Question

For a set of linear algebraic equations in matrix format,$\mathbf{Ax}\mathbf{-}\mathbf{y}$, for a unique solution to exist, $\mathbf{det}\mathbf{}\left(A\right)$ should be ________.

Short answer

Therefore, for a set of linear algebraic equations in matrix format, $Ax-y$, for a unique solution to exist, $det\left(A\right)$should be non-zero.

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.

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

In the equations (1),$A$, $x$ and $y$ may be real or complex.

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

The solution of $x$will be unique when $det\left(A\right)$ is non-zero.

For a set of linear algebraic equations in matrix format,$Ax-y$, for a unique solution to exist, $det\left(A\right)$should be non-zero.