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Class 12 RD Sharma Solutions – Chapter 8 Solution of Simultaneous Linear Equations – Exercise 8.1 | Set 2

  • Last Updated : 23 Jul, 2021

Question 8. (i) If A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}     , find A−1. Using A−1, solve the system of linear equations  x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7.

Solution:

Here,  

A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}

|A| = 1 (1 + 6) + 2 (2 – 0) + 0 (- 4 – 0)

= 7 + 4 + 0



= 11

Let Cij be the cofactors of the elements aij in A.

C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ - 2 & 1\end{vmatrix} = 7, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 0 & 1\end{vmatrix} = - 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 0 & - 2\end{vmatrix} = - 4

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 2 & 0 \\ - 2 & 1\end{vmatrix} = 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 2 \\ 0 & - 2\end{vmatrix} = 2

C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 2 & 0 \\ 1 & 3\end{vmatrix} = - 6, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 0 \\ 2 & 3\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 2 \\ 2 & 1\end{vmatrix} = 5

adj A = \begin{bmatrix}7 & - 2 & - 4 \\ 2 & 1 & 2 \\ - 6 & - 3 & 5\end{bmatrix}^T

\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}

A^{- 1} = \frac{1}{\left| A \right|}adj A



\frac{1}{11}\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}

Now, 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}

X = \frac{1}{11}\begin{bmatrix}70 + 16 - 42 \\ - 20 + 8 - 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 and z = 1.

(ii) If A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}     , find A−1 and hence solve the following system of equations: 

Solution:

Here,

A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}

|A| = 9 – 12 – 6



=  -9

Let Cij be the cofactors of the elements aij in A.

C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & 5 \\ 0 & 1\end{vmatrix} = 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 5 \\ 1 & 1\end{vmatrix} = 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 3 \\ 1 & 0\end{vmatrix} = - 3

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 4 & 2 \\ 0 & 1\end{vmatrix} = 4, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & - 4 \\ 1 & 0\end{vmatrix} = - 4

{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 4 & 2 \\ 3 & 5\end{vmatrix} = - 26, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 2 \\ 2 & 5\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & - 4 \\ 2 & 3\end{vmatrix} = 17

adj A = \begin{bmatrix}3 & 3 & - 3 \\ 4 & 1 & - 4 \\ - 26 & - 11 & 17\end{bmatrix}^T

\begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}

A^{- 1} = \frac{1}{\left| A \right|}adj A

A = \frac{1}{- 9}\begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}

Here,  



A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}     , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}      and B =  \begin{bmatrix}- 1 \\ 7 \\ 2\end{bmatrix}

X = A-1 B

X = \frac{1}{- 9}\begin{bmatrix}3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17\end{bmatrix}\begin{bmatrix}- 1 \\ 7 \\ 2\end{bmatrix}

X = \frac{1}{- 9}\begin{bmatrix}- 3 + 28 - 52 \\ - 3 + 7 - 22 \\ 3 - 28 + 34\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}- 27 \\ - 18 \\ 9\end{bmatrix}

Therefore x = 3, y = 2 and z = – 1.

(iii) If A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}     and B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}     , find AB. Hence, solve the system of equations: x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7.

Solution:

Here,

A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}     , 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}



\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}

11\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}

= 11 I

=> \left( \frac{1}{11}B \right)A = I

A^{- 1} = \frac{1}{11}B

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 and z = 1.

(iv) 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.

Solution:

 Here,  

A = \begin{bmatrix}1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1\end{bmatrix}

|A| = -3 + 4 + 0

= 1

Let Cij be the cofactors of the elements aij in A.

