Advertisements
Advertisements
प्रश्न
Solve the system of linear equations using the matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
उत्तर
`[(2,3,3),(1,-2,1),(3,-1,-2)] [(x),(y),(z)] = [(5),(-4),(3)]` AX = B
A = `[(2,3,3),(1,-2,1),(3,-1,-2)], X = [(x),(y),(z)]` or B = `[(5),(-4),(3)]`
Now, `abs A = [(2,3,3),(1,-2,1),(3,-1,-2)]`
`= 2 (4 + 1) - 3 (-2 - 3) + 3 (-1 + 6)`
`= 10 + 15 + 15 = 40 ne 0`
`therefore A^-1` exists and hence the given equation has a unique solution.
`A_11 = (-1)^(1 + 1) abs ((-2,1),(-1,-2)) = 4+1 = 5`
`A_12 = (-1)^(1 + 2) abs ((1,1),(3,-2)) = (-2 - 3) = 5`
`A_13 = (-1)^(1 + 3) abs ((1,-2),(3,-1)) = -1 + 6 = 5`
`A_21 = (-1)^(2 + 1) abs ((3,3),(-1,-2)) = -(-6 + 3) = 3`
`A_22 = (-1)^(2 + 2) abs ((2,3),(3,-2)) = -4 - 9 = -13`
`A_23 = (-1)^(2 + 3) abs((2,3),(3,-1)) = -(-2 - 9) = 11`
`A_31 = (-1)^ (3 + 1) abs ((3,3),(-2,1)) = 3 + 6 = 9`
`A_32 = (-1)^(3 + 2) abs ((2,3),(1,1)) = -(2 - 3) = 1`
`A_33 = (-1)^(3 + 3) abs ((2,3),(1, -2)) = -4-3 = -7`
`therefore A^-1 = 1/abs A (Adj A)`
`= 1/40 [(5,5,5),(3,-13,11),(9,1,-7)] = 1/40 [(5,3,9),(5,-13,1),(5,11,-7)]`
`X = A^-1 B`
`=> [(x),(y),(z)] = 1/40 [(5,3,9),(5,-13,1),(5,11,-7)] [(5),(-4),(3)]`
`= 1/40 [(25 - 12 + 27),(25 + 52 + 3),(25 - 44 - 21)]`
`= 1/40 [(40),(80),(-40)] = [(1),(2),(-1)]`
so, x = `1, y =2, or z = -1.`
APPEARS IN
संबंधित प्रश्न
Find the value of x, if
\[\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]
\[\begin{vmatrix}- a \left( b^2 + c^2 - a^2 \right) & 2 b^3 & 2 c^3 \\ 2 a^3 & - b \left( c^2 + a^2 - b^2 \right) & 2 c^3 \\ 2 a^3 & 2 b^3 & - c \left( a^2 + b^2 - c^2 \right)\end{vmatrix} = abc \left( a^2 + b^2 + c^2 \right)^3\]
If \[\begin{vmatrix}a & b - y & c - z \\ a - x & b & c - z \\ a - x & b - y & c\end{vmatrix} =\] 0, then using properties of determinants, find the value of \[\frac{a}{x} + \frac{b}{y} + \frac{c}{z}\] , where \[x, y, z \neq\] 0
Using determinants show that the following points are collinear:
(5, 5), (−5, 1) and (10, 7)
Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).
If the points (3, −2), (x, 2), (8, 8) are collinear, find x using determinant.
Prove that :
Prove that :
Prove that :
Prove that :
2x − y = 17
3x + 5y = 6
Given: x + 2y = 1
3x + y = 4
x+ y = 5
y + z = 3
x + z = 4
3x − y + 2z = 3
2x + y + 3z = 5
x − 2y − z = 1
2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2
Find the value of the determinant
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]
Write the value of \[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix} .\]
Write the cofactor of a12 in the following matrix \[\begin{bmatrix}2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7\end{bmatrix} .\]
If |A| = 2, where A is 2 × 2 matrix, find |adj A|.
If \[A = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}\]. Write the cofactor of the element a32.
If \[\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}\] , find the value of x.
If x ∈ N and \[\begin{vmatrix}x + 3 & - 2 \\ - 3x & 2x\end{vmatrix}\] = 8, then find the value of x.
The value of \[\begin{vmatrix}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{vmatrix}\]
The value of the determinant
If x, y, z are different from zero and \[\begin{vmatrix}1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z\end{vmatrix} = 0\] , then the value of x−1 + y−1 + z−1 is
Solve the following system of equations by matrix method:
3x + y = 7
5x + 3y = 12
Solve the following system of equations by matrix method:
6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10
Solve the following system of equations by matrix method:
3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0
Solve the following system of equations by matrix method:
x − y + z = 2
2x − y = 0
2y − z = 1
Solve the following system of equations by matrix method:
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
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.
A company produces three products every day. Their production on a certain day is 45 tons. It is found that the production of third product exceeds the production of first product by 8 tons while the total production of first and third product is twice the production of second product. Determine the production level of each product using matrix method.
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.
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.
`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`
Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:
If the system of equations x + λy + 2 = 0, λx + y – 2 = 0, λx + λy + 3 = 0 is consistent, then
Let P = `[(-30, 20, 56),(90, 140, 112),(120, 60, 14)]` and A = `[(2, 7, ω^2),(-1, -ω, 1),(0, -ω, -ω + 1)]` where ω = `(-1 + isqrt(3))/2`, and I3 be the identity matrix of order 3. If the determinant of the matrix (P–1AP – I3)2 is αω2, then the value of α is equal to ______.
