Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

Question:

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:
(i) $A=\left[\begin{array}{ll}5 & 20 \\ 0 & -1\end{array}\right]$

(ii) $A=\left[\begin{array}{cc}-1 & 4 \\ 2 & 3\end{array}\right]$

(iii) $A=\left[\begin{array}{ccc}1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2\end{array}\right]$

(iv) $A=\left[\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right]$

(v) $A=\left[\begin{array}{lll}0 & 2 & 6 \\ 1 & 5 & 0 \\ 3 & 7 & 1\end{array}\right]$

(vi) $A=\left[\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right]$

(vii) $A=\left[\begin{array}{cccc}2 & -1 & 0 & 1 \\ -3 & 0 & 1 & -2 \\ 1 & 1 & -1 & 1 \\ 2 & -1 & 5 & 0\end{array}\right]$

Solution:

(i)

$M_{11}=-1$

$M_{21}=20$

$C_{i j}=(-1)^{i+j} M_{i j}$

$C_{11}=(-1)^{1+1}(-1)=-1$

$C_{21}=(-1)^{1+2}(20)=-20$

$D=(-1 \times 5)-(20 \times 0)=-5$

(ii)

$M_{11}=3$

$M_{21}=4$

$C_{i j}=(-1)^{i+j} M_{i j}$

$C_{11}=(-1)^{1+1} M_{11}=3$

$C_{21}=(-1)^{2+1} M_{21}=-(4)=-4$

$D=(3 \times-1)-(4 \times 2)=-3-8=-11$

(iii)

$M_{11}=\left|\begin{array}{cc}-1 & 2 \\ 5 & 2\end{array}\right|=-2-10=-12$

$M_{21}=\left|\begin{array}{cc}-3 & 2 \\ 5 & 2\end{array}\right|=-6-10=-16$

$M_{31}=\left|\begin{array}{ll}-3 & 2 \\ -1 & 2\end{array}\right|=-6+2=-4$

$C_{11}=(-1)^{1+1} M_{11}=-12$

$C_{21}=(-1)^{2+1} M_{21}=-(-16)=16$

$C_{31}=(-1)^{3+1} M_{31}=-4$

$D=1(-12)+3(8-6)+2(20+3)=-12+6+46=40$

(iv)

$M_{11}=\left|\begin{array}{ll}b & c a \\ c & a b\end{array}\right|=a b^{2}-c^{2} a=a\left(b^{2}-c^{2}\right)$

$M_{21}=\left|\begin{array}{ll}a & b c \\ c & a b\end{array}\right|=a^{2} b-c^{2} b=b\left(a^{2}-c^{2}\right)$

$M_{31}=\left|\begin{array}{ll}a & b c \\ b & c a\end{array}\right|=a^{2} c-b^{2} c=c\left(a^{2}-b^{2}\right)$

$C_{11}=(-1)^{1+1} M_{11}=a\left(b^{2}-c^{2}\right)$

$C_{21}=(-1)^{2+1} M_{21}=-b\left(a^{2}-c^{2}\right)$

$C_{31}=(-1)^{3+1} M_{31}=c\left(a^{2}-b^{2}\right)$

$D=1 . a\left(b^{2}-c^{2}\right)-a(a b-c a)+b . c(c-b)$

$=a b^{2}-a c^{2}-a^{2} b+a^{2} c+c^{2} b-b^{2} c$

$=a^{2}(c-b)+b^{2}(a-c)+c^{2}(b-a)$

(v)

$M_{11}=\left|\begin{array}{ll}5 & 0 \\ 7 & 1\end{array}\right|=5-0=5$

$M_{21}=\left|\begin{array}{ll}2 & 6 \\ 7 & 1\end{array}\right|=2-42=-40$

$M_{31}=\left|\begin{array}{ll}2 & 6 \\ 5 & 0\end{array}\right|=0-30=-30$

$C_{11}=(-1)^{1+1} M_{11}=5$

$C_{21}=(-1)^{2+1} M_{21}=-(-40)$

$C_{31}=(-1)^{3+1} M_{31}=-30$

$D=0(5-0)-2(1-0)+6(7-15)=-2-48=-50$

(vi)

$M_{11}=\left|\begin{array}{ll}b & f \\ f & c\end{array}\right|=b c-f^{2}$

$M_{21}=\left|\begin{array}{ll}h & g \\ f & c\end{array}\right|=h c-f g$

$M_{31}=\left|\begin{array}{ll}h & g \\ b & f\end{array}\right|=h f-g b$

$C_{11}=(-1)^{1+1} M_{11}=b c-f^{2}$

$C_{21}=(-1)^{2+1} M_{11}=-(h c-f g)=f g-h c$

$C_{31}=(-1)^{3+1} M_{11}=h f-g b$

$D=a\left(b c-f^{2}\right)-h(h c-f g)+g(f h-b g)$

$=a b c-a f^{2}-h^{2} c+f g h+f g h-b g^{2}$

$=a b c+2 h f g-a f^{2}-b g^{2}-c h^{2}$

(vii)

$M_{11}=0(0-5)-1(0+1)-2(5-1)=-1-8=-9$

$M_{21}=-1(0-5)+1(5-1)=5+4=9$

$M_{31}=-1(0+10)+1(0+1)=-10+1=-9$

$M_{41}=-1(1-2)+1(0-1)=1-1=0$

$C_{11}=(-1)^{1+1} M_{11}=-9$

$C_{21}=(-1)^{2+1} M_{21}=(-1) \times 9$

$C_{31}=(-1)^{3+1} M_{31}=-9$

$C_{41}=(-1)^{4+1} M_{41}=0$

$D=2\left|\begin{array}{ccc}0 & 1 & -2 \\ 1 & -1 & 1 \\ -1 & 5 & 0\end{array}\right|+1\left|\begin{array}{ccc}-3 & 1 & -2 \\ 1 & -1 & 1 \\ 2 & 5 & 0\end{array}\right|-1\left|\begin{array}{ccc}-3 & 0 & 1 \\ 1 & 1 & -1 \\ 2 & -1 & 5\end{array}\right|$

$=-18-27+15=30$