C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & - 2 \\ - 1 & 1\end{vmatrix} = - 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}- 2 & - 2 \\ 0 & 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}- 2 & - 1 \\ 0 & - 1\end{vmatrix} = 2

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & 0 \\ - 1 & 1\end{vmatrix} = - 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 2 \\ 0 & - 1\end{vmatrix} = 1



C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & 0 \\ - 1 & - 2\end{vmatrix} = - 4, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 0 \\ - 2 & - 2\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 2 \\ - 2 & - 1\end{vmatrix} = 3

adj A = \begin{bmatrix}- 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3\end{bmatrix}^T

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

A^{- 1} = \frac{1}{\left| A \right|}adj A

\frac{1}{1}\begin{bmatrix}- 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3\end{bmatrix}

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

Now, X = A-1 B

X = \begin{bmatrix}- 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3\end{bmatrix}\begin{bmatrix}10 \\ 8 \\ 7\end{bmatrix}

X = \begin{bmatrix}- 30 + 16 + 14 \\ - 20 + 8 + 7 \\ - 40 + 16 + 21\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}0 \\ - 5 \\ - 3\end{bmatrix}



Therefore x = 0, y = – 5 and z = -3.

(v) If A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}     , B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}     , find BA and use this to solve the system of equations  y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17.

Solution:

We have,

A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}     , B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}

BA = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}

BA = \begin{bmatrix}2 + 4 + 0 & 2 - 2 + 0 & - 4 + 4 + 0 \\ 4 - 12 + 8 & 4 + 6 - 4 & - 8 - 12 + 20 \\ 0 - 4 + 4 & 0 + 2 - 2 & 0 - 4 + 10\end{bmatrix}

BA = \begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix}

BA = 6\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}

BA = 6I

B\left( \frac{1}{6}A \right) = I



B^{- 1} = \frac{1}{6}A

B^{- 1} = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}

Now, BX = C. So, we have

X = B-1 C

X = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}12 \\ - 6 \\ 24\end{bmatrix}

Therefore x = 2, y = – 1 and z = 4.

Question 9. The sum of three numbers is 2. If twice the second number is added to the sum of the first and third, the sum is 1. By adding the second and third numbers to five times the first number, we get 6. Find the three numbers by using matrices.

Solution:

Let the three numbers be x, y and z.  



According to the question,

x + y + z = 2

x + 2y + z = 1

5x + y + z = 6

The given system of equations can be written in matrix form as,

\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}

A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 1 \\ 5 & 1 & 1\end{bmatrix}     , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}      and B = \begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}

|A| = 1 + 4 – 9

= -4

Let Cij be the cofactors of the elements aij in A.

C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = 4, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 2 \\ 5 & 1\end{vmatrix} = - 9

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = - 4, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 5 & 1\end{vmatrix} = 4

C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = 1

adj A = \begin{bmatrix}1 & 4 & - 9 \\ 0 & - 4 & 4 \\ - 1 & 0 & 1\end{bmatrix}^T

\begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}

A^{- 1} = \frac{1}{\left| A \right|}adj A

\frac{1}{- 4}\begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}

X = A-1 B

X = \frac{1}{- 4}\begin{bmatrix}1 & 0 & - 1 \\ 4 & - 4 & 0 \\ - 9 & 4 & 1\end{bmatrix}\begin{bmatrix}2 \\ 1 \\ 6\end{bmatrix}

X = \frac{1}{- 4}\begin{bmatrix}2 + 0 - 6 \\ 8 - 4 + 0 \\ - 18 + 4 + 6\end{bmatrix}



\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 4}\begin{bmatrix}- 4 \\ 4 \\ - 8\end{bmatrix}

Therefore x = 1, y = – 1 and z = 2.

Question 10. 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 the first and second investment is Rs 190, short of the income from the third. Find the investment in each using the matrix method.

Solution:

Let the numbers are x, y, and z.

x + y + z = 10,000

0.1x + 0.12y + 0.15z = 1310

0.1x + 0.12y – 0.15z = – 190

The given system of equation can be written in matrix form as,

\begin{bmatrix}1 & 1 & 1 \\ 0 . 1 & 0 . 12 & 0 . 15 \\ - 0 . 1 & - 0 . 12 & 0 . 15\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}10000 \\ 1310 \\ 190\end{bmatrix}

Here,  

A = \begin{bmatrix}1 & 1 & 1 \\ 0 . 1 & 0 . 12 & 0 . 15 \\ - 0 . 1 & - 0 . 12 & 0 . 15\end{bmatrix}     , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}      and B = \begin{bmatrix}10000 \\ 1310 \\ 190\end{bmatrix}

|A| = 0.036 – 0.03 + 0

= 0.006

Let Cij be the cofactors of the elements aij in A.

