Name: How to Find Cofactor Matrix?
Uploaded: 2019-07-19
Duration: 4 min 54 s
Description: A cofactor is a number we get when we remove the column and row of a designated element in a matrix. In this video, we will learn to find the cofactor matrix...

Question

Show that $|\begin{array}{ccc} (a + b)^{2} & a c & b c \\ a c & (b + c)^{2} & a b \\ b c & a b & (c + a)^{2} \end{array}| = 2 a b c (a + b + c)^{2}$ .

Note: There is mistake in question. The correct question is as below.

Show that $|\begin{array}{ccc} (a + b)^{2} & a c & b c \\ a c & (b + c)^{2} & a b \\ b c & a b & (c + a)^{2} \end{array}| = 2 a b c (a + b + c)^{3}$ .

Accepted Answer

Consider, $|\begin{array}{ccc} (a + b)^{2} & a c & b c \\ a c & (b + c)^{2} & a b \\ b c & a b & (c + a)^{2} \end{array}|$

Multiply $R_{1}, R_{2}$ and $R_{3}$ by $c, a$ and $b$ respectively.

$= \frac{1}{a b c} |\begin{array}{lcc} c (a + b)^{2} & a c^{2} & b c^{2} \\ a^{2} c & a (b + c)^{2} & a^{2} b \\ b^{2} c & a b^{2} & b (c + a)^{2} \end{array}|$

Taking $c, a$ and $b$ common from $C_{1}, C_{2}$ and $C_{3}$ respectively.

$= |\begin{array}{ccc} (a + b)^{2} & c^{2} & c^{2} \\ a^{2} & (b + c)^{2} & a^{2} \\ b^{2} & b^{2} & (c + a)^{2} \end{array}|$

Apply $C_{1} \to C_{1} - C_{3}, C_{2} \to C_{2} - C_{3}$

$= |\begin{array}{ccc} (a + b)^{2} - c^{2} & 0 & c^{2} \\ 0 & (b + c)^{2} - a^{2} & a^{2} \\ b^{2} - (c + a)^{2} & b^{2} - (c + a)^{2} & (c + a)^{2} \end{array}|$

$= |\begin{array}{ccc} (a + b - c) (a + b + c) & 0 & c^{2} \\ 0 & (b + c - a) (b + c + a) & a^{2} \\ (b - c - a) (b + c + a) & (b - c - a) (b + c + a) & (c + a)^{2} \end{array}|$

Taking $(a + b + c)$ common from $C_{1} and C_{2}$ .

$= (a + b + c)^{2} |\begin{array}{ccc} a + b - c & 0 & c^{2} \\ 0 & b + c - a & a^{2} \\ b - c - a & b - c - a & (c + a)^{2} \end{array}|$

Apply $R_{3} \to R_{3} - (R_{1} + R_{2})$

$= (a + b + c)^{2} |\begin{array}{ccc} a + b - c & 0 & c^{2} \\ 0 & b + c - a & a^{2} \\ - 2 a & - 2 c & 2 a c \end{array}|$

Multiply $C_{1}$ and $C_{2}$ by $c$ and $a$ respectively.

$= \frac{(a + b + c)^{2}}{a c} |\begin{array}{ccc} a c + b c - c^{2} & 0 & c^{2} \\ 0 & b a + c a - a^{2} & a^{2} \\ - 2 a c & - 2 a c & 2 a c \end{array}|$

Apply $C_{1} \to C_{1} + C_{3}, C_{2} \to C_{2} + C_{3}$

$= \frac{(a + b + c)^{2}}{a c} |\begin{array}{ccc} a c + b c & c^{2} & c^{2} \\ a^{2} & a b + c a & a^{2} \\ 0 & 0 & 2 a c \end{array}|$

Taking $c$ and $a$ common from $R_{1}$ and $R_{2}$ respectively.

$= (a + b + c)^{2} |\begin{array}{ccc} a + b & c & c \\ a & b + c & a \\ 0 & 0 & 2 a c \end{array}|$

Expanding along $R_{3}$

$= (a + b + c)^{2} [0 - 0 + 2 a c {(a + b) (b + c) - a c}]$

$= (a + b + c)^{2} [2 a c (a b + a c + b^{2} + b c - a c)]$

$= 2 a b c (a + b + c)^{3}$

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE

Questions from Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 1: EXERCISE with Hints & Solutions

Learn and Prepare for any exam you want !