• What is the determinant of

• $A=\left(\begin{array}{lll}a & a^2 & a^3-1 \\ a^w & a^{2 \omega} & a^{3 \omega}-1 \\ a^{\omega^2} & a^{2 \omega^2} & a^{3 \omega z}-1\end{array}\right)$

• What $\omega$ is the cube root of unity

• Soln. And we known that

• $1 + \omega + {\omega}^2 = 0$

• $\therefore|A|$

• $= \left|\begin{array}{lll}a & a^2 & a^3 \\ a^\omega & a^{2 \omega} & a^{3 \omega} \\a^{\omega^2} & a^{2 \omega^2} & a^{3 \omega^2}\end{array}\right| - \left|\begin{array}{lll}a & a^2 & 1 \\a^{\omega} & a^{2 \omega} & 1 \\a^{\omega^2} & a^{2 \omega^2} & 1\end{array}\right|$

• $\left|A_1\right|=a \cdot a^\omega \cdot a^{\omega^2}\left|\begin{array}{ccc}1 & a & a^2 \\1 & a^\omega & a^{2 \omega} \\1 & a^{\omega^2} & a^{2 \omega^2}\end{array}\right|$

• $=a^{{1+\omega+\omega^2}{=0}}\left|\begin{array}{lll}1 & a & a^2 \\1 & a^\omega & a^2 \omega \\1 & a^{\omega^2} & a^{2 \omega^2}\end{array}\right|$

• $\therefore\left|A_1\right|=\left|\begin{array}{lll}1 & a & a^2 \\1 & a^\omega & a^{2 \omega} \\1 & a^{\omega^2} &a^{2\omega^2}\end{array}\right|$

• Now $|A_2| = \left|\begin{array}{lll} a & a^2 & 1 \\a^{\omega} & a^{2\omega} & 1 \\a^{\omega^2} & a^{2\omega^2} & 1\end{array}\right|$

• We can get $|A_2|$ from $|A_1|$ by first interchanging column with column , then interchanging column 1 and column 2

• $\therefore$ we get $| A_2|$ from $|A_1|$ by two interchanges,

• $|A_2| = (-1)^{2} |A_1|$

• or $|A| = |A_1| - | A_2|$

• = 0

• Any matrix whose determinant in 0 is called a singular matrix

• A matrix with non zero determinant is called a non singular matrix.

• If A is a non singular matrix we can find another non singular matrix B of the same order of A,

• such that, A. B = B.A = I

• where A and B are non matrix

• Question: How to obtain $A^{-1}$ given a n × n matrix A.

• We shall compute $A^{-1}$ with the help of $|A|$ and ( adj A)

• We know that $A(\operatorname{adj} A)=|A| I_n$

• $\therefore A \frac{(\operatorname{adi} A)}{|A|}=I_n$

• $\therefore$ We obtain a matrix $\frac{(\operatorname{adi} A)}{|A|} ∃ \frac{(\operatorname{adj} A)}{|A|}=I$.

• and parrallely $\frac{(\operatorname{adj} A)}{|A|} \cdot A=I_n$.

• Now question is can there be two inverses of a Matrix?

• Ans: NO.

• If possible, let $B$ and $C$ be two inverses of $A$.

• $\therefore(A \cdot B)=(B \cdot A)=I .$

• and $(A \cdot C)=(C \cdot A)=I .$

• $\therefore C=I \cdot C=(B \cdot A) \cdot C$

• $=B \cdot(A \cdot C)=B \cdot I=B$

• Result: what is the inverse of A.B. ?

• We know that $(A \cdot B)\left(B^{-1} \cdot A^{-1}\right)$

• $=A\left(B \cdot B^{-1}\right) \cdot A^{-1}=A I A^{-1}$

• $=A A^{-1}=I$

• $\therefore$ We can say by unique number of inverse that $(A B)^{-1}=B^{-1} A^{-1}$

• Result: If $A \& B$ are non singular matrix then (adj $A B$ )