C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 . 12 & 0 . 15 \\ - 0 . 12 & 0 . 15\end{vmatrix} = 0 . 036, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 . 1 & 0 . 15 \\ - 0 . 1 & 0 . 15\end{vmatrix} = - 0 . 03, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 . 1 & 0 . 12 \\ - 0 . 1 & - 0 . 12\end{vmatrix} = 0

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ - 0 . 12 & 0 . 15\end{vmatrix} = - 0 . 27, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ - 0 . 1 & 0 . 15\end{vmatrix} = 0 . 25, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ - 0 . 1 & - 0 . 12\end{vmatrix} = 0 . 02

C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 0 . 12 & 0 . 15\end{vmatrix} = 0 . 03, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 0 . 1 & 0 . 15\end{vmatrix} = - 0 . 05, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 0 . 1 & 0 . 12\end{vmatrix} = 0 . 02

adj A = \begin{bmatrix}0 . 036 & - 0 . 03 & 0 \\ - 0 . 27 & 0 . 25 & 0 . 02 \\ 0 . 03 & - 0 . 05 & 0 . 02\end{bmatrix}^T

\begin{bmatrix}0 . 036 & - 0 . 27 & 0 . 03 \\ - 0 . 03 & 0 . 25 & - 0 . 05 \\ 0 & 0 . 02 & 0 . 02\end{bmatrix}

A^{- 1} = \frac{1}{\left| A \right|}adj A



\frac{1}{0 . 006}\begin{bmatrix}0 . 036 & - 0 . 27 & 0 . 03 \\ - 0 . 03 & 0 . 25 & - 0 . 05 \\ 0 & 0 . 02 & 0 . 02\end{bmatrix}

X = A-1 B

X = \frac{1}{0 . 006}\begin{bmatrix}0 . 036 & - 0 . 27 & 0 . 03 \\ - 0 . 03 & 0 . 25 & - 0 . 05 \\ 0 & 0 . 02 & 0 . 02\end{bmatrix}\begin{bmatrix}10000 \\ 1310 \\ 190\end{bmatrix}

X = \frac{1}{0 . 006}\begin{bmatrix}360 - 353 . 7 + 5 . 7 \\ - 300 + 327 . 5 - 9 . 5 \\ 0 + 26 . 2 + 3 . 8\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1000}{6}\begin{bmatrix}12 \\ 18 \\ 30\end{bmatrix}

Therefore x = 2000, y = 3000 and z = 5000.

Thus, the three investments are of Rs 2000, Rs 3000 and Rs 5000, respectively.

Question 11. A company produces three products every day. Their production on a certain day is 45 tons. It is found that the production of the third product exceeds the production of the first product by 8 tons while the total production of the first and third product is twice the production of the second product. Determine the production level of each product using the matrix method.

Solution:

Let x, y and z be the production level of the first, second and third product, respectively .

x + y + z = 45 

– x + z = 8 

x – 2y + z = 0 

The given system of equations can be written in matrix form as,

\begin{bmatrix}1 & 1 & 1 \\ - 1 & 0 & 1 \\ 1 & - 2 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}45 \\ 8 \\ 0\end{bmatrix}

AX = B

A = \begin{bmatrix}1 & 1 & 1 \\ - 1 & 0 & 1 \\ 1 & - 2 & 1\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}   , B = \begin{bmatrix}45 \\ 8 \\ 0\end{bmatrix}

Now,

|A| = 2 + 2 + 2

= 6

Let Cij be the cofactors of the elements aij in A.



