If $A=\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)$

Then if $B=\left(\begin{array}{lll}k a_{11} & k a_{12} & k a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)$

Then $|B|=k|A|$

Question: What if all the elements of A are multiplied with $k$. How does it affect the determinant?

Soln. $B=\left(\begin{array}{lll}k a_{11} & k a_{12} & k a_{13} \\ k a_{21} & k a_{22} & k a_{23} \\ k a_{31} & k a_{32} & k a_{33}\end{array}\right)$

$=k a_{11}\left(k a_{22} \cdot k a_{33}-k a_{23} \cdot k a_{32}\right)$

$- k a_{12}\left(k a_{21} \cdot k a_{33}\right. - k a_{31} \cdot k a_{23})$

$+k a_{13}(k a_{21}. k a_{32}$ $-ka_{22} \cdot ka_{31})$

$=k^3\left(|B|=k^3|A|\right.$ if all element of A are multiple with k )

If two rows or column of a matrix are identical then the determinant of the matrix in 0

Consider $A=\left(\begin{array}{lll}a & b & c \\a & b & c \\ x & y & z\end{array}\right)$

$|A|=a(b z-c y)-b(a z-c x) +c(a y-b x)$

$=a b z-a c y-a b z+b c x+a c y-b c x =0$

Suppose $A$ is a given matrix. And we obtain a new matrix $B$ by interchanging two rows (or two column) then $|B|=-|A|$

Ex $\left(\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & k\end{array}\right)$

$|A|=a(e k-f h)-b(d k-f g) =c(d h-e g)$

Let B=$\left(\begin{array}{lll}g & h & k \\ d & e & f \\ a & b & f\end{array}\right)$

$\therefore|B|={g(e c}-b f) -h(d e-a f) +k(d b-a e)$

$\therefore |B| = - |A|$

Ex. $\left(\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & k\end{array}\right)$

$|A|=a(e k-f h)-b(d k-f g)+ e(d h - e g)$

$|A \cdot B|=|A||B|$

Illustration Let $A=\left(\begin{array}{ll}a & b \\c & d\end{array}\right)$ & $B=\left(\begin{array}{ll}m & n \\p & q\end{array}\right)$

$\therefore |A|=a d-b c$ & $|B|=m q-n p$

Now $A B=\left(\begin{array}{ll}a m+b p & a x+b q \\c m+d p & c x+d q\end{array}\right)$

$\therefore |A B|=(a m+b p)(c x+d q)-(c m+d p)(a x + bq)$

$=a m c x+b p c x+a m d q + b p d q -a x c m - a x d p - b c m q - b d p q$

Let us now consider

$|A| \cdot|B|=(a d-b c)(m q-n p)$

$=a d m q-b c m q-a d n p+b c n p$

Illustration

consider $A=\left|\begin{array}{ccc}a+k & b+m & c+n \\ d & e & f \\ x & y & z\end{array}\right|$

Claim $|A|$

$=\left|\begin{array}{lll}a & b & c \\d & e & f \\x & y & z\end{array}\right|+\left|\begin{array}{lll}k & m & n \\d & e & f \\x & y & z\end{array}\right|$

$|A|=(a+k)\left|\begin{array}{ll}e & f \\ y & z\end{array}\right|-(b+m)\left|\begin{array}{ll}d & f \\ x & z\end{array}\right|$

$+(c+n)\left|\begin{array}{ll}d & e \\ x & y\end{array}\right|$

$=a \left|\begin{array}{ll}e & f \\ y & z\end{array}\right|$ $-b\left|\begin{array}{ll}d & f \\ x & z\end{array}\right|$

$+c\left|\begin{array}{ll}d & e \\ x & y\end{array}\right|$

$=k \left|\begin{array}{ll}e & f \\ y & z\end{array}\right|$ $-m\left|\begin{array}{ll}d & f \\ x & z\end{array}\right|$

$+n\left|\begin{array}{ll}d & e \\ x & y\end{array}\right|$

$=\left|\begin{array}{lll}a & b & c \\ d & e & f \\ x & y & z\end{array}\right|+\left|\begin{array}{lll}k & m & n \\ d & n & n \\ x & y & z\end{array}\right|$

Find the determinant of

$\left(\begin{array}{lll}1 & a & a^2 \\1 & b & b^2 \\1 & c & c^2\end{array}\right)$

$\therefore|A|=1\left(b c^2-c b^2\right)-a(c^2-b^2) +a^2(c-b)$

Soln. $=b c(c-b)-a(c-b)(c+b) +a^2(c-b)$

$=(c-b)\left(b c-a c-a b+a^2\right)$

$=(c-b)(c(b-a)-a(b-a)$

$=(c-b)(b-a)(c-a)=((a-b)(b-c)(c-a)$

If in a matrix the $i^{th}$ rows in replaced by the sum of $i^{th}$ rite $j^{th}$ row then the determinant does not change

