Answer:
Step-by-step explanation:
Given the 3* 3 matrices [tex]\left[\begin{array}{ccc}0&4&1\\5&-3&0\\2&3&1\end{array}\right][/tex], 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;
[tex]= +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[/tex]
The determinant is 1 using the first row as co-factor
Similarly, using the second column [tex]\left[\begin{array}{c}4\\-3\\3\end{array}\right][/tex] 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;
[tex]= -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[/tex]
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.
