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$\left| \begin{array} { l l l } { 1 } & { 1 } & { - 1 } \\ { 0 } & { 1 } & { 2 } \\ { 1 } & { 1 } & { 0 } \end{array} \right| $
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det(\left(\begin{matrix}1&1&-1\\0&1&2\\1&1&0\end{matrix}\right))
Find the determinant of the matrix using the method of diagonals.
\left(\begin{matrix}1&1&-1&1&1\\0&1&2&0&1\\1&1&0&1&1\end{matrix}\right)
Extend the original matrix by repeating the first two columns as the fourth and fifth columns.
2=2
Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.
-1+2=1
Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.
2-1
Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.
1
Subtract 1 from 2.
det(\left(\begin{matrix}1&1&-1\\0&1&2\\1&1&0\end{matrix}\right))
Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).
det(\left(\begin{matrix}1&2\\1&0\end{matrix}\right))-det(\left(\begin{matrix}0&2\\1&0\end{matrix}\right))-det(\left(\begin{matrix}0&1\\1&1\end{matrix}\right))
To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.
-2-\left(-2\right)-\left(-1\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.
1
Add the terms to obtain the final result.