**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$$.

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