Question: Is the trace of square matrice the product of its diagonal entries ? Justify. Determine whether the set $$w = [ (a_1 , a_2 , a_3) \in R_3 : a_1 + 2a_2 – 3a_3 = 2]$$ is subspace or not of $$R^3$$ with justification.\

Solution:
First part:
No, the trace of square matrix is not the product of it’s diagonal entries. The trace of a square matrix of order $$m \times n$$ is the sum of its principal diagonal entries.
i.e; Tr(M) = $$M_{11} + M_{22} + M_{33} + .......+ M_{nn}$$
Second part:
Here, $$w = [ (a_1 , a_2 , a_3) \in R^3 : a_1 + 2a_2 – 3a_3 = 2]$$. We have to show that w is subspace of $$R^3$$
For this , we take $$(c_1 , c_2) \in R$$ and $$v_1 = (x_1 , x_2 , x_3)$$ and $$v_2 = (y_1 , y_2 , y_3) \in w$$
So, $$x_1 + 2x_2 – 3x_3 = -2$$ ……………(i)
$$y_1 + 2y_2 – 3y_3 = -2$$……………(ii)
since,
$$c_1v_1 + c_2v_2$$ \ = $$c_1(x_1 , x_2, x_3) + c_2(y_1 , y_2 , y_3)$$
= $$(c_1x_1 + c_2y_1, c_1x_2 + c_2y_2 , c_1x_3 + c_2y_3)$$
Which is a traid form.
By definition of w. \ $$(c_1x_1 + c_2y_1) + 2(c_1x_2 + c_2y_2) – 3(c_1x_3 + c_2y_3) = 2$$
Or $$(c_1x_1 + 2c_1x_2 – 3c_1x_3 ) + (c_2y_1 + 2c_2y_2 – 3c_2y_3) = 2$$
Or $$c_1(x_1 + 2x_2 – 3x_3 ) + c_2(y_1 + 2y_2 – 3y_3) = 2$$\ Or $$2c_1 + 2c_2 = 2 ≠ 0.c_1 + 0.c_2 = 0$$ Or $$c_1 + c_2 = 1$$.
Hence w is not subspace of $$R^3$$.
Alternatively
Given, $$w = [ (a_1 , a_2 , a_3) \in R^3 : a_1 + 2a_2 – 3a_3 = 2]$$.
If we choose
$$u = (a_1, a_2, a_3) = (1, 2, 1) \in R^3$$
$$v = (a_1, a_2, a_3) = (0, 1, 0)$$
Where, $$u, v \in w$$
For subspace,
$$v_{ss1}$$: if $$(a_1, a_2, a_3) = (0, 0, 0) \in R^3$$
$$= 0 + 2 \times 0 - 3 \times 0$$
$$= 0 + 0 + 0$$
$$= 0 \neq 2$$
$$v_{ss2}$$ :for $$u, v \in w$$
$$u + w = (1, 2, 1) + (0, 1, 0)$$
$$= (1, 3, 1)$$
Now, $$a_1 + 2a_2 -3a_3$$
$$=1 + 2\times 3 - 3 \times 1$$
$$= 1 + 6 -3$$
$$= 4 \neq 2$$
$$v_{ss3}$$:for $$c \in R\$$ cu = c(a_1, a_2, a_3) $$\ if we choose c = 3 and$$ (a_1, a_2, a_3) = (1, 2, 1) $$\$$ cu = 3(1, 2, 1) $$\$$ =(3, 6, 3) $$\ such that,$$ a_1 + 2a_2 - 3a_3 = 3 + 2 \times 6 - 3 \times3 $$\$$6$$\$$3(2)$$\ Hence$$ v_{ss3} $$holds. Hence the set w is not satisfied all the property of vector subspace hence w is not subspace of$$R^3.

Tags:

Categories:

Updated: