Advertisements
Advertisements
Question
Solve the following system of equations by matrix method:
x − y + z = 2
2x − y = 0
2y − z = 1
Advertisements
Solution
Here,
\[A = \begin{bmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{bmatrix}\]
\[\left| A \right| = \begin{vmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{vmatrix}\]
\[ = 1\left( 1 - 0 \right) + 1\left( - 2 - 0 \right) + 1(4 - 0)\]
\[ = 1 - 2 + 4\]
\[ = 3\]
\[ {\text{ Let }C}_{ij} {\text{ be the cofactors of the elements a }}_{ij}\text{ in }A\left[ a_{ij} \right].\text{ Then,}\]
\[ C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 0 \\ 2 & - 1\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 0 \\ 0 & - 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 0 & 2\end{vmatrix} = 4\]
\[ C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 1 \\ 2 & - 1\end{vmatrix} = 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 0 & - 1\end{vmatrix} = - 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 0 & 2\end{vmatrix} = - 2\]
\[ C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 1 \\ - 1 & 0\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 0\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = 1\]
\[adj A = \begin{bmatrix}1 & 2 & 4 \\ 1 & - 1 & - 2 \\ 1 & 2 & 1\end{bmatrix}^T \]
\[ = \begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A\]
\[ = \frac{1}{1}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\]
\[X = A^{- 1} B\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\begin{bmatrix}2 \\ 0 \\ 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}2 + 1 \\ 4 + 2 \\ 8 + 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{1}\begin{bmatrix}3 \\ 6 \\ 9\end{bmatrix}\]
\[ \Rightarrow x = \frac{3}{3}, y = \frac{6}{3}\text{ and }z = \frac{9}{3}\]
\[ \therefore x = 1, y = 2\text{ and }z = 3\]
APPEARS IN
RELATED QUESTIONS
If `|[2x,5],[8,x]|=|[6,-2],[7,3]|`, write the value of x.
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
Solve the system of linear equations using the matrix method.
2x – y = –2
3x + 4y = 3
Solve the system of linear equations using the matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Show that
\[\begin{vmatrix}\sin 10^\circ & - \cos 10^\circ \\ \sin 80^\circ & \cos 80^\circ\end{vmatrix} = 1\]
Evaluate
\[∆ = \begin{vmatrix}0 & \sin \alpha & - \cos \alpha \\ - \sin \alpha & 0 & \sin \beta \\ \cos \alpha & - \sin \beta & 0\end{vmatrix}\]
\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]
Prove that:
`[(a, b, c),(a - b, b - c, c - a),(b + c, c + a, a + b)] = a^3 + b^3 + c^3 -3abc`
Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]
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
Prove that :
5x + 7y = − 2
4x + 6y = − 3
x − 4y − z = 11
2x − 5y + 2z = 39
− 3x + 2y + z = 1
3x − y + 2z = 6
2x − y + z = 2
3x + 6y + 5z = 20.
x + 2y = 5
3x + 6y = 15
x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10
For what value of x, the following matrix is singular?
If the matrix \[\begin{bmatrix}5x & 2 \\ - 10 & 1\end{bmatrix}\] is singular, 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.
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 a, b, c are distinct, then the value of x satisfying \[\begin{vmatrix}0 & x^2 - a & x^3 - b \\ x^2 + a & 0 & x^2 + c \\ x^4 + b & x - c & 0\end{vmatrix} = 0\text{ is }\]
Solve the following system of equations by matrix method:
6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10
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
If \[A = \begin{bmatrix}1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1\end{bmatrix}\] , find A−1. Using A−1, solve the system of linear equations x − 2y = 10, 2x − y − z = 8, −2y + z = 7
If \[\begin{bmatrix}1 & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ - 1 \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\] , find x, y and z.
Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if \[\begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}\]= 0, then the system has
Find the inverse of the following matrix, using elementary transformations:
`A= [[2 , 3 , 1 ],[2 , 4 , 1],[3 , 7 ,2]]`
The value of x, y, z for the following system of equations x + y + z = 6, x − y+ 2z = 5, 2x + y − z = 1 are ______
Three chairs and two tables cost ₹ 1850. Five chairs and three tables cost ₹2850. Find the cost of four chairs and one table by using matrices
`abs (("a"^2, 2"ab", "b"^2),("b"^2, "a"^2, 2"ab"),(2"ab", "b"^2, "a"^2))` is equal to ____________.
If `alpha, beta, gamma` are in A.P., then `abs (("x" - 3, "x" - 4, "x" - alpha),("x" - 2, "x" - 3, "x" - beta),("x" - 1, "x" - 2, "x" - gamma)) =` ____________.
`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`
The number of values of k for which the linear equations 4x + ky + 2z = 0, kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is
If `|(x + a, beta, y),(a, x + beta, y),(a, beta, x + y)|` = 0, then 'x' is equal to
