第 12 课 NCERT 解决方案 - 数学第一部分 - 第 3 章矩阵 - 练习 3.2 |设置 2

第 3 章矩阵 - 练习 3.2 |设置 1

问题 11. 如果$x\left[\begin{array}{l} 2 \\ 3 \end{array}\right]+y\left[\begin{array}{c} -1 \\ 1 \end{array}\right]=\left[\begin{array}{l} 10 \\ 5 \end{array}\right]$ ，求 x 和 y 的值。

解决方案：

Given: $x\left[\begin{array}{l} 2 \\ 3 \end{array}\right]+y\left[\begin{array}{c} -1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{c} 2 x \\ 3 x \end{array}\right]+\left[\begin{array}{c} -y \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{c} 2 x-y \\ 3 x+y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right]$

Equating corresponding entries, we have

2x – y = 10           -(1)

3x + y = 5           -(2)

Adding eq.(1) and (2), we have 5x = 15 ⇒ x = 3

Putting x = 3 in eq.(2)

9 + y = 5 ⇒ y = -4

Therefore, x = 3 and y = -4

问题 12. 给定$3\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{cc} x & 0 \\ -1 & 2 w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right]$ ，求 x、y、z 和 w 的值。

解决方案：

Given: $3\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{cc} x & 0 \\ -1 & 2 w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right]$

$\Rightarrow\left[\begin{array}{ll} 3 x & 3 y \\ 3 z & 3 w \end{array}\right]=\left[\begin{array}{cc} x+4 & 6+x+y \\ -1+z+w & 2 w+3 \end{array}\right]$

Equating corresponding entries, we have

3x = x + 4 ⇒ 2x = 4 ⇒ x = 2

and 3y = 6 + x + y

⇒ 2y = 6 + 2

⇒ 2y = 8

⇒ y = 4

and 3z = -1 + z + w ⇒ 2z – w = – 1           -(1)

and 3w = 2w + 3 ⇒ w = 3

Putting w = 3 in eq(i), 2z – 3 = -1  

⇒ 2z = 2 ⇒ z = 1

Therefore, x = 2, y = 4, z = 1, w = 3

问题 13. 如果$F(x)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]$ ，证明 F(x) F(y) = F(x + y)。

解决方案：

$\begin{aligned} &\text { } F(x)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right], F(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &F(x+y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &F(x) F(y)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}$

$=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right]$

$=\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y+0 & -\cos x \sin y-\sin x \cos y+0 & 0 \\ \sin x \cos y+\cos x \sin y+0 & -\sin x \sin y+\cos x \cos y+0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

= F(x + y) 

= F(x) F(y) = F(x + y) 

问题 14. 证明

$(i) \left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right] \neq\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]$

$\text { (ii) }\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \neq\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]$

解决方案：

(i) L.H.S =$\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\\ =\left[\begin{array}{ll} 5(2)-1(3) & 5(1)-1(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{array}\right]\\ =\left[\begin{array}{cc} 10-3 & 5-4 \\ 12+21 & 6+28 \end{array}\right]\\ =\left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right] \ \ \ -(1)$

R.H.S = $\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{ll} 5 & -1 \\ 6 & 7 \end{array}\right]\\ =\left[\begin{array}{ll} 2(5)+1(6) & 2(-1)+1(7) \\ 3(5)+4(6) & 3(-1)+4(7) \end{array}\right]\\ =\left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right]\\ =\left[\begin{array}{ll} 16 & 5 \\ 39 & 25 \end{array}\right] \ \ \ -(2)$

Therefore, from (1) and (2), we get

$\text { }\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right] \neq\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]$

i.e. L.H.S. ≠ R.H.S

(ii) L.H.S = $\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]$

Multiply both the matrices 

$=\left[\begin{array}{lll} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\ 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\ 1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4) \end{array}\right]\\ =\left[\begin{array}{ccc} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right]$

R.H.S.= $\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]$

$\begin{array}{l} =\left[\begin{array}{ccc} -1(1)+1(0)+0(1) & (-1) 2+1(1)+0(1) & (-1) 3+1(0)+0(0) \\ 0(1)+(-1) 0+1(1) & (0) 2+1(-1)+1(1) & (0) 3+0(-1)+1(0) \\ 2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0) \end{array}\right] \\ =\left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \end{array}$

