Advertisements
Advertisements
प्रश्न
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that
Advertisements
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
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
Two schools A and B want to award their selected students on the values of sincerity, truthfulness and helpfulness. School A wants to award Rs x each, Rs y each and Rs z each for the three respective values to 3, 2 and 1 students, respectively with a total award money of Rs 1,600. School B wants to spend Rs 2,300 to award 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount of award for one prize on each value is Rs 900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for an award.
Find the inverse of the matrices (if it exists).
`[(2,-2),(4,3)]`
For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = 0.
Find the adjoint of the following matrix:
\[\begin{bmatrix}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{bmatrix}\]
Find the adjoint of the following matrix:
Verify that (adj A) A = |A| I = A (adj A) for the above matrix.
If \[A = \begin{bmatrix}- 4 & - 3 & - 3 \\ 1 & 0 & 1 \\ 4 & 4 & 3\end{bmatrix}\], show that adj A = A.
Find the inverse of the following matrix:
Find the inverse of the following matrix.
Find the inverse of the following matrix.
Given \[A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\], compute A−1 and show that \[2 A^{- 1} = 9I - A .\]
Let
\[F \left( \alpha \right) = \begin{bmatrix}\cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\text{ and }G\left( \beta \right) = \begin{bmatrix}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta\end{bmatrix}\]
Show that
If \[A = \begin{bmatrix}4 & 3 \\ 2 & 5\end{bmatrix}\], find x and y such that
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}1 & 2 & 0 \\ 2 & 3 & - 1 \\ 1 & - 1 & 3\end{bmatrix}\]
Find the inverse by using elementary row transformations:
\[\begin{bmatrix}3 & 0 & - 1 \\ 2 & 3 & 0 \\ 0 & 4 & 1\end{bmatrix}\]
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 a non-singular symmetric matrix, write whether A−1 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.
If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\] be such that \[A^{- 1} = k A,\] then find the value of k.
If \[A = \begin{bmatrix}3 & 1 \\ 2 & - 3\end{bmatrix}\], then find |adj A|.
If A satisfies the equation \[x^3 - 5 x^2 + 4x + \lambda = 0\] then A-1 exists if _____________ .
For non-singular square matrix A, B and C of the same order \[\left( A B^{- 1} C \right) =\] ______________ .
If A and B are invertible matrices, which of the following statement is not correct.
(a) 3
(b) 0
(c) − 3
(d) 1
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 \[A = \begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\], find A−1 and hence solve the system of linear equations 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3
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.
Find the value of x for which the matrix A `= [(3 - "x", 2, 2),(2,4 - "x", 1),(-2,- 4,-1 - "x")]` is singular.
If the equation a(y + z) = x, b(z + x) = y, c(x + y) = z have non-trivial solutions then the value of `1/(1+"a") + 1/(1+"b") + 1/(1+"c")` is ____________.
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 ____________.
For matrix A = `[(2,5),(-11,7)]` (adj A)' is equal to:
For A = `[(3,1),(-1,2)]`, then 14A−1 is given by:
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.
