Find the Determinant of the Resulting Matrix [[x,1],[2,x]]=-1

$[\begin{array}{c} x & 1 \\ 2 & x \end{array}] = - 1$

The matrix is in the most simplified form.

$[\begin{array}{c} x & 1 \\ 2 & x \end{array}] = - 1$

The identity matrix or unit matrix of size $2$ is the $2 \times 2$ square matrix with ones on the main diagonal and zeros elsewhere.

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Adding $- 1$ to the square matrix is the same as adding the matrix to $- 1$ times the $2 \times 2$ identity matrix.

$[\begin{array}{c} x & 1 \\ 2 & x \end{array}] - 1 = [\begin{array}{c} x & 1 \\ 2 & x \end{array}] - 1 [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Subtract the corresponding elements.

$[\begin{array}{c} x - 1 & 1 - 0 \\ 2 - 0 & x - 1 \end{array}] =$

Simplify each element of the matrix $[\begin{array}{c} x - 1 & 1 - 0 \\ 2 - 0 & x - 1 \end{array}]$ .

Simplify $1 - 0$ .

$[\begin{array}{c} x - 1 & 1 \\ 2 - 0 & x - 1 \end{array}] =$

Simplify $2 - 0$ .

$[\begin{array}{c} x - 1 & 1 \\ 2 & x - 1 \end{array}] =$

Set up the determinant by breaking it into smaller components.

Nothing further can be done with this topic. Please check the expression entered or try another topic.

$[\begin{array}{c} x & 1 \\ 2 & x \end{array}] = - 1$