Question:

Without expanding the determinant, prove that $\left|\begin{array}{cc}\mathrm{a}& a^2\mathrm{bc}\\ \mathrm{b}& b^2\mathrm{ca}\\ \mathrm{c}& c^2\mathrm{ab}\end{array}\right|$ =$\left|\begin{array}{cc}1& a^2{a}^3\\ 1& b^2{b}^3\\ 1& c^2{c}^3\end{array}\right|$

Answer:

Solution:

We can use the Laplace Expansion Theorem to prove this statement. According to the theorem, the determinant of a matrix is equal to the sum of the products of the elements of any row or column and its corresponding cofactor.

Therefore, we can expand the determinant using the first column of the given matrix and its corresponding cofactor.

$\left|\begin{array}{cc}\mathrm{a}& a^2\mathrm{bc}\\ \mathrm{b}& b^2\mathrm{ca}\\ \mathrm{c}& c^2\mathrm{ab}\end{array}\right|$

= a$\left|\begin{array}{cc}1& a^2\mathrm{bc}\\ \mathrm{b}& b^2\mathrm{ca}\\ \mathrm{c}& c^2\mathrm{ab}\end{array}\right|$

• b$\left|\begin{array}{cc}\mathrm{a}& a^{2}1\\ \mathrm{b}& b^2\mathrm{ca}\\ \mathrm{c}& c^2\mathrm{ab}\end{array}\right|$
• c$\left|\begin{array}{cc}\mathrm{a}& a^2\mathrm{bc}\\ \mathrm{b}& b^{2}1\\ \mathrm{c}& c^2\mathrm{ab}\end{array}\right|$

= a

Question:

Let A=$\left|\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right|$Verify that (i)$\begin{array}{}{\mathrm{[adj\; A]}}^{\mathrm{-1}}\end{array}$=adj$\begin{array}{}{\mathrm{(A)}}^{\mathrm{-1}}\end{array}$ (ii)${\displaystyle {\left(A^{-1}\right)}^{-1}}$=A

Answer:

Step 1: Calculate the adjugate of A

adj A = $\left|\begin{array}{ccc}3& 2& -1\\ 2& -1& -1\\ -1& -1& 1\end{array}\right|$

Step 2: Calculate the inverse of A

A^-1 = $\left|\begin{array}{ccc}3& 2& -1\\ 2& -1& -1\\ -1& -1& 1\end{array}\right|$

Step 3: Verify that (i) adj A^-1 = adj (A^-1)

adj A^-1 = $\left|\begin{array}{ccc}3& 2& -1\\ 2& -1& -1\\ -1& -1& 1\end{array}\right|$

adj (A^-1) = $\left|\begin{array}{ccc}3& 2& -1\\ 2& -1& -1\\ -1& -1& 1\end{array}\right|$

Since adj A^-1 = adj (A^-1), the first part of the statement is true.

Step 4: Verify that (ii) (A^-1)^-1 = A

(A^-1)^-1 =

Question:

Using properties of determinants, prove that: $\left|\begin{array}{ccc}\mathrm{\alpha }& {\alpha }^2& \mathrm{\beta +\gamma }\\ \mathrm{\beta }& {\beta }^2& \mathrm{\gamma +\alpha }\\ \mathrm{\gamma }& {\gamma }^2& \mathrm{\alpha +\beta }\end{array}\right|$ =(α−β)(β−γ)(γ−α)(α+β+γ)

Answer:

1. Expand the determinant using the Laplace Expansion along the last column:

$\left|\begin{array}{ccc}\mathrm{\alpha }& {\alpha }^2& \mathrm{\beta +\gamma }\\ \mathrm{\beta }& {\beta }^2& \mathrm{\gamma +\alpha }\\ \mathrm{\gamma }& {\gamma }^2& \mathrm{\alpha +\beta }\end{array}\right|=(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma })\left|\begin{array}{ccc}\mathrm{\alpha }& {\alpha }^2& \mathrm{\beta }\\ \mathrm{\beta }& {\beta }^2& \mathrm{\gamma }\\ \mathrm{\gamma }& {\gamma }^2& \mathrm{\alpha }\end{array}\right|$

2. Apply the rule of Sarrus:

$\left|\begin{array}{ccc}\mathrm{\alpha }& {\alpha }^2& \mathrm{\beta }\\ \mathrm{\beta }& {\beta }^2& \mathrm{\gamma }\\ \mathrm{\gamma }& {\gamma }^2& \mathrm{\alpha }\end{array}\right|=\mathrm{\alpha }{\alpha }^2+\mathrm{\beta }{\beta }^2+\mathrm{\gamma }{\gamma }^2+2\mathrm{\alpha }\mathrm{\beta }\mathrm{\gamma }-\mathrm{\alpha }{\beta }^2-\mathrm{\beta }{\gamma }^2-\mathrm{\gamma }{\alpha }^2$

3. Apply the distributive property: