Evaluate the determinant ( will pick brainiest)

Ответ:

cameron12502

14.03.2022

$\bf \begin{bmatrix}
x&1&-1\\x^2&x&x\\\boxed{0}&\boxed{x}&\boxed{1}
\end{bmatrix}
\begin{array}{llll}
\\\\\impliedby 
\end{array}\textit{we'll use this row for our \underline{cofactors}}$

why the one at the bottom? well, because it has the factors of 0 x and 1, since 0 will be a cofactor, anything times 0 is just 0, and we can zero out that minor that quickly.

also the 1, is quicker to just multiply the minor by 1.

just a quick jogging, the cofactors, are the factors we get from the row we use in the matrix, and the minors are the submatrix that comes from using that factor, exempting any elements on that same row and column.

also recall that we need to use the checkerboard sign, against the cofactor,

$\bf \begin{bmatrix}
+&-&+\\-&+&-\\+&-&+
\end{bmatrix}\impliedby \textit{and this is our \underline{checkerboard} signs}$

so let's use them

$\bf \begin{bmatrix}
&1&-1\\&x&x\\0&&
\end{bmatrix}\implies +(0)
\begin{bmatrix}
1&-1\\x&x
\end{bmatrix}\implies 0\\\\
-------------------------------\\\\
\begin{bmatrix}
x&&-1\\x^2&&x\\&x&
\end{bmatrix}\implies -(x)
\begin{bmatrix}
x&-1\\x^2&x
\end{bmatrix}\implies -(x)[x^2~-~(-x^2)]
\\\\\\
-(x)[x^2+x^2]\implies -(x)[2x^2]\implies -2x^3\\\\
-------------------------------$

$\bf \begin{bmatrix}
x&1&\\x^2&x&\\&&1
\end{bmatrix}\implies +(1)
\begin{bmatrix}
x&1\\x^2&x
\end{bmatrix}\implies +(1)[x^2~~-~~(x^2)]
\\\\\\
+(1)[0]\implies 0\\\\
-------------------------------\\\\
\textit{and now we add all the minors}
\\\\\\
0-2x^3+0\implies -2x^3$

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Evaluate the determinant ( will pick brainiest) $\left[\begin{array}{ccc}x& 1& -1\\x^2& x& x\\0& x& 1\end{array}\right]$ plz you

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Evaluate the determinant ( will pick brainiest) plz you

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Evaluate the determinant ( will pick brainiest) $\left[\begin{array}{ccc}x& 1& -1\\x^2& x& x\\0& x& 1\end{array}\right]$ plz you