Class 12 RD Sharma解决方案–第五章矩阵代数–练习5.2 |套装1

问题1(i)：计算以下总和： $\begin{bmatrix} 3 & -2\\ 1 & 4 \end{bmatrix} + \begin{bmatrix} -2 & 4\\ 1 & 3 \end{bmatrix}$ 。

解决方案：

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 2×2.

=>$\begin{bmatrix} 3-2 & -2+4 \\ 1+1 & 4+3 \end{bmatrix}$

=>$\begin{bmatrix} 1 & 2 \\ 2 & 7 \end{bmatrix}$

问题1(ii)：计算以下总和： $\begin{bmatrix} 2 & 1 & 3 \\ 0 & 3 & 5 \\ -1 & 2 & 5 \\ \end{bmatrix} + \begin{bmatrix} 1 & -2 & 3 \\ 2 & 6 & 1 \\ 0 & -3 & 1 \\ \end{bmatrix}$ 。

解决方案：

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 3×3.

=>$\begin{bmatrix} 2+1 & 1-2 & 3+3 \\ 0+2 & 3+6 & 5+1\\ -1+0 & 2-3 & 5+1 \\ \end{bmatrix}$

=>$\begin{bmatrix} 3 & -1 & 6 \\ 2 & 9 & 6 \\ -1 & -1 & 6 \\ \end{bmatrix}$

问题2：让A = $\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ ，B = $\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$和C = $\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$ 。查找以下每个：

(i)：2A – 3B

解决方案：

Both the matrices A and B are of the same order which is 2×2, hence the operation can be performed.

=> 2A =$2\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 6 & 4 \end{bmatrix}$

=> 3B =$3\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix}$

=> 2A – 3B =$\begin{bmatrix} 4-3 & 8-9 \\ 6+6 & 4-15 \end{bmatrix}$

=> 2A – 3B =$\begin{bmatrix} 1 & -1 \\ 12 & -11 \end{bmatrix}$

(ii)：B – 4C

解决方案：

Both the matrices B and C are of the same order which is 2×2, hence the operation can be performed.

=> B =$\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$

=> 4C =$4\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix}$

=> B – 4C =$\begin{bmatrix} 1+8 & 3-20 \\ -2-12 & 5-16 \end{bmatrix}$

=> B – 4C =$\begin{bmatrix} 9 & -17 \\ -14 & -11 \end{bmatrix}$

(iii)：3A – C

解决方案：

Both the matrices A and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =$3\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}$

=> C =$\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$

=> 3A – C =$\begin{bmatrix} 6+2 & 12-5 \\ 9-3 & 6-4 \end{bmatrix}$

=> 3A – C =$\begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

(iv)：3A -2B + 3C

解决方案：

The matrices A, B and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =$3\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}$

=> 2B =$2\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix}$

=> 3C =$3\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -6 & 15 \\ 9 & 12 \end{bmatrix}$

=> 3A – 2B + 3C =$\begin{bmatrix} 6-2-6 & 12-6+15 \\ 9+4+9 & 6-10+12 \end{bmatrix}$

=> 3A – 2B + 3C =$\begin{bmatrix} -2 & 21 \\ 22 & 8 \end{bmatrix}$

问题3：如果A = $\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$ ，B = $\begin{bmatrix} -1 & 0 & 2\\ 3 & 4 & 1 \end{bmatrix}$ ，C = $\begin{bmatrix} -1 & 2 & 3\\ 2 & 1 & 0 \end{bmatrix}$ ， 找：

(i)：A + B和B + C

解决方案：

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3.

B+C can be computed and is solved as follows:

=> B + C =$\begin{bmatrix} -1-1 & 0+2 & 2+3 \\ 3+2 & 4+1 & 1+0 \end{bmatrix}$

=> B + C =$\begin{bmatrix} -2 & 2 & 5 \\ 5 & 5 & 1 \end{bmatrix}$

(ii)：2B + 3A和3C – 4B

解决方案：

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3, and thus 2B + 3A can not be calculated.

