Question

Compute the determinants using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.
[ 0 4 1
5 −3 0
2 3 1 ]

Answer 1

Answer:

The determinant is 1

Step-by-step explanation:

Given the 3* 3 matrices $\left[\begin{array}{ccc}0&4&1\\5&-3&0\\2&3&1\end{array}\right]$, to compute the determinant using the first row means using the row values [0 4 1 ] to compute the determinant. Note that the signs on the values on the first row are +0, -4 and +1

Calculating the determinant;

$= +0\left[\begin{array}{cc}-3&0\\3&1\\\end{array}\right] -4\left[\begin{array}{cc}5&0\\2&1\\\end{array}\right] +1\left[\begin{array}{cc}5&-3\\2&3\\\end{array}\right] \\\\= 0 - 4[5(1)-2(0)] +1[5(3)-2(-3)]\\= 0 -4[5-0]+1[15+6]\\= 0-20+21\\= 1$

The determinant is 1 using the first row as co-factor

Similarly, using the second column $\left[\begin{array}{c}4\\-3\\3\end{array}\right]$ as the cofactor, the determinant will be expressed as shown;

Note that the signs on the values are -4, +(-3) and -3.

Calculating the determinant;

$= -4\left[\begin{array}{cc}5&0\\2&1\\\end{array}\right] -3\left[\begin{array}{cc}0&1\\2&1\\\end{array}\right] -3\left[\begin{array}{cc}0&1\\5&0\\\end{array}\right] \\\\= -4[5(1)-2(0)] - 3[0(1)-2(1)] -3[(0)-5(1)]\\= -4[5-0] -3[0-2]-3[0-5]\\= -20+6+15\\= -20+21\\= 1$

The determinant is also 1 using the second column as co factor.

It can be concluded that the same value of the determinant will be arrived at no matter the cofactor we choose to use.