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* |A^(-2)|\\\\=2^3* |A|^(-2)\\\\=(8)/(|A^(2)|)\\\\=(8)/(|Adj.A|)\\\\=(8)/(1)\\\\=8`

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

`\rightarrow|2 A^(-2)|=8`