第 12 类 RD Sharma 解——第 8 章联立线性方程组的解——练习 8.1 |设置 2

问题 8. (i) 如果 A = $\begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}$ ，求 A ^-1 。使用 A ^-1求解线性方程组 x - 2y = 10, 2x + y + 3z = 8, -2y + z = 7。

解决方案：

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 C_ij be the cofactors of the elements a_ij 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) 如果 A = $\begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}$ ，找到 A ^-1并因此求解以下方程组：

解决方案：

Here,

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

|A| = 9 – 12 – 6

=  -9

Let C_ij be the cofactors of the elements a_ij 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) 如果 A = $\begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}$和 B = $\begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}$ , 找到 AB。因此，求解方程组：x - 2y = 10、2x + y + 3z = 8 和 -2y + z = 7。

解决方案：

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) 如果 A = $\begin{bmatrix}1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1\end{bmatrix}$ ，求 A ^-1 。使用 A ^-1求解线性方程组：x - 2y = 10, 2x - y - z = 8, -2y + z = 7。

解决方案：

 Here,  

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

|A| = -3 + 4 + 0

= 1

Let C_ij be the cofactors of the elements a_ij 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) 如果 A = $\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$ , B = $\begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}$ , 找到 BA 并用它来求解方程组 y + 2z = 7, x - y = 3, 2x + 3y + 4z = 17。

解决方案：

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.

问题 9. 三个数的和是 2。如果第二个数的两倍加上第一个和第三个数的和，则和是 1。将第二个和第三个数加到第一个数的五倍，我们得到 6。使用矩阵求三个数。

解决方案：

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 C_ij be the cofactors of the elements a_ij 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.

问题 10. 以每年 10%、12% 和 15% 的利率将 10,000 卢比投入三项投资。总收入为 1310 卢比，第一次和第二次投资的总收入为 190 卢比，低于第三次投资的收入。使用矩阵方法找到每个的投资。

解决方案：

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 C_ij be the cofactors of the elements a_ij 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.

问题 11. 一家公司每天生产三种产品。他们某一天的产量是45吨。发现第三种产品的产量比第一种产品的产量高8吨，而第一种和第三种产品的总产量是第二种产品产量的两倍。使用矩阵法确定每种产品的生产水平。

解决方案：

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 C_ij be the cofactors of the elements a_ij 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.

问题 12. 三种商品 P、Q 和 R 的价格分别为每单位 Rs x、y 和 z。 A 购买 4 单位 R 并出售 3 单位 P 和 5 单位 Q。B 购买 3 单位 Q 并出售 2 单位 P 和 1 单位 R。C 购买 1 单位 P 并出售 4 单位 Q 和 6单位 R。在过程 A、B 和 C 中，分别赚取 6000 卢比、5000 卢比和 13000 卢比。如果卖出单位是正收益，买单位是负收益，用矩阵法求三种商品的单位价格。

解决方案：

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 C_ij be the cofactors of the elements a_ij 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.

问题 13. 一个居住区的管理委员会决定奖励一些成员（比如 x）诚实，一些（比如 y）帮助别人，还有一些（比如 z）监督工人保持殖民地整洁和干净的。全体获奖者之和为12。合作监督获奖者的三倍加上诚信获奖者的两倍之和为33。如果诚信监督获奖者人数之和为获奖者数量的两倍为了帮助他人，使用矩阵法，找出每个类别的获奖人数。除了这些价值观，即诚实、合作和监督之外，还表明了殖民地管理必须包括奖励的另一种价值观。

解决方案：

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.

问题 14。一所学校希望以 6,000 卢比的现金奖励奖励其学生的诚实、规律和勤奋的价值观。辛勤工作的奖金加上诚实的奖金是 11,000 卢比的三倍。诚实和努力工作的奖金是规律性奖金的两倍。以代数方式表示上述情况，并使用矩阵方法找到每个值的奖金。除了这些价值观，即诚实、规律和努力工作之外，还提出了学校必须在奖励中包含的另一种价值观。

解决方案：

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.