3C – 4B can be computed and is solved as follows:

=> 3C =$3\begin{bmatrix} -1 & 2 & 3\\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} -3 & 6 & 9\\ 6 & 3 & 0 \end{bmatrix}$

=> 4B =$4\begin{bmatrix} -1 & 0 & 2\\ 3 & 4 & 1 \end{bmatrix} = \begin{bmatrix} -4 & 0 & 8\\ 12 & 16 & 4 \end{bmatrix}$

=> 3C – 4B =$\begin{bmatrix} -3+4 & 6+0 & 9-8 \\ 6-12 & 3-16 & 10-4 \end{bmatrix}$

=> 3C – 4B =$\begin{bmatrix} 1 & 6 & 1 \\ -6 & -13 & 6 \end{bmatrix}$

问题4：设A = $\begin{bmatrix} -1 & 0 & 2 \\ 3 & 1 & 4 \end{bmatrix}$ ，B = $\begin{bmatrix} 0 & -2 & 5 \\ 1 & -3 & 1 \end{bmatrix}$和C = $\begin{bmatrix} 1 & -5 & 2 \\ 6 & 0 & -4 \end{bmatrix}$ 。计算2A – 3B + 4C。

解决方案：

The result can be computed since A, B and C are of the same order which is 2×3.

=> 2A =$2\begin{bmatrix} -1 & 0 & 2 \\ 3 & 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 0 & 4 \\ 6 & 2 & 8 \end{bmatrix}$

=> 3B =$3\begin{bmatrix} 0 & -2 & 5 \\ 1 & -3 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -6 & 15 \\ 3 & -9 & 3 \end{bmatrix}$

=> 4C =$4\begin{bmatrix} 1 & -5 & 2 \\ 6 & 0 & -4 \end{bmatrix} = \begin{bmatrix} 4 & -20 & 8 \\ 24 & 0 & -16 \end{bmatrix}$

=> 2A – 3B + 4C =$\begin{bmatrix} -2-0+4 & 0+6-20 & 4-15+8 \\ 6-3+24 & 2+9+0 & 8-3-16 \end{bmatrix}$

=> 2A – 3B + 4C =$\begin{bmatrix} 2 & -14 & -3 \\ 27 & 11 & -11 \end{bmatrix}$

问题5：如果A = diag(2，-5，9)，B = diag(1，1，-4)和C = diag(-6，3，4)，则找到：

(i)：A – 2B

解决方案：

In the given question A and B are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> A = diag(2, -5, 9)

=> 2B = 2. diag(1, 1, -4) = diag(2, 2, -8)

=> A – 2B = diag(2-2, -5-2, 9+8)

=> A – 2B = diag(0, -7, 17)

(ii)：B + C – 2A

解决方案：

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> B = diag(1, 1, -4)

=> C = diag(-6, 3, 4)

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> B + C – 2A = diag(1-6-4, 1+3+10, -4+4-18)

=> B + C – 2A = diag(-9, 14, -18)

(iii)：2A + 3B – 5C

解决方案：

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> 3B = 3. diag(1, 1, -4) = diag(3, 3, -12)

=> 5C = 5. diag(-6, 3, 4) = diag(-30, 15, 20)

=> 2A + 3B – 5C = diag(4+3+30, -10+3-15, 18-12-20)

=> 2A + 3B – 5C = diag(37, -22, -14)

问题6：给定矩阵A = $\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix}$ ，B = $\begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix}$和C = $\begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix}$ 。验证(A + B)+ C = A +(B + C)。

解决方案：

Given L.H.S :

=> (A + B) = $\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} + \begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix}$

=> (A + B) =$\begin{bmatrix} 2+9 & 1+7 & 1-1 \\ 3+3 & -1+5 & 0+4 \\ 0+2 & 2+1 & 4+6 \\ \end{bmatrix}$

=> (A + B) =$\begin{bmatrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \\ \end{bmatrix}$

=> (A + B) + C =$\begin{bmatrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \\ \end{bmatrix} + \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix}$

=> (A + B) + C =$\begin{bmatrix} 11+2 & 8-4 & 0+3 \\ 6+1 & 4-1 & 4+0 \\ 2+9 & 3+4 & 10+5 \\ \end{bmatrix}$

=> (A + B) + C =$\begin{bmatrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \\ \end{bmatrix}$

Given R.H.S :

=> (B + C) =$\begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix} + \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix}$

=> (B + C) =$\begin{bmatrix} 9+2 & 7-4 & -1+3 \\ 3+1 & 5-1 & 4+0 \\ 2+9 & 1+4 & 6+5 \\ \end{bmatrix}$

=> (B + C) =$\begin{bmatrix} 11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11 \\ \end{bmatrix}$

=> A + (B + C) =$\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} + \begin{bmatrix} 11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11 \\ \end{bmatrix}$

=> A + (B + C) =$\begin{bmatrix} 2+11 & 1+3 & 1+2 \\ 3+4 & -1+4 & 0+4 \\ 0+11 & 2+5 & 4+11 \\ \end{bmatrix}$

=> A + (B + C) =$\begin{bmatrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \\ \end{bmatrix}$

Hence R.H.S = L.H.S has been verified.