C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & 1 \\ - 2 & 1\end{vmatrix} = 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}- 1 & 1 \\ 1 & 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}- 1 & 0 \\ 1 & - 2\end{vmatrix} = 2

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ - 2 & 1\end{vmatrix} = - 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = 3

C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 0 & 1\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ - 1 & 1\end{vmatrix} = - 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ - 1 & 0\end{vmatrix} = 1

adj A = \begin{bmatrix}2 & 2 & 2 \\ - 3 & 0 & 3 \\ 1 & - 2 & 1\end{bmatrix}^T

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

A^{- 1} = \frac{1}{\left| A \right|}adj A

\frac{1}{6}\begin{bmatrix}2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1\end{bmatrix}

X = A-1 B

X = \frac{1}{6}\begin{bmatrix}2 & - 3 & 1 \\ 2 & 0 & - 2 \\ 2 & 3 & 1\end{bmatrix}\begin{bmatrix}45 \\ 8 \\ 0\end{bmatrix}

X = \frac{1}{6}\begin{bmatrix}90 - 24 + 0 \\ 90 + 0 + 0 \\ 90 + 24 + 0\end{bmatrix}



\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}66 \\ 90 \\ 114\end{bmatrix}

Therefore x = 11, y = 15 and z = 19.

Thus, the production level of first, second and third product is 11, 15 and 19, respectively.

Question 12. The prices of three commodities P, Q, and R are Rs x, y, and z per unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchase 3 units of Q and sells 2 units of P and 1 unit of R. C purchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B and C earn Rs 6000, Rs 5000 and Rs 13000 respectively. If selling the units is positive earning and buying the units is negative earnings, find the price per unit of three commodities by using the matrix method.

Solution:

The prices of three commodities P, Q and R are Rs x, Rs y and Rs z per unit, respectively.

3x + 5y – 4z = 6000

2x – 3y + z = 5000

-x + 4y + 6z = 13000

The given system of equations can be written in matrix form as follows:

\begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}

AX = B

Here,

A = \begin{bmatrix}3 & 5 & - 4 \\ 2 & - 3 & 1 \\ - 1 & 4 & 6\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}   , B = \begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}

Now,

|A| = -66 – 65 – 20

= -151

Let Cij be the cofactors of the elements aij in A.

C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 3 & 1 \\ 4 & 6\end{vmatrix} = - 22, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ - 1 & 6\end{vmatrix} = - 13, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 3 \\ - 1 & 4\end{vmatrix} = 5

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}5 & - 4 \\ 4 & 6\end{vmatrix} = - 46, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 4 \\ - 1 & 6\end{vmatrix} = 14, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 5 \\ - 1 & 4\end{vmatrix} = - 17

C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}5 & - 4 \\ - 3 & 1\end{vmatrix} = - 7, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 4 \\ 2 & 1\end{vmatrix} = - 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 5 \\ 2 & - 3\end{vmatrix} = - 19



adj A = \begin{bmatrix}- 22 & - 13 & 5 \\ - 46 & 14 & - 17 \\ - 7 & - 11 & - 19\end{bmatrix}^T

\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}

A^{- 1} = \frac{1}{\left| A \right|}adj A

\frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}

X = A-1 B

X = \frac{1}{- 151}\begin{bmatrix}- 22 & - 46 & - 7 \\ - 13 & 14 & - 11 \\ 5 & - 17 & - 19\end{bmatrix}\begin{bmatrix}6000 \\ 5000 \\ 13000\end{bmatrix}

X = \frac{1}{- 151}\begin{bmatrix}- 453000 \\ - 151000 \\ - 302000\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3000 \\ 1000 \\ 2000\end{bmatrix}

Therefore x = 3000, y = 1000 and z = 2000.

Thus, the prices of the three commodities P, Q and R are Rs 3000, Rs 1000 and Rs 2000 per unit, respectively.

Question 13. The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others, and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using the matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation, and supervision, suggest one more value which the management of the colony must include for awards.

Solution:

According to the question, we have

x + y + z = 12 

2x + 3(y +z) = 33 

x + z -2y = 0

The given system of equations can be written in matrix form as,

\begin{bmatrix}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}12 \\ 33 \\ 0\end{bmatrix}

AX = B

Here,

A = \begin{bmatrix}1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}    and B = \begin{bmatrix}12 \\ 33 \\ 0\end{bmatrix}



Now, |A| = 3.

