第12类NCERT解决方案-数学第I部分–第3章矩阵–练习3.1

问题1.在矩阵A = $\begin{vmatrix} 2&5&19&-7\\ 35&-2&5/2&12\\ \sqrt{3}&1&-5&17\\ \end{vmatrix}$ ， 写：

(i)矩阵的顺序

解决方案：

We can see that matrix contains 3 rows and 4 columns So, the order of this matrix is 3×4

(ii)要素数量

解决方案：

We know that number of elements in the matrix = product of number of rows and number of columns in matrix So, number of elements = 3 x 4 =12.

(iii)写出元素a ₁₃ ，a ₂₁ ，a ₃₃ ，a ₂₄ ，a ₂₃

解决方案：

a₁₃ = Element in first row and third column i.e, 19       

a₂₁ = Element in second row and first column i.e, 35    

a₃₃ = Element in third row and third column i.e, -5

a₂₄ = Element in second row and fourth column i.e, 12

a₂₃ = Element in second row and  third column i.e, 5/2

问题2.如果一个矩阵有24个元素，它可能具有的阶数是多少?如果它有13个元素，该怎么办?

解决方案：

We know that number of elements in the matrix is the product of number 

of rows and number of columns in the matrix .

If matrix has order mxn then number  of elements are mn in that matrix.

So we have to find the ordered pairs of natural number whose product is 24.  

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)

Hence possible orders are: 1×24, 24×1, 2×12, 12×2, 3×8, 8×3, 4×6, and 6×4

If matrix has 13 elements then ordered pairs will be (1, 13) and (13, 1)

Hence possible orders are: 1×13 and 13×1

问题3.如果一个矩阵有18个元素，它可能具有的阶数是多少?如果有5个元素，该怎么办?

解决方案：

We know that number of elements in the matrix is the product of number 

of rows and number of columns in the matrix .

If matrix has order mxn then number  of elements are mn in that matrix.

So we have to find the ordered pairs of natural number whose product is 18.  

The ordered pairs are:(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), and (6, 3)

Hence possible orders are: 1×18, 18×1, 2×9, 9×2, 3×6, and 6×3

If matrix has 5 elements then ordered pairs will be (1, 5) and (5, 1)  

Hence possible orders are: 1×5 and 5×1

问题4.构造一个2×2矩阵A = [a _ij ]，其元素由下式给出：

_(ⅰ)IJ =(I + ^J)2分之2

解决方案：

Elements in this 2×2 matrix = a₁₁ , a₁₂ ,a₂₁ ,a₂₂

a₁₁ => i = 1 and j = 1 => (1 + 1)²/2 = 4/2 = 2

a₁₂ => i = 1 and j = 2 => (1 + 2)²/2 = 9/2

a₂₁ => i = 2 and j = 1 => (2 + 1)²/2 = 9/2

a₂₂ => i = 2 and j = 2 =>(2 + 2)²/2 = 16/2 = 8

Resulatnt Matrix is:

$\begin{vmatrix} 2&9/2\\ 9/2&8\\ \end{vmatrix}$

(ii)一个_ij = i / j

解决方案：

Elements in this 2×2 matrix = a₁₁ , a₁₂ ,a₂₁ ,a₂₂

a₁₁ => i = 1 and j = 1 = 1/1 = 1

a₁₂ => i = 1 and j = 2 = 1/2

a₂₁ => i = 2 and j = 1 = 2/1 = 2

a₂₂ => i = 2 and j = 2 = 2/2 = 1

Resulatnt Matrix is:  

$\begin{vmatrix} 1&1/2\\ 2&1\\ \end{vmatrix}$

_(ⅲ)IJ =(I + ^2J)2分之2

解决方案：

Elements in this 2×2 matrix = a₁₁ , a₁₂ , a₂₁ , a₂₂

a₁₁ => i = 1 and j = 1 => (1 + 2 x 1)²/2 = 9/2

a₁₂ => i = 1 and j = 2 => (1 + 2 x 2)²/2 = 25/2

a₂₁ => i = 2 and j = 1 =>(2 + 2 x 1)²/2  = 16/2 = 8

a₂₂ => i = 2 and j = 2 =>(2 + 2 x 2)²/2 = 36/2 = 18

Resulatnt Matrix is:

$\begin{vmatrix} 9/2&25/2\\ 8&18\\ \end{vmatrix}$

问题5.构造一个3×4矩阵，其元素由下式给出：

(i) _ij = 1/2 {| -3i + j |}

解决方案：

Elements in this 3 x 4 matrix are  a₁₁ , a₁₂ , a₁₃ , a₁₄ , a₂₁ , a₂₂, a₂₃ , a₂₄  , a₃₁ , a₃₂ , a₃₃ , a₃₄

a₁₁ => i = 1 and j = 1 => 1/2 (|-3 x 1 + 1|) = 1

a₁₂ =>  i = 1 and j = 2 => 1/2 (|-3 x 1 + 2|) = 1/2

a₁₃ => i = 1 and j = 3 => 1/2 (|-3 x 1 + 3) = 0

a₁₄ => i = 1 and j = 4 => 1/2 (|-3 x 1 + 4|) = 1/2

a₂₁ => i = 2 and j = 1 => 1/2 (|-3 x 2 + 1|) = 5/2

a₂₂ => i = 2 and j = 2 => 1/2 (|-3 x 2 + 2|) = 2

a₂₃ => i = 2 and j = 3 => 1/2 (|-3 x 2 + 3|) = 3/2

a₂₄ => i = 2 and j = 4 => 1/2 (|-3 x 2 + 4|) = 1

a₃₁ => i = 3 and j = 1 => 1/2 (|-3 x 3 + 1|) = 4

a₃₂ => i = 3 and j = 2 => 1/2 (|-3 x 3 + 2|) = 7/2

a₃₃ => i = 3 and j = 3 => 1/2 (|-3 x 3 + 3|) = 3

a₃₄ => i = 3 and j = 4 => 1/2 (|-3 x 3 + 4|) = 5/2

Resultant matrix is:  

$\begin{vmatrix} 1&1/2&0&1/2\\ 5/2&2&3/2&1\\ 4&7/2&3&5/2\\ \end{vmatrix}$

(ii)a _ij = 2i – j

解决方案：

Elements in this 3 x 4 matrix are  a₁₁ , a₁₂ , a₁₃ , a₁₄ , a₂₁ , a₂₂ , a₂₃ , a₂₄  , a₃₁ , a₃₂ , a₃₃ , a₃₄

So,

a₁₁ => i = 1 and j = 1 => 2 x 1 – 1 = 1

a₁₂ => i = 1 and j = 2 => 2 x 1 – 2 = 0

a₁₃ => i = 1 and j = 3 => 2 x 1 – 3 = -1

a₁₄ => i = 1 and j = 4 => 2 x 1 – 4 = -2

a₂₁ => i = 2 and j = 1 => 2 x 2 – 1 = 3

a₂₂ => i = 2 and j = 2 => 2 x 2 – 2 = 2

a₂₃ => i = 2 and j = 3 => 2 x 2 – 3 = 1

a₂₄ => i = 2 and j = 4 => 2 x 2 – 4 = 0

a₃₁ => i = 3 and j = 1 => 2 x 3 – 1 = 5

a₃₂ => i = 3 and j = 2 => 2 x 3 – 2 = 4

a₃₃ => i = 3 and j = 3 => 2 x 3 – 3 = 3

a₃₄ => i = 3 and j = 4 => 2 x 3 – 4 = 2

Resultant matrix is: $\begin{vmatrix} 1&0&-1&-2\\ 3&2&1&0\\ 5&4&3&2\\ \end{vmatrix}$ 

问题6.从以下方程式中找到x，y和z的值：

(一世) $\begin{vmatrix} 4&3\\ x&5\\ \end{vmatrix} = \begin{vmatrix} y&z\\ 1&5\\ \end{vmatrix}$