问题7：如果X + Y =，则求矩阵X和Y $\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}$和X – Y = $\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$ 。

解决方案：

We know that (X + Y) + (X – Y) = 2X.

=> (X + Y) + (X – Y) =$\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$

=> 2X =$\begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9-1 \end{bmatrix}$

=> 2X =$\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$

=> X =$\dfrac {1}{2}\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$

=> X =$\begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$

Now Y = (X + Y) – X

=> Y =$\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$

=> Y =$\begin{bmatrix} 5-4 & 2-4 \\ 0-0 & 9-4 \end{bmatrix}$

=> Y =$\begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}$

问题8：如果Y =，则找到X $\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$和2X + Y = $\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$ 。

解决方案：

Given 2X + Y =$\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$

=> 2X +$\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$=$\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$

=> 2X =$\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$

=> 2X =$\begin{bmatrix} 1-3 & 0-2 \\ -3-1 & 2-4 \end{bmatrix}$

=> 2X =$\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$

=> X =$\dfrac{1}{2}\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}$

=> X =$\begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$

问题9：如果2X – Y =，则求矩阵X和Y $\begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1\end{bmatrix}$和X + 2Y = $\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}$ 。

解决方案：

We know that 2 (2X – Y) + (X + 2Y) = 4X – 2Y + X + 2Y = 5X .

=> 2 (2X – Y) =$2\begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1\end{bmatrix}$

=> 2 (2X -Y) =$\begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 2\end{bmatrix}$

=> 2 (2X – Y) + (X + 2Y) =$\begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 2\end{bmatrix} + \begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}$

=> 5X =$\begin{bmatrix} 12+3 & -12+2 & 0+5\\ -8-2 & 4+1 & 2-7\end{bmatrix}$

=> 5X =$\begin{bmatrix} 15 & -10 & 5\\ -10 & 5 & -5\end{bmatrix}$

=> X =$\dfrac{1}{5}\begin{bmatrix} 15 & -10 & 5\\ -10 & 5 & -5\end{bmatrix}$

=> X =$\begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1\end{bmatrix}$

As (X + 2Y) =$\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}$

=>$\begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1\end{bmatrix} + 2Y = \begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}$

=> 2Y =$\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix} -\begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1\end{bmatrix}$

=> 2Y =$\begin{bmatrix} 0 & 4 & 4\\ 0 & 0 & -6\end{bmatrix}$

=> Y =$\dfrac{1}{2}\begin{bmatrix} 0 & 4 & 4\\ 0 & 0 & -6\end{bmatrix}$

=> Y =$\begin{bmatrix} 0 & 2 & 2\\ 0 & 0 & -3\end{bmatrix}$

问题10：如果X – Y = $\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix}$和X + Y = $\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix}$ ，找到X和Y。

解决方案：

We know that (X + Y) + (X – Y) = 2X.

=> 2X =$\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix}$

=> 2X =$\begin{bmatrix} 1+3 & 1+5 & 1+1\\ 1-1 & 1+1 & 0+4 \\1+11 & 0+8 & 0+0 \end{bmatrix}$

=> 2X =$\begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\12 & 8 & 0 \end{bmatrix}$

=> X =$\dfrac {1}{2}\begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\12 & 8 & 0 \end{bmatrix}$

=> X =$\begin{bmatrix} 2 & 3 & 1\\ 0 & 1 & 2 \\6 & 4 & 0 \end{bmatrix}$

Also (X + Y) – (X -Y) = 2Y.

=> 2Y =$\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix}$

=> 2Y =$\begin{bmatrix} 3-1 & 5-1 & 1-1\\ -1-1 & 1-1 & 4-0 \\11-1 & 8-0 & 0-0 \end{bmatrix}$

=> 2Y =$\begin{bmatrix} 2 & 4 & 0\\ -2 & 0 & 4 \\10 & 8 & 0 \end{bmatrix}$

=> Y =$\dfrac{1}{2}\begin{bmatrix} 2 & 4 & 0\\ -2 & 0 & 4 \\10 & 8 & 0 \end{bmatrix}$

=> Y =$\begin{bmatrix} 1 & 2 & 0\\ -1 & 0 & 2 \\5 & 4 & 0 \end{bmatrix}$