Advertisements
Advertisements
प्रश्न
उत्तर
Here,
\[ A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[AB = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix} \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[ \Rightarrow AB = \begin{bmatrix}7 + 4 + 0 & 2 - 2 + 0 & - 6 + 6 + 0 \\ 14 - 2 - 12 & 4 + 1 + 6 & - 12 - 3 + 15 \\ 0 + 4 - 4 & 0 - 2 + 2 & 0 + 6 + 5\end{bmatrix}\]
\[ = \begin{bmatrix}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{bmatrix}\]
\[ \Rightarrow AB = 11\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[AB = 11 I_3 \]
\[ \Rightarrow \frac{1}{11}AB = I_3 \]
\[ \Rightarrow \left( \frac{1}{11}B \right)A = I_3 \]
\[ \Rightarrow A^{- 1} = \frac{1}{11}B\]
\[ \Rightarrow A^{- 1} = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\]
\[X = A^{- 1} B\]
\[X = \frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}70 + 16 - 42 \\ - 20 + 9 - 21 \\ - 40 + 16 + 35\end{bmatrix}\]
\[\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{11}\begin{bmatrix}44 \\ - 33 \\ 11\end{bmatrix}\]
\[ \therefore x = 4, y = - 3\text{ and }z = 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following determinant:
\[\begin{vmatrix}x & - 7 \\ x & 5x + 1\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]
\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]
\[\begin{vmatrix}0 & b^2 a & c^2 a \\ a^2 b & 0 & c^2 b \\ a^2 c & b^2 c & 0\end{vmatrix} = 2 a^3 b^3 c^3\]
Prove the following identity:
\[\begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} = \left( x + y + z \right)^3\]
Prove the following identity:
`|(a^3,2,a),(b^3,2,b),(c^3,2,c)| = 2(a-b) (b-c) (c-a) (a+b+c)`
Solve the following determinant equation:
If the points (x, −2), (5, 2), (8, 8) are collinear, find x using determinants.
Prove that :
9x + 5y = 10
3y − 2x = 8
x + 2y = 5
3x + 6y = 15
For what value of x, the following matrix is singular?
Write the value of the determinant
If \[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\] , then for any natural number, find the value of Det(An).
\[\begin{vmatrix}\log_3 512 & \log_4 3 \\ \log_3 8 & \log_4 9\end{vmatrix} \times \begin{vmatrix}\log_2 3 & \log_8 3 \\ \log_3 4 & \log_3 4\end{vmatrix}\]
If \[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & - 2 \\ 7 & 3\end{vmatrix}\] , then x =
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
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 \[\begin{bmatrix}1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\], find x, y and z.
The number of solutions of the system of equations
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
is
The number of solutions of the system of equations:
2x + y − z = 7
x − 3y + 2z = 1
x + 4y − 3z = 5
The existence of the unique solution of the system of equations:
x + y + z = λ
5x − y + µz = 10
2x + 3y − z = 6
depends on
The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13
If \[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\] ,find A–1 and hence solve the system of equations x – 2y = 10, 2x + y + 3z = 8 and –2y + z = 7.
x + y = 1
x + z = − 6
x − y − 2z = 3
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 A = `[(2, 0),(0, 1)]` and B = `[(1),(2)]`, then find the matrix X such that A−1X = B.
Show that if the determinant ∆ = `|(3, -2, sin3theta),(-7, 8, cos2theta),(-11, 14, 2)|` = 0, then sinθ = 0 or `1/2`.
If `|(2x, 5),(8, x)| = |(6, -2),(7, 3)|`, then value of x is ______.
If A = `[(1,-1,0),(2,3,4),(0,1,2)]` and B = `[(2,2,-4),(-4,2,-4),(2,-1,5)]`, then:
What is the nature of the given system of equations
`{:(x + 2y = 2),(2x + 3y = 3):}`
