Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Suppose, A = \[\begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\]
\[A \times B = \begin{bmatrix}1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4\end{bmatrix}\begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}\]
\[ = \begin{bmatrix}- 2 - 9 + 12 & 0 - 2 + 2 & 1 + 3 - 4 \\ 0 + 18 - 18 & 0 + 4 - 3 & 0 - 6 + 6 \\ - 6 - 18 + 24 & 0 - 4 + 4 & 3 + 6 - 8\end{bmatrix}\]
\[ = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
Since, A × B = I,
Now, the given system of equations is
x + 3z = 9
−x + 2y − 2z = 4
2x − 3y + 4z = −3
This can also be represented as,
Here, we can observe that \[\begin{bmatrix}1 & 0 & 3 \\ - 1 & 2 & - 2 \\ 2 & - 3 & 4\end{bmatrix} = A^T\]
So, \[A^T \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
Multiply the above expression by \[\left( A^T \right)^{- 1}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \left( A^T \right)^{- 1} \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \left( A^{- 1} \right)^T \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix} \]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = B^T \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix} \left[\text{ Using }(1) \right]\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}- 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2\end{bmatrix}^T \begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
\[ = \begin{bmatrix}- 2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & - 3 & - 2\end{bmatrix}\begin{bmatrix}9 \\ 4 \\ - 3\end{bmatrix}\]
\[ = \begin{bmatrix}- 18 + 36 - 18 \\ 0 + 8 - 3 \\ 9 - 12 + 6\end{bmatrix}\]
\[ = \begin{bmatrix}0 \\ 5 \\ 3\end{bmatrix}\]
Hence, x = 0, y = 5 and 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
Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
Evaluate the following determinant:
\[\begin{vmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{vmatrix}\]
If \[A = \begin{bmatrix}2 & 5 \\ 2 & 1\end{bmatrix} \text{ and } B = \begin{bmatrix}4 & - 3 \\ 2 & 5\end{bmatrix}\] , verify that |AB| = |A| |B|.
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:
\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}0 & x & y \\ - x & 0 & z \\ - y & - z & 0\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2\end{vmatrix}\]
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 the following:
\[\begin{vmatrix}0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0\end{vmatrix}\]
\[\begin{vmatrix}0 & b^2 a & c^2 a \\ a^2 b & 0 & c^2 b \\ a^2 c & b^2 c & 0\end{vmatrix} = 2 a^3 b^3 c^3\]
Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]
Show that x = 2 is a root of the equation
Solve the following determinant equation:
If \[a, b\] and c are all non-zero and
Using determinants show that the following points are collinear:
(3, −2), (8, 8) and (5, 2)
2x + 3y = 10
x + 6y = 4
2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11
Solve each of the following system of homogeneous linear equations.
x + y − 2z = 0
2x + y − 3z = 0
5x + 4y − 9z = 0
Write the value of the determinant
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]
If \[\begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0\]
If \[A = \begin{bmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{bmatrix}\]. Write the cofactor of the element a32.
If a > 0 and discriminant of ax2 + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]
If \[x, y \in \mathbb{R}\], then the determinant
2x − y + z = 0
3x + 2y − z = 0
x + 4y + 3z = 0
x + y − z = 0
x − 2y + z = 0
3x + 6y − 5z = 0
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
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
Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.
Solve the following system of equations x − y + z = 4, x − 2y + 2z = 9 and 2x + y + 3z = 1.
If the system of equations 2x + 3y + 5 = 0, x + ky + 5 = 0, kx - 12y - 14 = 0 has non-trivial solution, then the value of k is ____________.
A set of linear equations is represented by the matrix equation Ax = b. The necessary condition for the existence of a solution for this system is
In system of equations, if inverse of matrix of coefficients A is multiplied by right side constant B vector then resultant will be?
What is the nature of the given system of equations
`{:(x + 2y = 2),(2x + 3y = 3):}`
Choose the correct option:
If a, b, c are in A.P. then the determinant `[(x + 2, x + 3, x + 2a),(x + 3, x + 4, x + 2b),(x + 4, x + 5, x + 2c)]` is
The number of real values λ, such that the system of linear equations 2x – 3y + 5z = 9, x + 3y – z = –18 and 3x – y + (λ2 – |λ|z) = 16 has no solution, is ______.
Let the system of linear equations x + y + az = 2; 3x + y + z = 4; x + 2z = 1 have a unique solution (x*, y*, z*). If (α, x*), (y*, α) and (x*, –y*) are collinear points, then the sum of absolute values of all possible values of α is ______.
