Advertisements
Advertisements
Question
Solve the following by inversion method 2x + y = 5, 3x + 5y = −3
Solution
Matrix form of the given system of equations is
`[(2, 1),(3, 5)] [(x), (y)] = [(5),(-3)]`
This is of the form AX = B,
where A = `[(2, 1),(3, 5)]`, X = `[(x),(y)]` and B = `[(5),(-3)]`
To determine X, we have to find A−1
|A| = `[(2, 1),(3, 5)]`
= 10 − 3
= 7 ≠ 0
∴ A−1 exists.
Consider AA−1 = I
∴ `[(2, 1),(3, 5)]` A−1 = `[(1, 0),(0, 1)]`
Applying R2 → 2R2 − 3R1, we get
`[(2, 1),(0, 7)]` A−1 = `[(1, 0),(-3, 2)]`
Applying R1 → 7R1 – R2, we get
`[(14, 0),(0, 7)]` A−1 = `[(10, -2),(-3, 2)]`
Applying R1 → `(1/14)` R1 and R2 → `(1/7)` R2, we get
`[(1, 0),(0, 1)]` A−1 = `[(10/14, (-2)/14),((-3)/7, 2/7)]`
∴ A−1 = `1/7[(5, -1),(-3, 2)]`
Pre-multiplying AX = B by A−1, we get
A−1(AX) = A−1B
∴ (A−1A)X = A−1B
∴ IX = A−1B
∴ X = A−1B
∴ X = `1/7[(5, -1),(-3, 2)] [(5),(-3)]`
∴ `[(x),(y)] = 1/7[(25 + 3),(-15 - 6)]`
= `1/7[(28),(-21)]`
= `[(4),(-3)]`
∴ By equality of matrices, we get
x = 4, y = – 3
RELATED QUESTIONS
Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
Evaluate
\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.
Find the value of x, if
\[\begin{vmatrix}x + 1 & x - 1 \\ x - 3 & x + 2\end{vmatrix} = \begin{vmatrix}4 & - 1 \\ 1 & 3\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & - 3 & 2 \\ 4 & - 1 & 2 \\ 3 & 5 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\left( 2^x + 2^{- x} \right)^2 & \left( 2^x - 2^{- x} \right)^2 & 1 \\ \left( 3^x + 3^{- x} \right)^2 & \left( 3^x - 3^{- x} \right)^2 & 1 \\ \left( 4^x + 4^{- x} \right)^2 & \left( 4^x - 4^{- x} \right)^2 & 1\end{vmatrix}\]
Using properties of determinants prove that
\[\begin{vmatrix}x + 4 & 2x & 2x \\ 2x & x + 4 & 2x \\ 2x & 2x & x + 4\end{vmatrix} = \left( 5x + 4 \right) \left( 4 - x \right)^2\]
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:
If a, b, c are real numbers such that
\[\begin{vmatrix}b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a\end{vmatrix} = 0\] , then show that either
\[a + b + c = 0 \text{ or, } a = b = c\]
If the points (x, −2), (5, 2), (8, 8) are collinear, find x using determinants.
Using determinants, find the equation of the line joining the points
(3, 1) and (9, 3)
Prove that :
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1
x+ y = 5
y + z = 3
x + z = 4
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
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.
If a, b, c are non-zero real numbers and if the system of equations
(a − 1) x = y + z
(b − 1) y = z + x
(c − 1) z = x + y
has a non-trivial solution, then prove that ab + bc + ca = abc.
For what value of x, the following matrix is singular?
Let \[\begin{vmatrix}x^2 + 3x & x - 1 & x + 3 \\ x + 1 & - 2x & x - 4 \\ x - 3 & x + 4 & 3x\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\]
be an identity in x, where a, b, c, d, e are independent of x. Then the value of e is
The value of the determinant
Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12
Solve the following system of equations by matrix method:
x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9
Solve the following system of equations by matrix method:
2x + 6y = 2
3x − z = −8
2x − y + z = −3
Solve the following system of equations by matrix method:
8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5
Show that each one of the following systems of linear equation is inconsistent:
x + y − 2z = 5
x − 2y + z = −2
−2x + y + z = 4
The sum of three numbers is 2. If twice the second number is added to the sum of first and third, the sum is 1. By adding second and third number to five times the first number, we get 6. Find the three numbers by using matrices.
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.
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.
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A 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 ₹1,600. School B wants to spend ₹2,300 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is ₹900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.
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.
2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13
If A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A
Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.
`abs ((("b" + "c"^2), "a"^2, "bc"),(("c" + "a"^2), "b"^2, "ca"),(("a" + "b"^2), "c"^2, "ab")) =` ____________.
If c < 1 and the system of equations x + y – 1 = 0, 2x – y – c = 0 and – bx+ 3by – c = 0 is consistent, then the possible real values of b are
For what value of p, is the system of equations:
p3x + (p + 1)3y = (p + 2)3
px + (p + 1)y = p + 2
x + y = 1
consistent?
If `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to
If the system of linear equations
2x + y – z = 7
x – 3y + 2z = 1
x + 4y + δz = k, where δ, k ∈ R has infinitely many solutions, then δ + k is equal to ______.
If the following equations
x + y – 3 = 0
(1 + λ)x + (2 + λ)y – 8 = 0
x – (1 + λ)y + (2 + λ) = 0
are consistent then the value of λ can be ______.
