Advertisements
Advertisements
प्रश्न
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that
उत्तर
\[A = \begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 &- 1& 3 \end{bmatrix} \]
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{vmatrix} = \left( 1 \times 3 \right) - \left( 1 \times 9 \right) + \left( 1 \times - 5 \right) = 3 - 9 - 5 = - 11 \]
\[\text{ Since, }\left| A \right| \neq 0\]
\[\text{Hence, }A^{- 1}\text{ exists . }\]
Now,
\[ A^2 = \begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 &- 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 + 1 + 2 & 1 + 2 - 1 & 1 - 3 + 3\\1 + 2 - 6 & 1 + 4 + 3 & 1 - 6 - 9\\2 - 1 + 6 & 2 - 2 - 3 & 2 + 3 + 9 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 1\\ - 3 & 8 & - 14\\ 7 & - 3 & 14 \end{bmatrix}\]
\[\text{ and }A^3 = A^2 . A = \begin{bmatrix} 4 & 2 & 1\\ - 3 & 8 & - 14\\7 & - 3 & 14 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 + 2 + 2 & 4 + 4 - 1 & 4 - 6 + 3\\ - 3 + 8 - 28 & - 3 + 16 + 14 & - 3 - 24 - 42\\ 7 - 3 + 28 & 7 - 6 - 14 & 7 + 9 + 42 \end{bmatrix} = \begin{bmatrix} 8 & 7 & 1\\ - 23 & 27 & - 69\\ 32 & - 13 & 58 \end{bmatrix}\]
\[\text{ Now, }A^3 - 6 A^2 + 5A + 11 I_3 = \begin{bmatrix} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{bmatrix} - 6 \begin{bmatrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{bmatrix} + 5 \begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{bmatrix} + 11 \begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\]
\[ = \begin{bmatrix} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0\\ - 23 + 18 + 5 + 0 & 27 - 48 + 10 + 11 & - 69 + 84 - 15 + 0\\ 32 - 42 + 10 + 0 & - 13 + 18 - 5 + 0 & 58 - 84 + 15 + 11 \end{bmatrix} \]
\[ = \begin{bmatrix} 0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0 \end{bmatrix} = O (\text{ Null matrix})\]
\[\text{ Again, }A^3 - 6 A^2 + 5A + 11 I_3 = O\]
\[ \Rightarrow A^{- 1} \times \left( A^3 - 6 A^2 + 5A + 11 I_3 \right) = A^{- 1} \times O (\text{ Pre - multiplying both sides because }A^{- 1} exists) \]
\[ \Rightarrow \left( A^2 - 6A + 5 I_3 + 11 A^{- 1} \right) = 0\]
\[ \Rightarrow \begin{bmatrix} 4 & 2 & 1\\ - 3 & 8 - 14\\ 7 & - 3 & 14 \end{bmatrix} - 6\begin{bmatrix} 1 & 1 & 1\\1 & 2 &- 3\\2 & - 1 & 3 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} = - 11 A^{- 1} \]
\[ \Rightarrow \begin{bmatrix} 4 - 6 + 5 & 2 - 6 + 0 & 1 - 6 + 0\\ - 3 - 6 + 0 & 8 - 12 + 5 & - 14 + 18 + 0\\ 7 - 12 + 0 & - 3 + 6 + 0 & 14 - 18 + 5 \end{bmatrix} = - 11 A^{- 1} \]
\[ \Rightarrow A^{- 1} = - \frac{1}{11}\begin{bmatrix} 3 & - 4 & - 5\\- 9 & 1 & 4\\ - 5 & 3 & 1 \end{bmatrix} \]
APPEARS IN
संबंधित प्रश्न
For the matrix A = `[(1,1,1),(1,2,-3),(2,-1,3)]` show that A3 − 6A2 + 5A + 11 I = O. Hence, find A−1.
If x, y, z are nonzero real numbers, then the inverse of matrix A = `[(x,0,0),(0,y,0),(0,0,z)]` is ______.
Find the adjoint of the following matrix:
\[\begin{bmatrix}- 3 & 5 \\ 2 & 4\end{bmatrix}\]
Compute the adjoint of the following matrix:
\[\begin{bmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{bmatrix}\]
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
Find the inverse of the following matrix.
