![det ( \left[\begin{array}{cc}x&y\\a&z\end{array}\right] ) = xz-ay](/tpl/images/0519/5387/289ba.png)
![det (\left[\begin{array}{ccc}k&l&m\\o&p&q\\r&s&t\end{array}\right] )\\\\= k*det(\left[\begin{array}{cc}p&q\\s&t\end{array}\right] ) - l*det(\left[\begin{array}{cc}o&q\\r&t\end{array}\right] ) + m*det(\left[\begin{array}{cc}o&p\\r&s\end{array}\right] ) \\\\=k(pt-sq) - l(ot-rq) + m(os-rp)](/tpl/images/0519/5387/ccf51.png)
![\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right]](/tpl/images/0519/5387/ad238.png)
![det(\left[\begin{array}{ccc}s&-1&0\\-1&s&-1\\0&-1&1\end{array}\right])\\\\= s*det(\left[\begin{array}{cc}s&-1\\-1&1\end{array}\right] ) - (-1)*det(\left[\begin{array}{cc}-1&-1\\0&1\end{array}\right] ) + 0*det(\left[\begin{array}{cc}-1&s\\0&-1\end{array}\right] )\\\\= s(s(1)-(-1*-1)) - (-1)(-1*1 - (-1*0)) + 0\\= s(s - 1)) + 1(-1 + 0) \\=s^{2} -s-1](/tpl/images/0519/5387/603ea.png)
* (1 ±5 )Find the values of λ for which the determinant is zero.
A. ∣λ + 2 1 2 λ∣ 50.
B....
Mathematics, 21.02.2020 20:47 insaneshootermo
Find the values of λ for which the determinant is zero.
A. ∣λ + 2 1 2 λ∣ 50.
B. ∣λ − 5 1 3 λ − 5∣ 51.
C. ∣λ 0 0 2 λ + 1 1 0 2 λ∣ 52.
D. ∣λ 0 2 0 λ 2 1 3 λ − 2∣
Answers: 2
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