Therefore,

L.H.S. ≠ R.H.S.

i.e.$\text { }\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \neq\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]$

问题 15. 求 A ² – 5A + 6I，如果$A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\\\end{bmatrix}$

解决方案：

$\begin{aligned} &A^{2}-5 A+6 I=\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]-5\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]+6\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &=\left[\begin{array}{lll} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1-0 & 1-3+0 \end{array}\right]-\left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \end{aligned}$

$\left.\begin{array}{l} =\left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-\left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \end{array}\right]$

$\begin{array}{l} =\left[\begin{array}{ccc} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2+0+6 \end{array}\right] \\ =\left[\begin{array}{ccc} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{array}\right] \end{array}$

问题 16. 如果$A =\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\\\end{bmatrix}$ , 证明 A ³ – 6A ² + 7A + 2I = 0

解决方案：

$A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\\\end{bmatrix} A^{2} \\=A * A=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] \\ =\left[\begin{array}{lll} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{array}\right] \\ =\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]$

$6 A^{2} =6\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]=\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right] \\ A^{3} =A^{2} \times A \\ =\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] \\ =\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]$

$A^{3} - 6 A^{2}+7 A+2 I=\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right]+\left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right]+\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]\\ =\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

= 0 (Zero matrix)

= R.H.S.

Hence Proved

问题 17. 如果$A=\begin{bmatrix}3&-2\\4&-2\\\end{bmatrix} and \:I=\begin{bmatrix}1&0\\0&1\\\end{bmatrix}$ , 求 k 使得 A ² = kA – 2I

解决方案：

Given: 

$A=\left[\begin{array}{rr} 3 & -2 \\ 4 & -2 \end{array}\right] \text { and } I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ A^{2}=k A-2 I \Rightarrow\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=k\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ \Rightarrow\left[\begin{array}{cc} 9-8 & -6+4 \\ 12-8 & -8+4 \end{array}\right]=\left[\begin{array}{cc} 3 k & -2 k \\ 4 k & -2 k \end{array}\right]-\left[\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right]=\left[\begin{array}{ll} 3 k-2 & -2 k-0 \\ 4 k-0 & -2 k-2 \end{array}\right]$

Equating corresponding entries, we have 

3k – 2 = 1 

3k = 3  

k = 1

and 4k = 4 

k = 1 

and -4 = -2k – 2

2k = 2 

k = 1

Therefore, k = 1 

问题 18. 如果$A =\begin{bmatrix}0&-tan\frac{α}{2}\\tan\frac{α}{2}&0\\\end{bmatrix}$ I 是 2 阶单位矩阵，证明 I + A = (I – A) $\begin{bmatrix}cosα&-sinα\\sinα&cosα\\\end{bmatrix}$

解决方案：

$\begin{array}{l} \text { L.H.S. } I+A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{cc} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right] \\ \text { Now, } I-A=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right] \\ \text { R.H.S. }=(I-A)\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right]\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \end{array}$

$\begin{aligned} &=\left[\begin{array}{ccc} \cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\ -\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha \\ \end{array}\right]\\ &\text {} \end{aligned}$

$=\left[\begin{array}{ccc} \cos \alpha \cos \frac{\alpha}{2}+\sin \alpha \sin \frac{\alpha}{2}{\cos \frac{\alpha}{2}} & \frac{\alpha \sin \alpha \cos \frac{\alpha}{2}+\cos \alpha \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \ \\ \hline \frac{-\cos \alpha \sin \frac{\alpha}{2}+\sin \alpha \cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \frac{\sin \alpha \sin \frac{\alpha}{2}+\cos \alpha \cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \end{array}\right]$

$\begin{aligned} &=\left[\begin{array}{cc} \frac{\cos \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} & \frac{-\sin \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} \\ \frac{\sin \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} & \frac{\cos \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} \end{array}\right]=\left[\begin{array}{ccc} \frac{\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \frac{-\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \\ \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \frac{\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \end{array}\right]=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right]\end{aligned}$

L.H.S. = R.H.S.

