To find the rank of the given matrix, we can use the method of row reduction (Gaussian elimination) to transform the matrix into its row echelon form. The rank is then the number of non-zero rows in the echelon form.
The given matrix is:
$ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & -1 \\ 3 & 1 & 1 \end{bmatrix} $We apply elementary row operations to simplify the matrix:
Performing these operations:
$ \begin{bmatrix} 1 & 1 & 1 \\ 1-1 & -1-1 & -1-1 \\ 3-3(1) & 1-3(1) & 1-3(1) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -2 \\ 0 & -2 & -2 \end{bmatrix} $Now, subtract Row 2 from Row 3 ($R_3 \leftarrow R_3 - R_2$):
$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -2 \\ 0-0 & -2-(-2) & -2-(-2) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & -2 \\ 0 & 0 & 0 \end{bmatrix} $The matrix is now in row echelon form. The number of non-zero rows determines the rank.
There are 2 non-zero rows.
Therefore, the rank of the matrix is 2.
Let A be a skew-symmetric matrix of order 3.
What is the value of det(4A4) - det(3A3) + det(2A2) - det(A) + det(-I) where I is the identity matrix of order 3?
The system of linear equations
x + 2y + z = 4, 2x + 4y + 2z = 8 and 3x + 6y + 3z = 10 has
An ordered pair $(\alpha, \beta)$ for which the system of linear equations
$\alpha x + (\beta+1)y + z = 2$
$2\alpha x + (\beta+2)y + z = 3$
$\alpha x + \beta y + 2z = 2$ has a unique solution, is
Let A and B be two non zero square matrics and AB and BA both are defined. It means