Illustration -

$\text { consider } A=\left(\begin{array}{lll}a & b & c \\m & N & p \\x & y & z\end{array}\right)$

Soln. let $B= R_1\leftrightarrow R_1+R_2\left(\begin{array}{ccc}a+m & b+n & c+p \\m & n & p \\x & y & z\end{array}\right)$

Then $|B|=|A|$

$\ |B|=\left|\begin{array}{lll}a & b & c \\m & n & p \\x & y & z\end{array}\right|+\left|\begin{array}{ccc}m & n & p \\m & n & p \\x & y & z\end{array}\right|$

$=|A|+0$

$\therefore|B|=|A|$ Since two rows are identical

More generalization

If $R_i \longleftrightarrow R_i+k R_j$

Then also $|B|=|A|$

or If $B=\left|\begin{array}{ccc}a+k m & b+k n & c+k p \\ m & n & b \\ x & y & z\end{array}\right|$

Then $|B|=|A|$

This is because

$\left|\begin{array}{ccc}a+k m & b+k n & c+k p \\ m & n & p \\ x & y & z \end{array}\right|$

$=\left|\begin{array}{lll}a & b & c \\m & n & p \\x & y & z\end{array}\right| +\left|\begin{array}{ccc}k m & k n & k p \\m & n & p \\x & y & z\end{array}\right|$

$=|A|+k\left|\begin{array}{lll}m & n & p \\m & n & p \\x & y & z\end{array}\right|$

$\therefore|B|=|A|$

Find the determinant of

$\left|\begin{array}{lll}1 & b c & a(b+c) \\1 & c a & b(c+a) \\1 & a b & c(a+b)\end{array}\right|$

Soln. = $\left|\begin{array}{lll} 1 & b c & a b+a c \\1 & c a & b c+b a \\1 & a b & c a+c b\end{array}\right|$

= $\left|\begin{array}{lll} 1 & b c & b c+a b+a c \\1 & c a & c a+b c+b a \\1 & a b & a b+c a+c b\end{array}\right|$

$c_3 \leftarrow c_2+c_3$

$=(a b+b c+c a)\left|\begin{array}{lll}1 & b c & 1 \\1 & c a & 1 \\ 1 & a b & 1\end{array}\right|$

It determinant is 0 since it has two identical columns

Therefore the determinant of the original matrix in 0

Find the determinants of

$\left|\begin{array}{lll}x+y+2z & x & y \\z & y+z+2 x & y \\z & x & z+x+2 y\end{array}\right|$

Soln. $S1: C_1 \leftarrow C_1+C_2$

$\left|\begin{array}{lll}2 x+y+2 z & x & y \\2 x+y+2 z & y+z+2 x & y \\x+z & x & z+x+2 y\end{array}\right|$

$S2: c_1 \leftarrow C_1+C_3$

$\left|\begin{array}{lll}2 x+2 y+2 z & x & y \\2 x+2 y+2 z & y+z+2 x & y \\2x+2 z+2 y & x & z+x+2 y\end{array}\right|$

$= (2x+2y+2z)\left|\begin{array}{lll}1 & x & y \\1 & y+z+2x & y \\1 & x & z+x+2 y\end{array}\right|$

$R_2 \leftarrow R_2-R_1$

$2(x+y+z)\left|\begin{array}{lll}1 & x & y \\0 & y+z+x & 0 \\1 & x & z+x+2 y\end{array}\right|$

$R_3 \longleftarrow R_3-R_1 \quad \therefore$ Determinant

$2(x+y+z)\left|\begin{array}{lll}1 & x & y \\0 & x+y+z & 0 \\0 & 0 & x+y+ z\end{array}\right|$

$2(x+y+z)^3$

What is the determinant of

$\left|\begin{array}{lll}2 & 7 & 65 \\3 & 8 & 75 \\5 & 9 & 86\end{array}\right|$

Soln. $= \left|\begin{array}{lll}2 & 7 & 9 \times 7+2 \\3 & 8 & 9 \times 8+3 \\5 & 9 & 9 \times 9+5\end{array}\right|$

$= \left|\begin{array}{lll}2 & 7 & 9 \times 7 \\3 & 8 & 9 \times 8 \\5 & 9 & 9 \times 9 \end{array}\right|+\left|\begin{array}{lll}2 & 7 & 2 \\3 & 8 & 3 \\5 & 9 & 5\end{array}\right|$

$\therefore$ Answer is 0