And adj A is given by \begin{bmatrix}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{bmatrix}   .

A^{-1} = \frac{1}{|A|}(adj A) = \frac{1}{3}\begin{bmatrix}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{bmatrix}

X = A-1 B

X = \frac{1}{3}\begin{bmatrix}9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1\end{bmatrix}\begin{bmatrix}12 \\ 33 \\ 0\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3 \\ 4 \\ 5\end{bmatrix}

Therefore x= 3, y = 4 and z = 5.

Therefore, the number of awardees for Honesty, Cooperation and Supervision are 3, 4, and 5 respectively.

One more value which the management of the colony must include for awards may be Sincerity.

Question 14. A school wants to award its students for the values of Honesty, Regularity, and Hard work with a total cash award of Rs 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using the matrix method. Apart from these values, namely, Honesty, Regularity, and Hard work, suggest one more value which the school must include for awards.

Solution:

Let the award money given for Honesty, Regularity and Hard work be x, y and z respectively.

x + y + z = 6,000

x + 3 z = 11,000

x − 2y + z = 0

The given system of equations can be written in matrix form as,

\begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6000 \\ 11000 \\ 0\end{bmatrix}

AX = B

Here, A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & 3 \\ 1 & -2 & 1\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}    and B = \begin{bmatrix}6000 \\ 11000 \\ 0\end{bmatrix}

Now, |A| = 6.

And adj A is given by \begin{bmatrix}6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1\end{bmatrix}   .



A^{-1} = \frac{1}{|A|}(adj A) = \frac{1}{6}\begin{bmatrix}6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1\end{bmatrix}

X = A-1 B

X = \frac{1}{6}\begin{bmatrix}6 & -3 & 3 \\ 2 & 0 & -2 \\ -2 & 3 & -1\end{bmatrix}\begin{bmatrix}6000 \\ 11000 \\ 0\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}500 \\ 2000 \\ 3500\end{bmatrix}

Therefore x = 500, y = 2000 and z = 3500.

Thus, award money given for Honesty, Regularity and Hard work are Rs 500, Rs 2000 and Rs 3500 respectively.

School can include sincerity for awards.

Question 15. Two institutions decided to award their employees for the three values of resourcefulness, competence, and determination in the form of prices at the rate of Rs. x, y, and z respectively per person. The first institution decided to award respectively 4, 3, and 2 employees with total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y, and z.

Solution:

According to the question, we have

4x + 3y + 2z = 37000

5x + 3y + 4z = 47000

x + y + z = 12000

We can express these equations as AX = B where, A = \begin{bmatrix}4 & 3 & 2 \\ 5 & 3 & 4 \\ 1 & 1 & 1\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}    and B = \begin{bmatrix}37000 \\ 47000 \\ 12000\end{bmatrix}

|A| = 4 (3 – 4) – 3 (5 – 4) + 2 (5 – 3)

= -4 – 3 + 4 

=  -3 

adj A = \begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix}

A^{- 1} = \frac{1}{\left| A \right|}adj A = - \frac{1}{3}\begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix}

X = A-1 B

X = -\frac{1}{3}\begin{bmatrix}- 1 & - 1 & 6 \\ - 1 & 2 & - 6 \\ 2 & - 1 & - 3\end{bmatrix} \begin{bmatrix}37000 \\ 47000 \\ 12000\end{bmatrix}



X = -\frac{1}{3}\begin{bmatrix}- 37000 - 47000 + 72000 \\ - 37000 + 94000 - 72000 \\ 74000 - 47000 - 36000\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{3}\begin{bmatrix}- 12000 \\ - 15000 \\ - 9000\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}4000 \\ 5000 \\ 3000\end{bmatrix}

So, x = 4000 , y = 5000 and z = 3000.

Question 16. Two factories decided to award their employees three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations, and (c) keeping calm in tense situations, at the rate of ₹ x, ₹ y, and ₹ z per person respectively. The first factory decided to honor respectively 2, 4 and 3 employees with total prize money of ₹ 29000. The second factory decided to honor respectively 5, 2 and 3 employees with the prize money of ₹ 30500. If the three prizes per person together cost ₹ 9500, then

(i) represent the above situation by matrix equation and form linear equation using matrix multiplication.

(ii) Solve these equations by the matrix method.

Solution:

According to question,

2x + 3y + 4z = 29000 

5x + 2y + 3z = 30500 

x + y + z = 9500 

The given system of equations can be written in matrix form as,

\begin{bmatrix}2 & 3 & 4 \\ 5 & 2 & 3 \\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}29000 \\ 30500 \\ 9500\end{bmatrix}

\begin{bmatrix}- 2 & - 1 & 0 \\ 2 & - 1 & 0 \\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}- 9000 \\ 2000 \\ 9500\end{bmatrix}

\begin{bmatrix}- 2 & - 1 & 0 \\ - 2 & 1 & 0 \\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}- 9000 \\ - 2000 \\ 9500\end{bmatrix}

\begin{bmatrix}- 4 & 0 & 0 \\ - 2 & 1 & 0 \\ 3 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}- 11000 \\ - 2000 \\ 11500\end{bmatrix}

\begin{bmatrix}1 & 0 & 0 \\ - 2 & 1 & 0 \\ 3 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2750 \\ - 2000 \\ 11500\end{bmatrix}

\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}2750 \\ 3500 \\ 3250\end{bmatrix}

Therefore x = 2750, y = 3500 and z = 3250.

Question 17. Two schools A and B want to award their selected students with the values of sincerity, truthfulness, and helpfulness. School A wants to award ₹x each, ₹y each, and ₹z each for the three respective values to 3, 2, and 1 student respectively, with total award money of ₹1,600. School B wants to spend ₹2,300 to award its 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 ₹900, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award.

Solution:



Let the award money given for sincerity, truthfulness and helpfulness be ₹x, ₹y and ₹z respectively.

x + y + z = 900       

3x + 2y + z = 1600

4x + y + 3z = 2300 

The above system of equations can be written in matrix form as,

\begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}900 \\ 1600 \\ 2300\end{bmatrix}

, where C = \begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}    and D = \begin{bmatrix}900 \\ 1600 \\ 2300\end{bmatrix}

Now, |C| = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{vmatrix}

= 5 – 5 – 5

= -5

Let Cij be the cofactors of the elements aij in A.

C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 1 \\ 4 & 3\end{vmatrix} = - 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 4 & 1\end{vmatrix} = - 5

C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 4 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 4 & 1\end{vmatrix} = 3

C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = - 1

adj C = \begin{bmatrix}5 & - 5 & - 5 \\ - 2 & - 1 & 3 \\ - 1 & 2 & - 1\end{bmatrix}^T

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

C^{- 1} = \frac{1}{\left| C \right|}adj C

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

X = C-1 D

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\begin{bmatrix}900 \\ 1600 \\ 2300\end{bmatrix}



\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}4500 - 3200 - 2300 \\ - 4500 - 1600 + 4600 \\ - 4500 + 4800 - 2300\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 1000 \\ - 1500 \\ - 2000\end{bmatrix}

Therefore, x = -1000/-5, y = -1500/- 5 and z = -2000/-5

So, x = 200, y = 300 and z = 400.

Hence, the award money for each value of sincerity, truthfulness and helpfulness is ₹200, ₹300 and ₹400.

One more value which should be considered for award hardwork.

Question 18. Two schools P and Q want to award their selected students on the values of Discipline, Politeness, and Punctuality. The school P wants to award ₹x each, ₹y each, and ₹z each the three respectively values to its 3, 2, and 1 students with total award money of ₹1,000. School Q wants to spend ₹1,500 to award its 4, 1, and 3 students on the respective values (by giving the same award money for three values as before). If the total amount of awards for one prize on each value is ₹600, using matrices, find the award money for each value. Apart from the above three values, suggest one more value for awards.

