Difference Between Rows and Columns 

Ans. The collection of horizontal lines are called rows. A row is denoted as ‘i’. The collection of vertical lines are called columns.The column is denoted as ‘j’. 

For example, Consider an array of [a]i*j=3*3. It means in this case an array has 3 rows as well as 3 columns.

[a]_3*3 = \([ \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33} \end{matrix} ]\)

Ans. In the table, data is stored in the form of rows and tables. The stub in the table describes the row while the caption describes the column. In the row, data is arranged from left to right. On the other hand, data is arranged from top to bottom. The representation of data in the table is as follows:

Ans. A determinant is the collection of elements in the form of matrices. Basically, a matrix that contains elements in the form of rows and columns is called a determinant. Only square matrices have determinants. A determinant is denoted as del(A) or |A| or Δ, where A is a Matrix. A determinant is equal to the subtraction of the corresponding elements. Let us consider an example: 

if [A] = \([ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{matrix} ]\) 

 then its determinant is,

|A| = | a₁₁ a₂₂ - a₂₁ a₁₂  |

Ans. We have given matrix of 2*2 = | 2 4 -1 4 | 

The determinant of given matrix is = 2(4) - 4(-1) = 4 + 4 = 8 

Ans. We have given matrix [A]= \([ \begin{matrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{matrix} ]\)

In this matrix, the third column has two elements and is zero, so it is easy to expand the array by third column C3 .

After expanding and evaluating with C3 ,

|A| = 4 (-1 - 12)-0 +0

|A| = 4 (-13)

|A| = -52 

Ans. Let us consider a determinant |A|= \([ \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} ]\)

Expanding the given determinant using the Row Method, expanding through the first row.( Also, it can be expanded using Column Method, we will get the same result)

We get,

|A| = = a₁ (b₂ c₃ – b₃ c₂ ) – a₂ (b₁ c, – b₃ c, ) + a₃ (b₁ c₂ – b₂c₁ ) 

Now, Interchanging first and third rows, then the new determinant obtained is given by

|A|=Δ=| c₁ c₂ c₃

b₁ b₂ b₃ 

a₁ a₂ a₃ |

Expanding along third Row R₃, we get,

Δ1 = a₁ (c₂ b₃ – b₂ c₃ ) – a₂ (c₁ b₃ – c₃ b₁ ) + a₃ (b₂ c₁ – b₁ c₂ )

= – [a₁ (b₂ c₃ – b₃ c₂ ) – a₂ (b₁ c₃ – b₃ c, ) + a₃ (b₁ c₂ – b₂ c₁ )] 

Clearly Δ1= – Δ 

Hence, property 2 is verified.

Ans. Given, \([ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} ]\) \([ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & -1 & -2 \end{matrix} ]\)

Now, Multiply the given matrices, 

\([ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ]\)

In Matrix form, the given system of equation x-y+2z=1, 2y-3z=1, 3x–2y+4z=2 written as below,

\([ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} ]\) \([ \begin{matrix} x \\ y \\ z \end{matrix} ]\) = \([ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} ]\)

\([ \begin{matrix} x \\ y \\ z \end{matrix} ]\) = \([ \begin{matrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{matrix} ]\) \([ \begin{matrix} 1 \\ 1 \\ 2 \end{matrix} ]\)

\([ \begin{matrix} x \\ y \\ z \end{matrix} ]\) = \([ \begin{matrix} -2 & 0 & 2 \\ 9 & 2 & -6 \\ 6 & 1 & -4 \end{matrix} ]\)

\([ \begin{matrix} x \\ y \\ z \end{matrix} ]\) = \([ \begin{matrix} 0 \\ 5 \\ 3 \end{matrix} ]\) [x [-2+0+2

Compare the values with x,y,z coefficients, then 

 x=0 , y=5 , z= 3

Ans. Given, [A]= \([ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} ]\)

A2 = A. A = \([ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} ]\)\([ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} ]\)

A2 = \([ \begin{matrix} 7 & 12 \\ 4 & 7 \\ \end{matrix} ]\)

4A = 4 \([ \begin{matrix} 7 & 12 \\ 4 & 7 \\ \end{matrix} ]\)

4A =  \([ \begin{matrix} 8 & 12 \\ 4 & 8 \\ \end{matrix} ]\)

Now, A2- 4A + I = O 

\([ \begin{matrix} 7 & 12 \\ 4 & 7 \\ \end{matrix} ]\) - \([ \begin{matrix} 8 & 12 \\ 4 & 8 \\ \end{matrix} ]\) + \([ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} ]\) = \([ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} ]\)

Therefore, A2- 4A + I = 0, 

then,

A2- 4A = -I 

Multiplying both sides with A-1 (|A|≠0)

A.A(A-1) - 4A. A-1 = - I. A-1 

A.(A.A-1) - 4I = - A-1 

A.I - 4I = - A-1 

A-1 = 4I - A

A-1 = \([ \begin{matrix} 4 & 0 \\ 0 & 4 \\ \end{matrix} ]\) - \([ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} ]\)

A-1 = \([ \begin{matrix} 2 & 3 \\ -1 & 2 \\ \end{matrix} ]\)

Ans. Given, \([ \begin{matrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{matrix} ]\)

Using Row Method, Applying R1 → R1 + R2 to Δ, we get 

Δ= \([ \begin{matrix} x+y+z & y+z+x & z+x+y \\ z & x & y \\ 1 & 1 & 1 \end{matrix} ]\)

Since, the elements of R1 and R3 are proportional, then Δ = 0.