• $=(\operatorname{adj} B)(\operatorname{adj} A) \text {. }$

• Proof consider $(A B)(\operatorname{adj} B)(\operatorname{adj} A)$

• $=A(B(a d j B))(a d j A)$

• $=A(|B| I)(\operatorname{adj} A)$

• $=|B|(A \cdot I \cdot(\operatorname{adj} A))$

• $=|B|(A \cdot(\operatorname{adj} A \mid)=|B||A| I$

• Since (adj $A B$ ) in such that

• $(A B)($ adj $A B)=|A B| I$

• $=|A||B| I$

• $\therefore$ We find that

• $(A B)(\operatorname{adj} A B)=(A B)(\operatorname{adj} B)(\operatorname{adj} A)$

• $=|A||B| I .$

• $\therefore$ By Multiplying both sides with $(A B)^{-1}$

• we get $( adj A B)$ = $(\operatorname{adj} B)(\operatorname{adj} A)$

• Compute the inverse of
• $\begin{aligned} & A=\left(\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right) \\ &|A|=1 \cdot(2 \cdot 4-(-3)(-2)) \\ &+3(-1)(-3)-2 \cdot 2) \\ &=(8-6)+3(3-4) \\ &=2-3=-1 \neq 0 \end{aligned}$
• $\therefore A$ is non-singular

• $A=\left(\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right)$

• $\therefore A_{11}=4.2-(-3)(-2)-8-6=2 \\ A_{12}=(-1)^{1+2}(0.4-(-3) .3)=-9$

• $A_{13}=0 .(-2)-3.2=-6$

• $A_{21}=-4-2(-2)=0$

• $A_{22}=4.1-3.2=-2$

• $A_{23}=(-1)^{2+3}\left(1 \cdot(-2)-(-1)(3)\right)=-(-2+3) =-1$

• $A_ {31}=-1, ~ A_ {32} =3 ~$ and $~ A_{33}=2$

$\begin{aligned} & \therefore(\operatorname{adj} A)=\left(\begin{array}{ccc} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{array}\right) \\ & \therefore A^{-1}=\left(\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right) \ & \end{aligned}$

• Verify that $A \cdot A^{-1}=I_3$ $A^{-1} \cdot A=I_3$

• Q: What is the determinant of $A^{-1}$

• we know that $A \cdot A^{-1}=I$

• Soln. $\begin{aligned} \therefore\left|A \cdot A^{-1}\right| & =|I|=1 \ \therefore|A|\left|A^{-1}\right| & =1 \ \text { or }\left|A^{-1}\right| & =\frac{1}{|A|} \end{aligned}$

• Ex what is the inverse of

• $A=\left(\begin{array}{ccc}1 & 0 & 0 \\0 & \cos \alpha & \sin \alpha \\0 & \sin \alpha & -\cos \alpha\end{array}\right)$

• so we first compute the cofactor A.

• $A_{11}=\cos \alpha(-\cos \alpha)-\sin \alpha \cdot \sin \alpha$

• $=-\cos ^2 \alpha-\sin ^2 \alpha =0 \cdot 1$

• $A_{12}=0 \cdot(-\cos \alpha)-0(\sin \alpha)=0$

• $A_{13}=0 \cdot \sin \alpha-0 \cos \alpha=0$

• $A_{21}=0(-\cos \alpha)-0(\sin \alpha)=0$

• $A_{22}=-\cos \alpha-0=-\cos \alpha$

• In a similar way:

• $A_{23} =(-1)^{2+3}(1 \cdot \sin \alpha-0) =$

• $-\sin \alpha$

• $A_{31} =0 \cdot \sin \alpha-0 \sin \alpha=0$

• $A_{32} =(-1)(1 \cdot \sin \alpha-0)= -\sin \alpha$

• $A_{33} = 1 - \cos \alpha - 0$

• $=\cos \alpha .$

• $\therefore(\operatorname{adj} A)$ $=\left(\begin{array}{ccc}-1 & 0 & 0 \\0 & -\cos \alpha & -\sin \alpha \\0 & -\sin \alpha & \cos \alpha\end{array}\right)$

• $\therefore A \text { (adj } A)$

• $=\left[\begin{array}{ccc}1 & 0 & 0 \\0 & \operatorname{cos} \alpha & \sin \alpha \\0 & \sin \alpha & -\cos \alpha\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & 0 \\0 & -\cos \lambda & -\sin \alpha \\0 & -\sin \alpha & \cos \alpha\end{array}\right]$