Solution: 

Let the award money given for Discipline, Politeness and Punctuality be ₹x, ₹y and ₹z respectively.

x + y + z = 600 

3x + 2y + z = 1000    

4x + y + 3z = 1500

The above system of equations can be written in matrix form AX = B as

\begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}600 \\ 1000 \\ 1500\end{bmatrix}

Here, A = \begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}    and B = \begin{bmatrix}600 \\ 1000 \\ 1500\end{bmatrix}

Now,  

|A| = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{vmatrix}

= 5 – 5 – 5

= -5

adj A = \begin{bmatrix}5 & - 5 & - 5 \\ - 2 & - 1 & 3 \\ - 1 & 2 & - 1\end{bmatrix}^T

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

A^{- 1} = \frac{1}{\left| A \right|}adj A

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

X = A-1 B

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\begin{bmatrix}600 \\ 1000 \\ 1500\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}3000 - 2000 - 1500 \\ - 3000 - 1000 + 3000 \\ - 3000 + 3000 - 1500\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 500 \\ - 1000 \\ - 1500\end{bmatrix}

=> x = -500/- 5, y = -1000/- 5 and z = -1500/- 5

Therefore x = 100, y = 200 and z = 300.

Hence, the award money for each value of Discipline, Politeness and Punctuality is ₹100, ₹200 and ₹300.

​One more value which should be considered for award is Honesty.



Question 19. Two schools P and Q want to award their selected students on the values of Tolerance, Kindness, and Leadership. The school P wants to award ₹x each, ₹y each, and ₹z each for the three respective values to 3, 2, and 1 student respectively with total award money of ₹2,200. School Q wants to spend ₹3,100 to award its 4, 1, and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each value is ₹1,200, using matrices, find the award money for each value.

Apart from these three values, suggest one more value which should be considered for award.

Solution:

​​Let the award money given for Tolerance, Kindness and Leadership be ₹x, ₹y and ₹z respectively.

 x + y + z = 1200

3x + 2y + z = 2200     

4x + y + 3z = 3100

The above system of equations can be written in matrix form AX = B as

\begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}

where, A = \begin{bmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{bmatrix}   , X = \begin{bmatrix}x \\ y \\ z\end{bmatrix}    and B = \begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}

Now,  

|A| = \begin{vmatrix}1 & 1 & 1 \\ 3 & 2 & 1 \\ 4 & 1 & 3\end{vmatrix}

= 5 – 5 – 5

= -5

adj A = \begin{bmatrix}5 & - 5 & - 5 \\ - 2 & - 1 & 3 \\ - 1 & 2 & - 1\end{bmatrix}^T

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

A^{- 1} = \frac{1}{\left| A \right|}adj A

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

X = A-1 B

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}5 & - 2 & - 1 \\ - 5 & - 1 & 2 \\ - 5 & 3 & - 1\end{bmatrix}\begin{bmatrix}1200 \\ 2200 \\ 3100\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}6000 - 4400 - 3100 \\ - 6000 - 2200 + 6200 \\ - 6000 + 6600 - 3100\end{bmatrix}

\begin{bmatrix}x \\ y \\ z\end{bmatrix} = - \frac{1}{5}\begin{bmatrix}- 1500 \\ - 2000 \\ - 2500\end{bmatrix}



=> x = – 1500/-5, y = – 2000/-5 and z = -2500/-5

Therefore x = 300, y = 400 and z = 500.

Hence, the award money for each value of Tolerance, Kindness and Leadership is ₹300, ₹400 and ₹500.

​One more value which should be considered for award is Honesty.

Question 20. A total amount of ₹7000 is deposited in three different saving bank accounts with annual interest rates 5%, 8%, and 8.5% respectively. The total annual interest from these three accounts is ₹550. Equal amounts have been deposited in the 5% and 8% saving accounts. Find the amount deposited in each of the three accounts, with the help of matrices.

Solution:

​​​Let the amount deposited in each of the three accounts be ₹ x, ₹ x and ₹ y respectively. 

