Question

[0 1 2 3]
[1 0 3 0]
[2 3 0 1]
[3 0 1 2]

The determinant of matrix (4×4)​

Answer 1

Answer:

Explanation:

We will select first row [0 1 2 3] and find cofactors for each element

then find determinant by multipying element with its co factor and with -ve sign at even position..

For 0 we won't find cofactor since finally multiplying with 0 will give 0 as result so of no use

For 1

$\left[\begin{array}{ccc}1&3&0\\2&0&1\\3&1&2\end{array}\right] \\1*[0*2 - 1*1] +3[1*3-2*2]+0[2*1-3*0]=-4$

For 2

$\left[\begin{array}{ccc}1&0&0\\2&3&1\\3&0&2\end{array}\right] \\1*[3*2-1*0]+0[3-4]+0[0-9]=6$

For 3

$\left[\begin{array}{ccc}1&0&3\\2&3&0\\3&0&1\end{array}\right] \\1*[3*1-0*0]+0*[0-2]+3[2*0-3*3]=-24$

Final determinant

=0*[...] - 1*[-4] + 2*[6] - 3*[-24] = 88

Hope it's correct.....

Hope it helps :-)