find the determinant of the matrices below by inspection.give your reason in each case
C=
row1 a b c
row2 0 e f
row3 0 0 d
and
D=
row 1 3 0 0
row 2 0 -2 0
row3 0 0 -1
plse assist on how to do the above by inspection?
Well. The first term in the determinant for the first matrix is multiplying a, by the determinant of the diagonal matrix:
e f
0 d
Reconising that the zero is there should allow you to 'by inspection' rhyme off the first term as aed. Then the 2nd and third elements of the first column are 0, meaning that the 2nd two terms are 0. Determinant is aed.
In the 2nd matrix, you realise that in the firt term you are multiplying 3 by the diagonal matrix:
-2 0
0 -1
You should realise straight away that your first term is 3(-2)(-1) = 6, and again the 2nd and third elements of the first column are 0, meaning that the 2nd two terms are 0. Determinant is 6.
Both answered here: http://www.mathhelpforum.com/math-he...nantqnine.html
(It looks like you, mathseek and trythis all have the same problems. Try reading each others posts first)
Hello, Faz!
Evaluate by "minors", going down the first column.Find the determinant of the matrices below by inspection.
$\displaystyle (1)\;\;C\;=\;\begin{bmatrix}a & b & c \\ 0 & e & f \\ 0 & 0 & d\end{bmatrix}$
$\displaystyle C \;=\;a\cdot\begin{bmatrix}e&f\\0&d\end{bmatrix} - 0\cdot\begin{bmatrix}b&c\\0&d\end{bmatrix} + 0\cdot\begin{bmatrix}b&c\\e&f\end{bmatrix} \;=\;a(ed-0) - 0 + 0 \;=\;aed$
Evaluate by "minors", going across the first row.$\displaystyle (2)\;\;D\;=\;\begin{bmatrix}3 & 0 & 0 \\ 0 & \text{-}2 & 0 \\ 0 & 0 & \text{-}1 \end{bmatrix}$
$\displaystyle D \;=\;3\cdot\begin{bmatrix}\text{-}2&0\\0&\text{-}1 \end{bmatrix} - 0\cdot \begin{bmatrix}0&0\\0&\text{-}1\end{bmatrix} + 0 \cdot\begin{bmatrix}0&\text{-}2\\0&0\end{bmatrix} \;=\;3(2-0) - 0 + 0 \;=\;6$