Question

3x3 matrix A. r1(3 -2 0) r2(0 1 1) r3(2 -1 0). don't calculate A^-1 or raise any matrix to a power first. calculate det(2A^-2)

Answer 1

Answer with explanation:

For, a 3 × 3, matrix

$r_{1}=(3,-2,0)\\\\r_{2}=(0,1,1)\\\\r_{3}=(2,-1,0)$

which are entries of First, Second and Third Row Respectively.

So, if written in the form of Matrix (A)

 $A=\left[\begin{array}{ccc}3&-2&0\\0&1&1\\2&-1&0\end{array}\right]$

⇒Adjoint A= Transpose of Cofactor of A

$a_{11}=1,a_{12}=2,a_{13}=-2\\\\a_{21}=0,a_{22}=0,a_{23}=-1\\\\a_{31}=-2,a_{32}=- 3,a_{33}=3\\\\Adj.A=\left[\begin{array}{ccc}1&0&-2\\2&0&-3\\-2&-1&3\end{array}\right]$

⇒≡ |Adj.A|=1 ×(0-3) -2×(-2-0)

            = -3 +4

            =1     --------(1)

⇒For, a Matrix of Order, 3 × 3,

| Adj.A |=| A|²---------(2)

$|2 A^{-2}|=2^3\times |A^{-2}|\\\\=2^3\times |A|^{-2}\\\\=\frac{8}{|A^{2}|}\\\\=\frac{8}{|Adj.A|}\\\\=\frac{8}{1}\\\\=8$

                                              --------------------------------------------(Using 1 and 2)

$\rightarrow|2 A^{-2}|=8$