解决方案：

We can compare or equate both the matrices because both are equal 

So on equating both the matrices we get 

x = 1; y = 4; z = 3

(ii) $\begin{vmatrix} x+y&2\\ 5+z&xy\\ \end{vmatrix} = \begin{vmatrix} 6&2\\ 5&8\\ \end{vmatrix}$

解决方案：

We can compare or equate both the matrices because both are equal

So, on equating both the matrices. we get

x + y = 6         -(1)

5 + z = 5         -(2)

xy = 8         -(3)

Now, we can solve these equations   

z = 0 from eq(2)

x = 6 – y         -(4)

Now putting value of x from eq(4) in eq(3)

 (6 – y)(y) = 8

6y – y² = 8

y² – 6y + 8 = 0         -(5)

Now we have to factorize this equation

(y – 4)(y – 2) = 0

either y – 4 = 0 or y – 2 = 0

so, y = 2 or y = 4

Put these values in eq(4) we get 

x = 4 and x = 2

Therefore, the value of  x = 2 , y = 4 , z = 0

(iii) $\begin{vmatrix} x+y+z\\ x+z\\ y+z\\ \end{vmatrix} = \begin{vmatrix} 9\\ 5\\ 7\\ \end{vmatrix}$

解决方案：

We can compare or equate both the matrices because both are equal

So, on equating both the matrices, we get

x + y + z = 9          -(1) 

x + z = 5          -(2) 

y + z = 7          -(3) 

If we put the value of eq(2) in eq(1)

 we get, 5 + y = 9

y = 4 

On putting value of y in eq(3)

4 + z = 7

z = 3

On putting value of z in eq(2)

x + 3 = 5

x = 2

So, the value of  x = 2; y = 4; z = 3

问题7.从等式中找到a，b，c和d的值：

$\begin{vmatrix} a-b&2a+c\\ 2a-b&3c+d\\ \end{vmatrix} = \begin{vmatrix} -1&5\\ 0&13\\ \end{vmatrix}$

解决方案：

We can compare or equate both the matrices because both are equal

So, on equating both the matrices, we get

a – b = -1          -(1)   

2a – b = 0          -(2) 

2a + c= 5          -(3) 

3c + d = 13          -(4) 

On solving eq(1) and eq(2) we get 

a = 1

On putting a = 1 in eq(3) we get 

c = 3

On putting a = 1 in eq(2) we get 

b = 2

On putting c = 3 in eq(4) we get 

d = 4

So, the value of  a = 1; b = 2; c = 3; d = 4

问题8。A = [a _ij ] _mxn是一个方矩阵，如果

(A)m n(C)m = n(D)这些都不是

解决方案：

This will be square matrix if number of rows = number of columns

So , m = n is correct option.

Hence, the option answer is C.

问题9.给定的x和y值中的哪个使下面的矩阵对相等$\begin{vmatrix} 3x+7&5\\ y+1&2-3x\\ \end{vmatrix}, \begin{vmatrix} 0&y-2\\ 8&4\\ \end{vmatrix}$

(A)x = -1/3，y = 7(B)找不到(C)y = 7，x = -2/3(D)x = -1/3，y = -2/3

解决方案：

We can compare or equate both the matrices because both are equal

So on equating both the matrices we get;

3x + 7 = 0           -(1)

y + 1 = 8           -(2)

2 – 3x = 4           -(3)

y – 2 = 5           -(4)

From eq(2) and eq(4) we get same value of y i.e, y=7

but on solving eq(1) we get value of x = -7/3 and on solving eq(3) we get value of x = -2/3

Both the values of x are different for the value of y. So, it is not possible to find.

Hence, the correct option is B

问题10.每个条目为0或1的3×3阶所有可能矩阵的数目为：

(A)27(B)18(C)81(D)512

解决方案：

We know that number of elements in a matrix of order mxn is mn. 

So number of elements in matrix of 3 x 3 is 9.

For each element we have two choices either 0 or 1

So, total number of possible matrices of order 3 x 3 with each entry 0 or 1 = 2⁹ = 512

Correct option is D