Advertisements
Advertisements
Question
Solve the system of linear equations using the matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Advertisements
Solution
`[(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, |A| = `|(2,3,3),(1,-2,1),(3,-1,-2)|`
= 2(4 + 1) − 3(−2 − 3) + 3(−1 + 6)
= 10 + 15 + 15
= 40 ≠ 0
∴ A−1 exists and hence the given equation has a unique solution.
A11 = `(-1)^(1 + 1) |(-2,1),(-1,-2)|`
= 4 + 1
= 5
A12 = `(-1)^(1 + 2) |(1,1),(3,-2)|`
= (−2 − 3)
= 5
A13 = `(-1)^(1 + 3) |(1,-2),(3,-1)|`
= −1 + 6
= 5
A21 = `(-1)^(2 + 1) |(3,3),(-1,-2)|`
= −(−6 + 3)
= 3
A22 = `(-1)^(2 + 2) |(2,3),(3,-2)|`
= −4 − 9
= −13
A23 = `(-1)^(2 + 3) |(2,3),(3,-1)|`
= −(−2 − 9)
= 11
A31 = `(-1)^ (3 + 1) |(3,3),(-2,1)|`
= 3 + 6
= 9
A32 = `(-1)^(3 + 2) |(2,3),(1,1)|`
= −(2 − 3)
= 1
A33 = `(-1)^(3 + 3) |(2,3),(1, -2)|`
= −4 − 3
= −7
∴ A−1 = `1/|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−1B
⇒ `[(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)]`
∴ x = 1, y = 2 and z = −1
APPEARS IN
RELATED QUESTIONS
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
For what value of x the matrix A is singular?
\[ A = \begin{bmatrix}1 + x & 7 \\ 3 - x & 8\end{bmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sqrt{23} + \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} + \sqrt{46} & 5 & \sqrt{10} \\ 3 + \sqrt{115} & \sqrt{15} & 5\end{vmatrix}\]
Evaluate the following:
\[\begin{vmatrix}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{vmatrix}\]
\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]
Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]
\[\begin{vmatrix}1 + a & 1 & 1 \\ 1 & 1 + a & a \\ 1 & 1 & 1 + a\end{vmatrix} = a^3 + 3 a^2\]
Prove the following identity:
\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\]
Show that
Using determinants show that the following points are collinear:
(3, −2), (8, 8) and (5, 2)
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 area of the triangle with vertices (−3, 5), (3, −6), (7, 2).
Find values of k, if area of triangle is 4 square units whose vertices are
(−2, 0), (0, 4), (0, k)
x − 2y = 4
−3x + 5y = −7
Prove that :
Prove that :
Prove that
2x + y − 2z = 4
x − 2y + z = − 2
5x − 5y + z = − 2
Find the value of the determinant
\[\begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}\]
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
If ω is a non-real cube root of unity and n is not a multiple of 3, then \[∆ = \begin{vmatrix}1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1\end{vmatrix}\]
The value of \[\begin{vmatrix}5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^6\end{vmatrix}\]
Solve the following system of equations by matrix method:
5x + 2y = 3
3x + 2y = 5
Solve the following system of equations by matrix method:
2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6
Show that the following systems of linear equations is consistent and also find their solutions:
x − y + z = 3
2x + y − z = 2
−x −2y + 2z = 1
Show that each one of the following systems of linear equation is inconsistent:
4x − 2y = 3
6x − 3y = 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
Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17
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
3x + y − 2z = 0
x + y + z = 0
x − 2y + z = 0
If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\], find x, y and z.
If A = `[[1,1,1],[0,1,3],[1,-2,1]]` , find A-1Hence, solve the system of equations:
x +y + z = 6
y + 3z = 11
and x -2y +z = 0
If ` abs((1 + "a"^2 "x", (1 + "b"^2)"x", (1 + "c"^2)"x"),((1 + "a"^2) "x", 1 + "b"^2 "x", (1 + "c"^2) "x"), ((1 + "a"^2) "x", (1 + "b"^2) "x", 1 + "c"^2 "x"))`, then f(x) is apolynomial of degree ____________.
If A = `[(1,-1,0),(2,3,4),(0,1,2)]` and B = `[(2,2,-4),(-4,2,-4),(2,-1,5)]`, then:
