Advertisements
Advertisements
प्रश्न
Solve the following system of equations by matrix method:
x + y + z = 6
x + 2z = 7
3x + y + z = 12
Advertisements
उत्तर
x + y + z = 6
x + 2z = 7
3x + y + z = 12
AX = B
`A = [(1,1,1),(1,0,2),(3,1,1)], X = [(x),(y),(z)], B = [(6),(7),(12)]`
X = A−1B
`A^-1 = 1/|A| . adj(A)`
`A = [(1,1,1),(1,0,2),(3,1,1)]`
Use cofactor expansion along the first row:
`|A| = 1 . |(0,2),(1,1)|-1. |(1,2),(3,1)| +1.|(1,0),(3,1)|`
Now calculate each 2 × 2 determinant:
`|(0,2),(1,1)| = 0(1)-2(1) = -2`
`|(1,2),(3,1)| = 1(1) -2(3) = 1-6=-5`
`|(1,0),(3,1)| = 1(1) - 0(3) = 1`
∣A∣ = 1(−2)−1 (−5) + 1(1) = −2 + 5 + 1 = 4
We already computed these minors earlier, but here’s the full cofactor matrix:
`Cof(A) = [(-2,5,1),(0,0,-2),(2,-1,-1)]`
adj(A) = Cof(A)T = `[(-2,0,2),(5,0,-1),(1,-2,-1)]`
`A^-1 = 1/4 . [(-2,0,2),(5,0,-1),(1,-2,-1)]`
X = A−1B = `1/4 [(-2,0,2),(5,0,-1),(1,-2,-1)] . [(6),(7),(12)]`
(−2) (6) + (0) (7) + (2) (12) = −12 + 0 + 24 = 12
(5) (6) + (0) (7) + (−1) (12) = 30 + 0 −12 = 18
(1) (6) + (−2) (7) + (−1) (12) = 6 − 14 − 12 = −20
`X = 1/4 . [(12),(18),(-20)] = [(3),(4.5),(-5)]`
x = 2, y = 1, z = 3
APPEARS IN
संबंधित प्रश्न
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Examine the consistency of the system of equations.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solve the system of the following equations:
`2/x+3/y+10/z = 4`
`4/x-6/y + 5/z = 1`
`6/x + 9/y - 20/x = 2`
Evaluate the following determinant:
\[\begin{vmatrix}\cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ\end{vmatrix}\]
\[∆ = \begin{vmatrix}\cos \alpha \cos \beta & \cos \alpha \sin \beta & - \sin \alpha \\ - \sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{vmatrix}\]
Find the value of x, if
\[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & 5 \\ 8 & 3\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^2 \\ 3^2 & 4^2 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 & 7^2\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b + c & a^2 \\ b & c + a & b^2 \\ c & a + b & c^2\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{vmatrix}\]
\[\begin{vmatrix}0 & b^2 a & c^2 a \\ a^2 b & 0 & c^2 b \\ a^2 c & b^2 c & 0\end{vmatrix} = 2 a^3 b^3 c^3\]
\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]
Solve the following determinant equation:
Solve the following determinant equation:
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
Find the value of \[\lambda\] so that the points (1, −5), (−4, 5) and \[\lambda\] are collinear.
Using determinants, find the area of the triangle with vertices (−3, 5), (3, −6), (7, 2).
Prove that :
Prove that :
3x + y = 5
− 6x − 2y = 9
If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.
Write the value of the determinant \[\begin{vmatrix}2 & - 3 & 5 \\ 4 & - 6 & 10 \\ 6 & - 9 & 15\end{vmatrix} .\]
If \[\begin{vmatrix}3x & 7 \\ - 2 & 4\end{vmatrix} = \begin{vmatrix}8 & 7 \\ 6 & 4\end{vmatrix}\] , find the value of x.
If \[∆_1 = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix}, ∆_2 = \begin{vmatrix}1 & bc & a \\ 1 & ca & b \\ 1 & ab & c\end{vmatrix},\text{ then }\]}
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant
The other factor in the value of the determinant is
If ω is a non-real cube root of unity and n is not a multiple of 3, then \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\]
Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3
Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?
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 + 2z = 0
5x + 3y − z = 0
x + 5y − 5z = 0
x + y + z = 0
x − y − 5z = 0
x + 2y + 4z = 0
2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0
Write the value of `|(a-b, b- c, c-a),(b-c, c-a, a-b),(c-a, a-b, b-c)|`
`abs ((("b" + "c"^2), "a"^2, "bc"),(("c" + "a"^2), "b"^2, "ca"),(("a" + "b"^2), "c"^2, "ab")) =` ____________.
If `|(x + 1, x + 2, x + a),(x + 2, x + 3, x + b),(x + 3, x + 4, x + c)|` = 0, then a, b, care in
