Question

If $2s = a + b + c$ then \left| {\begin{array}{*{20}{c}} {{a^2}}&{{{\left( {s - a} \right)}^2}}&{{{\left( {s - a} \right)}^2}} \\\ {{{\left( {s - b} \right)}^2}}&{{b^2}}&{{{\left( {s - b} \right)}^2}} \\\ {{{\left( {s - c} \right)}^2}}&{{{\left( {s - c} \right)}^2}}&{{c^2}} \end{array}} \right| is equal to
A. $2s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)$
B. $2{s^3}\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)$
C. $2{s^2}\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)$
D. $\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)$

Answer 1

To find the value of a given determinant, first we will convert it into the simplest form. Then, we will perform column and row operations to simplify the determinant.

Complete step-by-step solution:
In this problem, given that $2s = a + b + c$. Let us say a given determinant is denoted by $D$. So, we need to find the value of D = \left| {\begin{array}{*{20}{c}} {{a^2}}&{{{\left( {s - a} \right)}^2}}&{{{\left( {s - a} \right)}^2}} \\\ {{{\left( {s - b} \right)}^2}}&{{b^2}}&{{{\left( {s - b} \right)}^2}} \\\ {{{\left( {s - c} \right)}^2}}&{{{\left( {s - c} \right)}^2}}&{{c^2}} \end{array}} \right|.
To convert into the simplest form, let us assume $s - a = p,\;s - b = q$ and $s - c = r$. Then, $p + q = s - a + s - b = 2s - a - b = c\quad \left[ {\because 2s = a + b + c} \right] \\\ q + r = s - b + s - c = 2s - b - c = a\quad \left[ {\because 2s = a + b + c} \right] \\\ r + p = s - c + s - a = 2s - a - c = b\quad \left[ {\because 2s = a + b + c} \right] \\\ p + q + r = s - a + s - b + s - c = 3s - \left( {a + b + c} \right) = 3s - 2s = s\quad \left[ {\because 2s = a + b + c} \right] \\\$
Now we can write the given determinant as D = \left| {\begin{array}{*{20}{c}} {{{\left( {q + r} \right)}^2}}&{{p^2}}&{{p^2}} \\\ {{q^2}}&{{{\left( {r + p} \right)}^2}}&{{q^2}} \\\ {{r^2}}&{{r^2}}&{{{\left( {p + q} \right)}^2}} \end{array}} \right|.
Now we will perform column operations to simplify the above determinant.
Applying ${C_1} \to {C_1} - {C_2}$ and ${C_2} \to {C_2} - {C_3}$, we get
D = \left| {\begin{array}{*{20}{c}} {{{\left( {q + r} \right)}^2} - {p^2}}&{{p^2} - {p^2}}&{{p^2}} \\\ {{q^2} - {{\left( {r + p} \right)}^2}}&{{{\left( {r + p} \right)}^2} - {q^2}}&{{q^2}} \\\ {{r^2} - {r^2}}&{{r^2} - {{\left( {p + q} \right)}^2}}&{{{\left( {p + q} \right)}^2}} \end{array}} \right|
Use the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ and simplify the above determinant, we get
D = \left| {\begin{array}{*{20}{c}} {\left( {q + r - p} \right)\left( {q + r + p} \right)}&0&{{p^2}} \\\ {\left( {q - r - p} \right)\left( {q + r + p} \right)}&{\left( {r + p - q} \right)\left( {r + p + q} \right)}&{{q^2}} \\\ 0&{\left( {r - p - q} \right)\left( {r + p + q} \right)}&{{{\left( {p + q} \right)}^2}} \end{array}} \right|
Let us take the factor $\left( {p + q + r} \right)$ common out from the first and second column. Therefore, we get
D = {\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {\left( {q + r - p} \right)}&0&{{p^2}} \\\ {\left( {q - r - p} \right)}&{\left( {r + p - q} \right)}&{{q^2}} \\\ 0&{\left( {r - p - q} \right)}&{{{\left( {p + q} \right)}^2}} \end{array}} \right|
Applying ${R_3} \to {R_3} - \left( {{R_1} + {R_2}} \right)$, we get
D = {\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {\left( {q + r - p} \right)}&0&{{p^2}} \\\ {\left( {q - r - p} \right)}&{\left( {r + p - q} \right)}&{{q^2}} \\\ {0 - \left( {q + r - p + q - r - p} \right)}&{\left( {r - p - q} \right) - \left( {0 + r + p - q} \right)}&{{{\left( {p + q} \right)}^2} - \left( {{p^2} + {q^2}} \right)} \end{array}} \right|
Simplify the above determinant, we get
D = {\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {\left( {q + r - p} \right)}&0&{{p^2}} \\\ {\left( {q - r - p} \right)}&{\left( {r + p - q} \right)}&{{q^2}} \\\ {2p - 2q}&{ - 2p}&{2pq} \end{array}} \right|\quad \left[ {\because {{\left( {p + q} \right)}^2} = {p^2} + 2pq + {q^2}} \right]