2x + y = 7000    

26x + 17y = 110000

The above system of equations can be written in matrix form AX = B as,

\begin{bmatrix}2 & 1 \\ 26 & 17\end{bmatrix}\binom{x}{y} = \binom{7000}{110000}

where, A = \begin{bmatrix}2 & 1 \\ 26 & 17\end{bmatrix}   , X = \binom{x}{y}    and B = \binom{7000}{110000}

Now,

|A| = \begin{vmatrix}2 & 1 \\ 26 & 17\end{vmatrix}

= 34 – 26

= 8

adj A = \begin{bmatrix}17 & - 26 \\ - 1 & 2\end{bmatrix}^T

\begin{bmatrix}17 & - 1 \\ - 26 & 2\end{bmatrix}

A^{- 1} = \frac{1}{\left| A \right|}adj A

\frac{1}{8}\begin{bmatrix}17 & - 1 \\ - 26 & 2\end{bmatrix}

X = A-1 B

\binom{x}{y} = \frac{1}{8}\begin{bmatrix}17 & - 1 \\ - 26 & 2\end{bmatrix}\binom{7000}{110000}

\binom{x}{y} = \frac{1}{8}\binom{119000 - 110000}{ - 182000 + 220000}

\binom{x}{y} = \frac{1}{8}\binom{9000}{38000}

=> x = \frac{9000}{8}    and y = \frac{38000}{8}

Therefore x = 1125 and y = 4750.

Hence, the amount deposited in each of the three accounts is ₹1125, ₹1125 and ₹4750.

Question 21. A shopkeeper has 3 varieties of pens ‘A’, ‘B’ and ‘C’. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of ‘A’ variety 3 pens of ‘B’ variety and 2 pens of ‘C’ variety for Rs 60. While Shikha purchased 6 pens of ‘A’ variety, 2 pens of ‘B’ variety, and 3 pens of ‘C’ variety for Rs 70. Using the matrix method, find the cost of each variety of pens.

Solution:

Suppose there are 3 varieties of pen A, B and C

A + B + C = 21

4A + 3B + 2C = 60

6A + 2B + 3C = 70

\begin{bmatrix}1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3\end{bmatrix}\begin{bmatrix}A \\ B \\ C\end{bmatrix} = \begin{bmatrix}21 \\ 60 \\ 70\end{bmatrix}

where, P = \begin{bmatrix}1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3\end{bmatrix}   , Q = \begin{bmatrix}21 \\ 60 \\ 70\end{bmatrix}

|P| = -5

Therefore, we get X = P-1 Q

adj P = \begin{bmatrix}5 & 0 & - 10 \\ - 1 & - 3 & 4 \\ - 1 & 2 & - 1\end{bmatrix}^T

\begin{bmatrix}5 & - 1 & - 1 \\ 0 & - 3 & 2 \\ - 10 & 4 & - 1\end{bmatrix}

P^{- 1} = \frac{1}{- 5}\begin{bmatrix}5 & - 1 & - 1 \\ 0 & - 3 & 2 \\ - 10 & 4 & - 1\end{bmatrix}

So, X = \frac{1}{- 5}\begin{bmatrix}5 & - 1 & - 1 \\ 0 & - 3 & 2 \\ - 10 & 4 & - 1\end{bmatrix}\begin{bmatrix}21 \\ 60 \\ 70\end{bmatrix}

\frac{1}{- 5}\begin{bmatrix}105 - 60 - 70 \\ 0 - 180 + 140 \\ - 210 + 240 - 70\end{bmatrix}

\frac{1}{- 5}\begin{bmatrix}- 25 \\ - 40 \\ - 40\end{bmatrix}

Therefore, X = \begin{bmatrix}5 \\ 8 \\ 8\end{bmatrix}

Therefore, cost of variety of pens for A, B and C is Rs 5, Rs 8 and Rs 8 respectively.

Attention reader! Don’t stop learning now. Join the First-Step-to-DSA Course for Class 9 to 12 students , specifically designed to introduce data structures and algorithms to the class 9 to 12 students




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