For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]
\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]
If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]
If \[A = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}\] , verify that \[A^2 - 4 A + I = O,\text{ where }I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\text{ and }O = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\] . Hence, find A−1.
If \[A = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\], find the value of \[\lambda\] so that \[A^2 = \lambda A - 2I\]. Hence, find A−1.
Show that the matrix, \[A = \begin{bmatrix}1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1\end{bmatrix}\] satisfies the equation, \[A^3 - A^2 - 3A - I_3 = O\] . Hence, find A−1.
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\] and hence find A−1.
prove that \[A^{- 1} = A^3\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}7 & 1 \\ 4 & - 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}5 & 2 \\ 2 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}2 & 5 \\ 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1\end{bmatrix}\]
If adj \[A = \begin{bmatrix}2 & 3 \\ 4 & - 1\end{bmatrix}\text{ and adj }B = \begin{bmatrix}1 & - 2 \\ - 3 & 1\end{bmatrix}\]
If A is symmetric matrix, write whether AT is symmetric or skew-symmetric.
If \[A = \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\text{ and }A \left( adj A = \right)\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}\], then find the value of k.
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
For non-singular square matrix A, B and C of the same order \[\left( A B^{- 1} C \right) =\] ______________ .
Let \[A = \begin{bmatrix}1 & 2 \\ 3 & - 5\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\] and X be a matrix such that A = BX, then X is equal to _____________ .
If \[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{bmatrix},\text{ then aI + bA + 2 }A^2\] equals ____________ .
If \[\begin{bmatrix}1 & - \tan \theta \\ \tan \theta & 1\end{bmatrix} \begin{bmatrix}1 & \tan \theta \\ - \tan \theta & 1\end{bmatrix} - 1 = \begin{bmatrix}a & - b \\ b & a\end{bmatrix}\], then _______________ .
If A is an invertible matrix, then det (A−1) is equal to ____________ .
Using matrix method, solve the following system of equations:
x – 2y = 10, 2x + y + 3z = 8 and -2y + z = 7
If A = `[(x, 5, 2),(2, y, 3),(1, 1, z)]`, xyz = 80, 3x + 2y + 10z = 20, ten A adj. A = `[(81, 0, 0),(0, 81, 0),(0, 0, 81)]`
If A = `[(0, 1, 3),(1, 2, x),(2, 3, 1)]`, A–1 = `[(1/2, -4, 5/2),(-1/2, 3, -3/2),(1/2, y, 1/2)]` then x = 1, y = –1.
If A and B are invertible matrices, then which of the following is not correct?
`("aA")^-1 = 1/"a" "A"^-1`, where a is any real number and A is a square matrix.
The value of `abs (("cos" (alpha + beta),-"sin" (alpha + beta),"cos" 2 beta),("sin" alpha, "cos" alpha, "sin" beta),(-"cos" alpha, "sin" alpha, "cos" beta))` is independent of ____________.
If A = [aij] is a square matrix of order 2 such that aij = `{(1"," "when i" ≠ "j"),(0"," "when" "i" = "j"):},` then A2 is ______.
For matrix A = `[(2,5),(-11,7)]` (adj A)' is equal to:
If `abs((2"x", -1),(4,2)) = abs ((3,0),(2,1))` then x is ____________.
If for a square matrix A, A2 – A + I = 0, then A–1 equals ______.
| To raise money for an orphanage, students of three schools A, B and C organised an exhibition in their residential colony, where they sold paper bags, scrap books and pastel sheets made by using recycled paper. Student of school A sold 30 paper bags, 20 scrap books and 10 pastel sheets and raised ₹ 410. Student of school B sold 20 paper bags, 10 scrap books and 20 pastel sheets and raised ₹ 290. Student of school C sold 20 paper bags, 20 scrap books and 20 pastel sheets and raised ₹ 440. |
Answer the following question:
- Translate the problem into a system of equations.
- Solve the system of equation by using matrix method.
- Hence, find the cost of one paper bag, one scrap book and one pastel sheet.
