN
संबंधित प्रश्न
Solve the system of linear equations using the matrix method.
5x + 2y = 3
3x + 2y = 5
Evaluate the following determinant:
\[\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a & b & c \\ a + 2x & b + 2y & c + 2z \\ x & y & z\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\]
\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]
Prove the following identities:
\[\begin{vmatrix}y + z & z & y \\ z & z + x & x \\ y & x & x + y\end{vmatrix} = 4xyz\]
Show that x = 2 is a root of the equation
Using determinants show that the following points are collinear:
(3, −2), (8, 8) and (5, 2)
Prove that :
3x + y = 19
3x − y = 23
2x + 3y = 10
x + 6y = 4
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
6x + y − 3z = 5
x + 3y − 2z = 5
2x + y + 4z = 8
x − y + z = 3
2x + y − z = 2
− x − 2y + 2z = 1
Write the value of the determinant
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]
Write the value of the determinant
Write the value of \[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix} .\]
The value of the determinant
If \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}
\[\begin{vmatrix}\log_3 512 & \log_4 3 \\ \log_3 8 & \log_4 9\end{vmatrix} \times \begin{vmatrix}\log_2 3 & \log_8 3 \\ \log_3 4 & \log_3 4\end{vmatrix}\]
The determinant \[\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ca & c - a & ab - a^2\end{vmatrix}\]
Solve the following system of equations by matrix method:
5x + 2y = 3
3x + 2y = 5
Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
Show that each one of the following systems of linear equation is inconsistent:
4x − 5y − 2z = 2
5x − 4y + 2z = −2
2x + 2y + 8z = −1
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 the three respectively 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 three values as before). If the total amount of awards for one prize on 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.
Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹2,200. School Q wants to spend ₹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.
A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.
Three chairs and two tables cost ₹ 1850. Five chairs and three tables cost ₹2850. Find the cost of four chairs and one table by using matrices
If the system of equations x + ky - z = 0, 3x - ky - z = 0 & x - 3y + z = 0 has non-zero solution, then k is equal to ____________.
In system of equations, if inverse of matrix of coefficients A is multiplied by right side constant B vector then resultant will be?
If `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to
Let A = `[(i, -i),(-i, i)], i = sqrt(-1)`. Then, the system of linear equations `A^8[(x),(y)] = [(8),(64)]` has ______.
