Advertisements
Advertisements
प्रश्न
Find the inverse of the following matrix, using elementary transformations:
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
Advertisements
उत्तर
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
AA-1 =I
`[[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]] A^(-1) = [[1 , 0 , 0 ],[0 , 1 , 0],[0 , 0 ,1]]`
R2 → R2 - R1
R3 → R3 - R1
`[[2 , 3 , 1 ],[0 , 1 , 0],[1, 4 ,1]] A^(-1) = [[1 , 0 , 0 ],[-1 , 1 , 0],[-1 , 0 ,1]]`
R1 ↔ R3
`[[1 , 4 , 1 ],[0 , 1 , 0],[2 , 3 ,1]] A^(-1) = [[-1 , 0 , 1 ],[-1 , 1 , 0],[1 , 0 ,0]]`
R3 → R3 - 2R1
`[[1 , 4 , 1 ],[0 , 1 , 0],[0 , -5 ,-1]] A^(-1) = [[-1 , 0 , 1 ],[-1 , 1 , 0],[3 , 0 ,-2]]`
R1 → R1 - 4R2
R3 → R3 - 5R2
`[[1 , 0 , 1 ],[0 , 1 , 0],[0 , 0 ,-1]] A^(-1) = [[3 , -4 , 1 ],[-1 , 1 , 0],[-2 , 5 ,-2]]`
`R_1 ->R_1 +R_3`
`[[1 , 0 , 0 ],[0 , 1 , 0],[0 , 0 ,-1]] A^(-1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[-2 , 5 ,-2]]`
`R_3 -> -R_3`
`[[1 , 0 , 0 ],[0 , 1 , 0],[0 , 0 ,1]] A^(-1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[2 , -5 ,2]]`
`I .A^(1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[2 , -5 ,2]]`
` ⇒A^(1) = [[1 , 1 , -1 ],[-1 , 1 , 0],[2 , -5 ,2]]`
संबंधित प्रश्न
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Solve the system of linear equations using the matrix method.
5x + 2y = 4
7x + 3y = 5
Solve the system of linear equations using the matrix method.
5x + 2y = 3
3x + 2y = 5
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method.
Evaluate
\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}^2 .\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3\end{vmatrix}\]
Without expanding, prove that
\[\begin{vmatrix}a & b & c \\ x & y & z \\ p & q & r\end{vmatrix} = \begin{vmatrix}x & y & z \\ p & q & r \\ a & b & c\end{vmatrix} = \begin{vmatrix}y & b & q \\ x & a & p \\ z & c & r\end{vmatrix}\]
Solve the following determinant equation:
Solve the following determinant equation:
Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?
Using determinants, find the value of k so that the points (k, 2 − 2 k), (−k + 1, 2k) and (−4 − k, 6 − 2k) may be collinear.
An automobile company uses three types of steel S1, S2 and S3 for producing three types of cars C1, C2and C3. Steel requirements (in tons) for each type of cars are given below :
| Cars C1 |
C2 | C3 | |
| Steel S1 | 2 | 3 | 4 |
| S2 | 1 | 1 | 2 |
| S3 | 3 | 2 | 1 |
Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.
Solve each of the following system of homogeneous linear equations.
2x + 3y + 4z = 0
x + y + z = 0
2x − y + 3z = 0
Find the real values of λ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions
\[2 \lambda x - 2y + 3z = 0\]
\[ x + \lambda y + 2z = 0\]
\[ 2x + \lambda z = 0\]
For what value of x, the following matrix is singular?
Find the value of the determinant \[\begin{vmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{vmatrix}\].
Let \[f\left( x \right) = \begin{vmatrix}\cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x\end{vmatrix}\] \[\lim_{x \to 0} \frac{f\left( x \right)}{x^2}\] is equal to
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 + z = 6
x + 2z = 7
3x + y + z = 12
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.
Two factories decided to award their employees for three values of (a) adaptable tonew techniques, (b) careful and alert in difficult situations and (c) keeping clam in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ₹ 29000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then
i) represent the above situation by matrix equation and form linear equation using matrix multiplication.
ii) Solve these equation by matrix method.
iii) Which values are reflected in the questions?
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 total amount of ₹7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8% and \[8\frac{1}{2}\] % respectively. The total annual interest from these three accounts is ₹550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.
2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0
x + y = 1
x + z = − 6
x − y − 2z = 3
Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.
The number of real value of 'x satisfying `|(x, 3x + 2, 2x - 1),(2x - 1, 4x, 3x + 1),(7x - 2, 17x + 6, 12x - 1)|` = 0 is