Let us take the number $2$ common out from the third row. Therefore, we get
D = 2{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {\left( {q + r - p} \right)}&0&{{p^2}} \\\ {\left( {q - r - p} \right)}&{\left( {r + p - q} \right)}&{{q^2}} \\\ {p - q}&{ - p}&{pq} \end{array}} \right|
Applying ${R_1} \to {R_1} - \left( {\dfrac{p}{q}} \right){R_3}$, we get
D = 2{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {\left( {q + r - p} \right) - \dfrac{p}{q}\left( {p - q} \right)}&{0 - \dfrac{p}{q}\left( { - p} \right)}&{{p^2} - \dfrac{p}{q}\left( {pq} \right)} \\\ {\left( {q - r - p} \right)}&{\left( {r + p - q} \right)}&{{q^2}} \\\ {p - q}&{ - p}&{pq} \end{array}} \right|
Simplify the above determinant, we get
D = 2{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {q + r - \dfrac{{{p^2}}}{q}}&{\dfrac{{{p^2}}}{q}}&0 \\\ {\left( {q - r - p} \right)}&{\left( {r + p - q} \right)}&{{q^2}} \\\ {p - q}&{ - p}&{pq} \end{array}} \right|
Applying ${R_2} \to {R_2} - \left( {\dfrac{q}{p}} \right){R_3}$, we get
D = 2{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {q + r - \dfrac{{{p^2}}}{q}}&{\dfrac{{{p^2}}}{q}}&0 \\\ {\left( {q - r - p} \right) - \dfrac{q}{p}\left( {p - q} \right)}&{\left( {r + p - q} \right) - \dfrac{q}{p}\left( { - p} \right)}&{{q^2} - \dfrac{q}{p}\left( {pq} \right)} \\\ {p - q}&{ - p}&{pq} \end{array}} \right|
Simplify the above determinant, we get
D = 2{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {q + r - \dfrac{{{p^2}}}{q}}&{\dfrac{{{p^2}}}{q}}&0 \\\ { - r - p + \dfrac{{{q^2}}}{p}}&{r + p}&0 \\\ {p - q}&{ - p}&{pq} \end{array}} \right|
We can see that in the third column two elements are zero. So, let us expand the above determinant along the third column. Therefore, we get
D = 2pq{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {q + r - \dfrac{{{p^2}}}{q}}&{\dfrac{{{p^2}}}{q}} \\\ { - r - p + \dfrac{{{q^2}}}{p}}&{r + p} \end{array}} \right|
Applying ${C_1} \to {C_1} + {C_2}$, we get
D = 2pq{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {q + r - \dfrac{{{p^2}}}{q} + \dfrac{{{p^2}}}{q}}&{\dfrac{{{p^2}}}{q}} \\\ { - r - p + \dfrac{{{q^2}}}{p} + r + p}&{r + p} \end{array}} \right|
Simplify the above determinant, we get
D = 2pq{\left( {p + q + r} \right)^2}\left| {\begin{array}{*{20}{c}} {q + r}&{\dfrac{{{p^2}}}{q}} \\\ {\dfrac{{{q^2}}}{p}}&{r + p} \end{array}} \right|
We can see that now there are two rows and two columns in the above determinant. Let us expand this determinant. Therefore, we get
$D = 2pq{\left( {p + q + r} \right)^2}\left[ {\left( {q + r} \right)\left( {r + p} \right) - \left( {\dfrac{{{p^2}}}{q}} \right)\left( {\dfrac{{{q^2}}}{p}} \right)} \right]$
Simplify the above expression, we get
$D = 2pq{\left( {p + q + r} \right)^2}\left[ {qr + qp + {r^2} + rp - pq} \right] \\\ \Rightarrow D = 2pq{\left( {p + q + r} \right)^2}\left[ {qr + {r^2} + rp} \right] \\\ \Rightarrow D = 2pq{\left( {p + q + r} \right)^2}r\left[ {q + r + p} \right] \\\ \Rightarrow D = 2pqr{\left( {p + q + r} \right)^3} \\\$
Now we have $s - a = p,\;s - b = q,\;s - c = r$ and $p + q + r = s$. Therefore, we get
$D = 2\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right){s^3} \\\ \Rightarrow D = 2{s^3}\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right) \\\$
Therefore, we get \left| {\begin{array}{*{20}{c}} {{a^2}}&{{{\left( {s - a} \right)}^2}}&{{{\left( {s - a} \right)}^2}} \\\ {{{\left( {s - b} \right)}^2}}&{{b^2}}&{{{\left( {s - b} \right)}^2}} \\\ {{{\left( {s - c} \right)}^2}}&{{{\left( {s - c} \right)}^2}}&{{c^2}} \end{array}} \right| = 2{s^3}\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right).

Note: \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\\ {{b_1}}&{{b_2}}&{{b_3}} \\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| is calculated as ${a_1}\left( {{b_2}{c_3} - {b_3}{c_2}} \right) - {a_2}\left( {{b_1}{c_3} - {b_3}{c_1}} \right) + {a_3}\left( {{b_1}{c_2} - {b_2}{c_1}} \right)$. We can perform row operations as well as column operations to convert the given determinant into simplest form.