• $=\left[\begin{array}{ccc}-1 & 0 & 0 \\0 & -\cos ^2 \alpha-\sin^2 \alpha & -\cos \alpha \cdot \sin \alpha +\cos \alpha \cdot \sin \alpha \\0 & -\cos \alpha \cdot \sin \alpha + \cos\alpha \cdot \sin \alpha & -\sin^2 \alpha - \cos^2 \alpha \end{array}\right]$

• $\begin{aligned}\therefore A(\operatorname{adj} A) & =\left[\begin{array}{ccc}-1 & 0 & 0 \\0 & -1 & 0 \\0 & 0 & -1\end{array}\right] \text { i.e }(-1) I_3 & =|A| I_3 .\end{aligned}$

• $\therefore$ The inverse of $A$ is

• $\begin{aligned}& \frac{(\operatorname{adj} A)}{|A|}=\left(\begin{array}{ccc}-1 & 0 & 0 \\0 & -\cos \alpha & -\sin \alpha \\0 & -\sin \alpha & \cos \alpha\end{array}\right) \end{aligned}$

• $\begin{aligned}& =\left(\begin{array}{ccc}1 & 0 & 0 \\0 & \cos \alpha & \sin \alpha \\0 & \sin \alpha & -\cos \alpha \end{array}\right)^{-1} \&\end{aligned}$

• Ex : Show that the matrix $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$ satisfies the equation $A^2-4 A-5 I_3=C$ where $O$ in the 0 -mamix of orden $3 \times 3$.

• Also, find out the inverse 8 the above matrix.

• since

• $A=\left[\begin{array}{lll}1 & 2 & 2 \\2 & 1 & 2 \\2 & 2 & 1\end{array}\right]$

• $A^2=A \cdot A =\left[\begin{array}{lll}1 & 2 & 2 \\2 & 1 & 2 \\2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\2 & 1 & 2 \\2 & 2 & 1\end{array}\right] =\left[\begin{array}{lll}9 & 8 & 8 \\8 & 9 & 8 \\8 & 8 & 9\end{array}\right]$

• $\begin{aligned} & \therefore A^2-4 A \\ & =\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-4\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right] \\ & =\left[\begin{array}{ccc}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]=5 I_3 \\ & \therefore A^2-4 A=5 I \\ & \therefore A^2-4 A-5 I=0_{3 \times 3}\end{aligned}$

• $\therefore \quad A^2-4 A=5 I$

• Pre-multiplying by $A^{-1}$

• $A^{-1}\left(A^2-4 A\right)=5 \cdot A^{-1} \cdot I \quad \text { or } \quad A-4 I=5 \cdot A^{-1}$

• $\therefore A^{-1} \cdot \frac{A-4 \cdot I}{5}$

• $= \frac{1}{5}\left(\begin{array}{lll}1 & 2 & 2 \\2 & 1 & 2 \\2 & 2 & 1\end{array}\right) -\left(\begin{array}{lll}4 & 0 & 0 \\0 & 4 & 0 \\0 & 0 & 4\end{array}\right)$

• $=\frac{1}{5}\left(\begin{array}{ccc}-3 & 2 & 2 \\2 & -3 & 2 \\2 & 2 & -3\end{array}\right)$

• $=\left(\begin{array}{ccc}-3 / 5 & 2 / 5 & 2 / 5 \\2 / 5 & -3 / 5 & 2 / 5 \\2 / 5 & 2 / 5 & -3 / 5\end{array}\right)$

• Verify that $A \cdot A^{-1}=A^{-1} A$ $=I$.

• What is determinant if (adj $(\operatorname{adj} A))$

• when $|A|$ is given

• Soln. we know $|A(\operatorname{adj} A)|$

• $\begin{aligned}& =|A||(\operatorname{ad} j A)|=|| A\left|I_n\right| \& =\left|\begin{array}{cccc} |A| & & & \\& |A| & & \\& 0 & & |A|\end{array}\right|_{\operatorname{n × n}} \&\end{aligned}$

• $=|A|^n$

• $\therefore|A||(\operatorname{adi} A)|=|A|^n$

• $\therefore \mid \operatorname{adj} A)\left.|=| A\right|^{n-1}$

• $\therefore|\operatorname{adj}(\operatorname{adj} A)|=\left(|A|^{n-1}\right)^{n-1}$

• Therefore given a matrix it are can compute $\left|A^{-1}\right|$ $| (adj A)|$ and also $|(adj (adj){A}) |$