English

If a − 1 = ⎡ ⎢ ⎣ 3 − 1 1 − 15 6 − 5 5 − 2 2 ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ 1 2 − 2 − 1 3 0 0 − 2 1 ⎤ ⎥ ⎦ , Find ( a B ) − 1 . - Mathematics

Advertisements
Advertisements

Question

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

Solution

We know that (AB)1 = B1 A1
\[B = \begin{bmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{bmatrix}\]
\[ B^{- 1} = \frac{1}{\left| B \right|}Adj . B\]
Now, 
\[\left| B \right| = \begin{vmatrix}1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1\end{vmatrix}\]
\[ = 1\left( 3 + 0 \right) + 1\left( 2 - 4 \right)\]
\[ = 1\]
\[\text{ Now, to find Adj . B}\]
\[B_{11} = \left( - 1 \right)^{1 + 1} \left( 3 \right) = 3\]
\[ B_{12} = \left( - 1 \right)^{1 + 2} \left( - 1 \right) = 1\]
\[ B_{13} = \left( - 1 \right)^{1 + 3} \left( 2 \right) = 2\]
\[B_{21} = \left( - 1 \right)^{2 + 1} \left( 2 - 4 \right) = 2\]
\[ B_{22} = \left( - 1 \right)^{2 + 2} \left( 1 \right) = 1 \]
\[ B_{23} = \left( - 1 \right)^{2 + 3} \left( - 2 \right) = 2\]
\[B_{31} = \left( - 1 \right)^{3 + 1} \left( 6 \right) = 6\]
\[ B_{32} = \left( - 1 \right)^{3 + 2} \left( - 2 \right) = 2\]
\[ B_{33} = \left( - 1 \right)^{3 + 3} \left( 3 + 2 \right) = 5\]
Therefore, 
\[Adj . B = \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix}\]
Thus, 
\[ B^{- 1} = \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix} . \]
\[ \left( AB \right)^{- 1} = B^{- 1} A^{- 1} \]
\[ = \begin{bmatrix}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{bmatrix}\begin{bmatrix}3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2\end{bmatrix}\]
\[ = \begin{bmatrix}9 - 30 + 30 & - 3 + 12 - 12 & 3 - 10 + 12 \\ 3 - 15 + 10 & - 1 + 6 - 4 & 1 - 5 + 4 \\ 6 - 30 + 25 & - 2 + 12 - 10 & 2 - 10 + 10\end{bmatrix}\]
\[ = \begin{bmatrix}9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2\end{bmatrix}\]
\[\text{ Hence,} \left( AB \right)^{- 1} = \begin{bmatrix}9 & - 3 & 5 \\ - 2 & 1 & 0 \\ 1 & 0 & 2\end{bmatrix}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 25]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 36 | Page 25

RELATED QUESTIONS

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.


Verify A(adj A) = (adj A)A = |A|I.

`[(1,-1,2),(3,0,-2),(1,0,3)]`


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

`[(1,0,0),(3,3,0),(5,2,-1)]`


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

`[(1,-1,2),(0,2,-3),(3,-2,4)]`


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


If A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` verify that A3 − 6A2 + 9A − 4I = 0 and hence find A−1.


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:

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

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

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

If \[A = \begin{bmatrix}4 & 5 \\ 2 & 1\end{bmatrix}\] , then show that \[A - 3I = 2 \left( I + 3 A^{- 1} \right) .\]


Given  \[A = \begin{bmatrix}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{bmatrix}, B^{- 1} = \begin{bmatrix}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}\] . Compute (AB)−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.


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

\[\text{ If }A = \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix},\text{ find }A^{- 1}\text{ and show that }A^{- 1} = \frac{1}{2}\left( A^2 - 3I \right) .\]

Find the inverse by using elementary row transformations:

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


Find the inverse by using elementary row transformations:

\[\begin{bmatrix}1 & 6 \\ - 3 & 5\end{bmatrix}\]


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:

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


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 \[A = \begin{bmatrix}3 & 1 \\ 2 & - 3\end{bmatrix}\], then find |adj A|.


If A is an invertible matrix, then which of the following is not true ?


If A is an invertible matrix of order 3, then which of the following is not true ?


If \[A = \begin{bmatrix}3 & 4 \\ 2 & 4\end{bmatrix}, B = \begin{bmatrix}- 2 & - 2 \\ 0 & - 1\end{bmatrix},\text{ then }\left( A + B \right)^{- 1} =\]


If B is a non-singular matrix and A is a square matrix, then det (B−1 AB) is equal to ___________ .


For non-singular square matrix A, B and C of the same order \[\left( A B^{- 1} C \right) =\] ______________ .


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 \[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2\end{bmatrix},\text{ then aI + bA + 2 }A^2\] 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 _______________ .


Find A−1, if \[A = \begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\] . Hence solve the following system of linear equations:x + 2y + 5z = 10, x − y − z = −2, 2x + 3y − z = −11


An amount of Rs 10,000 is put into three investments at the rate of 10, 12 and 15% per annum. The combined income is Rs 1310 and the combined income of first and  second investment is Rs 190 short of the income from the third. Find the investment in each using matrix method.

 

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.


If A, B be two square matrices such that |AB| = O, then ____________.


Find x, if `[(1,2,"x"),(1,1,1),(2,1,-1)]` is singular


If A = `[(0, 1),(0, 0)]`, then A2023 is equal to ______.


Read the following passage:

Gautam buys 5 pens, 3 bags and 1 instrument box and pays a sum of ₹160. From the same shop, Vikram buys 2 pens, 1 bag and 3 instrument boxes and pays a sum of ₹190. Also, Ankur buys 1 pen, 2 bags and 4 instrument boxes and pays a sum of ₹250.

Based on the above information, answer the following questions:

  1. Convert the given above situation into a matrix equation of the form AX = B. (1)
  2. Find | A |. (1)
  3. Find A–1. (2)
    OR
    Determine P = A2 – 5A. (2)

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×