Question

\( \operatorname{det}(A)=-4 \cdot\left|\begin{array}{cc}0 & 2 \\ 2 & m\end{array}\right|-2 \cdot\left|\begin{array}{cc}1 & 2 \\ 0 & m\end{array}\right|+1 \cdot\left|\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right| \)

          \( \operatorname{det}(A)=-4 \cdot\left|\begin{array}{cc}0 & 2 \\ 2 & m\end{array}\right|-2 \cdot\left|\begin{array}{cc}1 & 2 \\ 0 & m\end{array}\right|+1 \cdot\left|\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right| \)

det(A)=-4 ·|
0 2
2 m
|-2 ·|
1 2
0 m
|+1 ·|
1 0
0 2
|

Added by Sharon R.

Differential Equations and Linear Algebra

Stephen W. Goode, Scott A. Annin 4th Edition

Chapter 3

Instant Answer

Solved by Expert David Mccaslin

Step 1

\[ \left|\begin{array}{cc}0 & 2 \\ 2 & m\end{array}\right| = (0 \cdot m) - (2 \cdot 2) = 0 - 4 = -4 \] \[ \left|\begin{array}{cc}1 & 2 \\ 0 & m\end{array}\right| = (1 \cdot m) - (2 \cdot 0) = m - 0 = m \] \[ \left|\begin{array}{ll}1 & 0 \\ 0 & Show more…

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\( \operatorname{det}(A)=-4 \cdot\left|\begin{array}{cc}0 & 2 \\ 2 & m\end{array}\right|-2 \cdot\left|\begin{array}{cc}1 & 2 \\ 0 & m\end{array}\right|+1 \cdot\left|\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right| \)