Question

For an$\mathbf{N}\mathbf{\times }\mathbf{N}$square matrix$\mathbf{A}$, in$\mathbf{(}\mathbf{N}\mathbf{-}\mathbf{1}\mathbf{)}$steps, the technique of Gauss elimination can transform into an ________ matrix.

Short answer

Therefore, for an $N\times N$square matrix $A$, in $\left(N-1\right)$steps, the technique of Gauss elimination can transform into an upper triangular matrix.

Step-by-step solution

Determine the formula of linear equation.

Write the formula of linear algebraic equation.

$\mathbf{Ax}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$……. (1)

Here, $x$and $y$are $N$ vectors and $A$ is $N\times N$ square matrix.

Determine the answer.

Write equation (1) in matrix form.

$\left[\begin{array}{ccc}{A}_{11}& \dots & A_{1N}\\ -& -& -\\ -& -& -\\ A_{N1}& \dots & A_{NN}\end{array}\right]\left[\begin{array}{c}{x}_1\\ -\\ -\\ x_N\end{array}\right]-\left[\begin{array}{c}{y}_1\\ -\\ -\\ y_N\end{array}\right]=0$

The objective is to find $x$when $A$and $y$are given.

The $x$ can be found easily when matrix $A$ is upper triangular, with non-zero diagonal elements, that is,

$\left[\begin{array}{ccc}{A}_{11}& \dots & A_{1N}\\ -& -& -\\ 0& -& -\\ 0& 0& A_{NN}\end{array}\right]\left[\begin{array}{c}{x}_1\\ -\\ x_{N-1}\\ x_N\end{array}\right]-\left[\begin{array}{c}{y}_1\\ -\\ y_{N-1}\\ y_N\end{array}\right]=0$

The last equation is solved as,

$X_N=\frac{y_N}{A_{NN}}$

Next to last equation is solved as,

$x_{N-1}=\frac{y_{N-1}-A_{N-1,N}{x}_N}{A_{N-1,N-1}}$

So, if the matrix$A$is not an upper triangular matrix, then transform it to equivalentequations with an upper triangular matrix.

This transformation is called as gauss eliminations given by the $N-1$ steps.

Therefore, for an $N\times N$square matrix $A$, in $N-1$ steps, the technique of Gauss elimination can transform into an upper triangular matrix.