问题 15。两家机构决定以卢比的价格奖励其员工的机智、能力和决心这三个价值观。 x、y 和 z 分别为每人。第一个机构决定分别奖励4、3和2名员工，总价为卢比。 37000和第二机构决定分别奖励5名、3名和4名员工，总价款为卢比。 47000。如果每人的所有三个价格加起来为卢比。 12000 然后使用矩阵方法找到 x、y 和 z 的值。

解决方案：

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.

问题 16. 两家工厂决定授予其员工三个价值观：(a) 适应新技术，(b) 在困难的情况下小心谨慎，以及 (c) 在紧张的情况下保持冷静，以 ₹ x, ₹ y 的比率, 和 ₹ 每人分别。一厂决定分别奖励2、4、3名员工，总奖金为29000卢比。二厂决定分别奖励5、2、3名员工，奖金为30500卢比。如果每人三项奖金加起来花费9500卢比，然后

(i) 用矩阵方程表示上述情况，用矩阵乘法形成线性方程。

(ii) 用矩阵法求解这些方程。

解决方案：

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.

问题 17. A 和 B 两所学校希望以真诚、真实和乐于助人的价值观来奖励他们选择的学生。学校 A 希望分别向 3 名、2 名和 1 名学生分别奖励 3、2 和 1 名学生，每人 ₹x、每人 y 和 ₹z，总奖金为 ₹1,600。学校 B 希望花费 2,300 卢比来奖励其 4、1 和 3 名学生各自的价值观（通过与以前一样给予三个价值观相同的奖励）。如果每个值的一个奖项的总奖励金额为 900 卢比，则使用矩阵找到每个值的奖励金额。除了这三个值之外，建议另外一个值，应该考虑奖励。

解决方案：

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 C_ij be the cofactors of the elements a_ij 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.

问题 18. P 和 Q 两所学校希望根据纪律、礼貌和守时的价值观来奖励他们选定的学生。学校 P 想要向其 3、2 和 1 名学生分别奖励 ₹x、每个 ₹y 和 ₹z，这三个值分别为 ₹1,000。 Q 学校希望花费 1,500 卢比来奖励其 4、1 和 3 名学生各自的价值观（通过与以前一样为三个价值观提供相同的奖励）。如果每个值的一个奖项的总奖励金额为 600 卢比，则使用矩阵找到每个值的奖励金额。除了以上三个值，建议多一个奖励值。

解决方案：

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.

问题 19。两所学校 P 和 Q 想要奖励他们选择的学生宽容、善良和领导力的价值观。学校 P 希望分别向 3、2 和 1 名学生分别奖励 3、2 和 1 名学生，奖金总额为 2,200 卢比。 Q 学校希望花费 3,100 卢比来奖励其 4、1 和 3 名学生各自的价值观（通过向学校 P 的三个价值观提供相同的奖励）。如果每个值的一个奖项的总奖励金额为 1,200 卢比，则使用矩阵找到每个值的奖励金额。

除了这三个值之外，建议另外一个值，应该考虑奖励。

解决方案：

​​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.

问题 20. 总计 7000 卢比存入三个不同的储蓄银行账户，年利率分别为 5%、8% 和 8.5%。这三个账户的年利息总额为 550 卢比。相等的金额已存入 5% 和 8% 的储蓄账户。在矩阵的帮助下，找出三个账户中每个账户的存款金额。

解决方案：

​​​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.

问题 21. 一位店主有 3 种钢笔“A”、“B”和“C”。 Meenu 以 21 卢比的价格购买了每种品种的 1 支钢笔。Jeevan 以 60 卢比的价格购买了 4 支“A”品种、3 支“B”品种和 2 支“C”品种。而 Shikha 购买了 6 支“A”品种' 品种，2 支'B' 品种，3 支'C' 品种，售价 70 卢比。使用矩阵法，找出每种钢笔的成本。

解决方案：

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.