If adj[ \(\begin{bmatrix} 1 & 0 & 2\\ -1 & 1 & -2\\ 0 & 2 & 1 \end{bmatrix}\)= \(\begin{bmatrix} 5 & m & -2\\ 1 & 1 & 0\\ -2 & -2 & n \end{bmatrix}\) then m+n =
2
-3
5
-5
To solve the problem, we need to find the value of \( m+n \) from the given condition that the adjugate of a matrix is equal to another matrix.
1. Understanding the Adjugate:
The adjugate (adjoint) of a matrix is the transpose of the cofactor matrix. Given that: \[ \text{adj} \left( \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix} \right) = \begin{bmatrix} 5 & m & -2 \\ 1 & 1 & 0 \\ -2 & -2 & n \end{bmatrix} \] we compute the cofactor matrix and then compare it with the given matrix to find \( m \) and \( n \).
2. Finding the Cofactors:
Let’s denote the original matrix as \( A \). Compute cofactors of \( A \):
3. Matching Cofactor Transpose to Adjugate:
In the adjugate matrix (which is the transpose of the cofactor matrix):
From the matrix, we can now read off the unknowns:
From position (1,2): \( m = 4 \)
From position (3,3): \( n = 1 \)
4. Calculate \( m + n \):
\[ m + n = 4 + 1 = 5 \]
Final Answer: \( {5} \)
Let \( z \) satisfy \( |z| = 1, \ z = 1 - \overline{z} \text{ and } \operatorname{Im}(z)>0 \)
Then consider:
Statement-I: \( z \) is a real number
Statement-II: Principal argument of \( z \) is \( \dfrac{\pi}{3} \)
Then:
A determinant can be defined in many ways for a square matrix.
The first and most simple way is to formulate the determinant by taking into account the top-row elements and the corresponding minors. Take the first element of the top row and multiply it by its minor, then subtract the product of the second element and its minor. Continue to alternately add and subtract the product of each element of the top row with its respective min or until all the elements of the top row have been considered.
For example let us consider a 1×1 matrix A.
A=[a1…….an]
Read More: Properties of Determinants
Second Method to find the determinant:
The second way to define a determinant is to express in terms of the columns of the matrix by expressing an n x n matrix in terms of the column vectors.
Consider the column vectors of matrix A as A = [ a1, a2, a3, …an] where any element aj is a vector of size x.
Then the determinant of matrix A is defined such that
Det [ a1 + a2 …. baj+cv … ax ] = b det (A) + c det [ a1+ a2 + … v … ax ]
Det [ a1 + a2 …. aj aj+1… ax ] = – det [ a1+ a2 + … aj+1 aj … ax ]
Det (I) = 1
Where the scalars are denoted by b and c, a vector of size x is denoted by v, and the identity matrix of size x is denoted by I.
Read More: Minors and Cofactors
We can infer from these equations that the determinant is a linear function of the columns. Further, we observe that the sign of the determinant can be interchanged by interchanging the position of adjacent columns. The identity matrix of the respective unit scalar is mapped by the alternating multi-linear function of the columns. This function is the determinant of the matrix.