मराठी

Find the Matrix X for Which [ 3 2 7 5 ] X [ − 1 1 − 2 1 ] = [ 2 − 1 0 4 ] - Mathematics

Advertisements
Advertisements

प्रश्न

Find the matrix X for which 

\[\begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix} X \begin{bmatrix}- 1 & 1 \\ - 2 & 1\end{bmatrix} = \begin{bmatrix}2 & - 1 \\ 0 & 4\end{bmatrix}\]

 

Advertisements

उत्तर

\[Let A = \begin{bmatrix} 3 & 2\\7 & 5 \end{bmatrix}, B = \begin{bmatrix} - 1 & 1\\ - 2 & 1 \end{bmatrix} \text{ and }C = \begin{bmatrix} 2 & - 1\\0 & 4 \end{bmatrix}\]
Now,
\[\left| A \right| = \begin{vmatrix} 3 & 2\\7 & 5 \end{vmatrix} = 15 - 14 = 1 \]
\[\left| B \right| = \begin{vmatrix} - 1 & 1\\ - 2 & 1 \end{vmatrix} = - 1 + 2 = 1 \]
\[\text{ Since, }\left| A \right| \neq 0\text{ and }\left| B \right| \neq 0\]
\[\text{ Hence, A & B are invertible, so }A^{- 1}\text{ and }B^{- 1}\text{ exist }. \]
Cofactors of matrix A are
\[ A_{11} = 5 A_{12} = - 7 A_{21} = - 2 A_{22} = 3 \]
Now, 
\[adj A = \begin{bmatrix} 5 & - 7\\ - 2 & 3 \end{bmatrix}T = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix} \]
Cofactors of matrix B are
\[ B_{11} = 1 B_{12} = 2 B_{21} = - 1 B_{22} = - 1\]
Now, 
\[adj B = \begin{bmatrix} 1 & 2\\ - 1 & - 1 \end{bmatrix}^T = \begin{bmatrix} 1 & - 1\\ 2 & - 1 \end{bmatrix}\]
\[ B^{- 1} = \frac{1}{\left| B \right|}adj B = \begin{bmatrix} 1 & - 1\\ 2 & - 1 \end{bmatrix} \]
The given equation becomes AXB = C
\[ \Rightarrow \left( A^{- 1} A \right)X\left( B B^{- 1} \right) = A^{- 1} C B^{- 1} \]
\[ \Rightarrow \left( I \right)X\left( I \right) = A^{- 1} C B^{- 1} \]
\[ \Rightarrow X = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\begin{bmatrix} 2 & - 1\\0 & 4 \end{bmatrix}\begin{bmatrix} 1 & - 1\\ 2 & - 1 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\begin{bmatrix} 2 - 2 & - 2 + 1\\ 0 + 8 & 0 - 4 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\begin{bmatrix} 0 & - 1\\ 8 & - 4 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} - 16 & 3\\ 24 & - 5 \end{bmatrix}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [पृष्ठ २४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 32 | पृष्ठ २४

संबंधित प्रश्‍न

Find the inverse of the matrices (if it exists).

`[(1,0,0),(0, cos alpha, sin alpha),(0, sin alpha, -cos alpha)]`


Let A = `[(3,7),(2,5)]` and B = `[(6,8),(7,9)]`. Verify that (AB)−1 = B−1A−1.


If A = `[(3,1),(-1,2)]` show that A2 – 5A + 7I = 0. Hence, find A–1.


For the matrix A = `[(3,2),(1,1)]` find the numbers a and b such that A2 + aA + bI = 0.


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:
\[\begin{bmatrix}a & b \\ c & d\end{bmatrix}\]

Verify that (adj A) A = |A| I = A (adj A) for the above matrix.

Compute the adjoint of the following matrix:

\[\begin{bmatrix}1 & 2 & 5 \\ 2 & 3 & 1 \\ - 1 & 1 & 1\end{bmatrix}\]

Verify that (adj A) A = |A| I = A (adj A) for the above matrix.


Compute the adjoint of the following matrix:

\[\begin{bmatrix}2 & 0 & - 1 \\ 5 & 1 & 0 \\ 1 & 1 & 3\end{bmatrix}\]

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.


If \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\] , show that adj A = 3AT.


Find A (adj A) for the matrix  \[A = \begin{bmatrix}1 & - 2 & 3 \\ 0 & 2 & - 1 \\ - 4 & 5 & 2\end{bmatrix} .\]


Find the inverse of the following matrix:

\[\begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]

Find the inverse of the following matrix:

\[\begin{bmatrix}a & b \\ c & \frac{1 + bc}{a}\end{bmatrix}\]

Find the inverse of the following matrix.

\[\begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\]

Find the inverse of the following matrix and verify that \[A^{- 1} A = I_3\]

\[\begin{bmatrix}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{bmatrix}\]

For the following pair of matrix verify that \[\left( AB \right)^{- 1} = B^{- 1} A^{- 1} :\]

\[A = \begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix}\text{ and }B \begin{bmatrix}4 & 6 \\ 3 & 2\end{bmatrix}\]


Given \[A = \begin{bmatrix}2 & - 3 \\ - 4 & 7\end{bmatrix}\], compute A−1 and show that \[2 A^{- 1} = 9I - A .\]


If \[A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\], show that 

\[A^2 - 5A + 7I = O\].  Hence, find A−1.

If \[A = \begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\].
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\]  and hence find A−1.

If \[A = \frac{1}{9}\begin{bmatrix}- 8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & - 8 & 4\end{bmatrix}\],
prove that  \[A^{- 1} = A^3\]

\[\text{ If }A^{- 1} = \begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{bmatrix},\text{ find }\left( AB \right)^{- 1} .\]

Find the adjoint of the matrix \[A = \begin{bmatrix}- 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1\end{bmatrix}\]  and hence show that \[A\left( adj A \right) = \left| A \right| I_3\]. 


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\]


If A is a square matrix, then write the matrix adj (AT) − (adj A)T.


If \[S = \begin{bmatrix}a & b \\ c & d\end{bmatrix}\], then adj A is ____________ .


If A is a singular matrix, then adj A is ______.


If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is _____________ .


If \[A^2 - A + I = 0\], then the inverse of A is __________ .


If \[A = \begin{bmatrix}2 & 3 \\ 5 & - 2\end{bmatrix}\]  be such that \[A^{- 1} = kA\], then k equals ___________ .


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)]`


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 `abs((2"x", -1),(4,2)) = abs ((3,0),(2,1))` then x is ____________.


A and B are invertible matrices of the same order such that |(AB)-1| = 8, If |A| = 2, then |B| is ____________.


If A is a square matrix of order 3 and |A| = 5, then |adj A| = ______.


If A = `[(2, -3, 5),(3, 2, -4),(1, 1, -2)]`, find A–1. Use A–1 to solve the following system of equations 2x − 3y + 5z = 11, 3x + 2y – 4z = –5, x + y – 2z = –3


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:

  1. Translate the problem into a system of equations.
  2. Solve the system of equation by using matrix method.
  3. Hence, find the cost of one paper bag, one scrap book and one pastel sheet.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×