Hence, Proved. 

问题 19。信托基金有 30,000 卢比，必须投资于两种不同类型的债券。第一个债券每年支付 5% 的利息，第二个债券每年支付 7% 的利息。使用矩阵乘法，确定如何在两种债券中划分 ₹30,000。如果信托基金必须获得以下年度总利息：

(a) 1800 卢比

(b) 2000 卢比

解决方案：

Let invested in the first bond = Rs x 

Then, the sum of money invested in the second bond = ₹(30000 – x)

It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.

Thus, in order to obtain an annual total interest of ₹1800, we get:

$\begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}5/100\\7/100\end{bmatrix}=1800$

⇒ 5x/100 + 7(30000 − x)/100 = 1800

⇒ 5x + 210000 -7x = 180000

⇒ 210000 -2x = 180000

⇒ 2x = 210000 – 180000

⇒ 2x = 30000

⇒ x = 15000

Therefore, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹15000 in the first bond and the remaining ₹15000 in the second bond.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:

X = $\begin{bmatrix}x&30000-x\end{bmatrix}$

Hence, the interest obtained after one year can be expressed in matrix representation as:

$\begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}5/100\\7/100\end{bmatrix}=2000$

⇒ 5x/100 + 7(30000 − x)/100 = 2000

⇒ 5x + 210000 − 7x = 200000

⇒ 210000 − 2x = 200000

⇒ 2x = 210000 – 200000

⇒ 2x = 10000

⇒ x = 5000

Therefore, in order to obtain an annual total interest of ₹2000, the trust fund should invest ₹5000 in the first bond and the remaining ₹(30000 − 5000) = ₹25000 in the second bond.

问题20。某学校的书店有10打化学书，8打物理书，10打经济学书。它们的售价分别为 80 卢比、60 卢比和 40 卢比。使用矩阵代数找出书店从出售所有书籍中将获得的总金额。

解决方案：

Let the number of books as 1 × 3 matrix = B = $\begin{bmatrix}10 dozen&8dozen&10dozen\\10*12=120&8*12=96&10*12=120\\\end{bmatrix}$

Let the selling prices of each book is a 3 × 1 matrix S = $\begin{bmatrix}80\\60\\40\end{bmatrix}$

Therefore, Total amount received by selling all books = BS = $\begin{bmatrix}120&96&120\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}$

$\begin{bmatrix}120(80)&96(60)&120(40)\end{bmatrix}=\begin{bmatrix}9600&5760&4800\end{bmatrix}=\begin{bmatrix}20160\end{bmatrix}$

Therefore, Total amount received by selling all the books = Rs 20,160

假设 X、Y、Z、W 和 P 分别是 2 × n、3 × k、2 × p、n × 3 和 p × k 阶矩阵。在练习 21 和 22 中选择正确的答案。

问题 21. 对 n、k 和 p 的限制，以便定义 PY + WY：

(A) k = 3, p = n (B) k 是任意的, p = 2

(C) p 是任意的，k = 3 (D) k = 2，p = 3

解决方案：

Since, Matrices P and Y are of the orders p × k and 3 × k respectively.

Therefore, matrix PY will be defined if k = 3.

Then, PY will be of the order p × k = p × 3.

Matrices W and Y are of the orders n × 3 and 3 × k = 3 × 3 respectively.

As, the number of columns in W is equal to the number of rows in Y, Matrix WY is well-defined and is of the order n × 3.

Matrices PY and WY can be added only when their orders are the same.

Therefore, PY is of the order p × 3 and WY is of the order n × 3.

Thus, we must have p = n.

Therefore, k = 3 and p = n are the restrictions on n, k and p so that PY + WY will be defined.

Therefore, answer is (A)

问题 22. 如果 n = p，则矩阵 7X – 5Z 的阶数为：

(A) p × 2 (B) 2 × n

(C) n × 3 (D) p × n

解决方案：

Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of order 2 × p = 2 × n               -(∵ p = n)

Then, Matrix 5Z is also of the same order.

Now, both the matrices 7X and 5Z are of the order 2 × n.

Thus, matrix 7X – 5Z is well- defined and is of the order 2 × n.

Therefore, answer is (B)