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Chapter 10: Problem 60

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\ (x-1)^{2} & x^{2} & (x+1)^{2} \\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right|=-8 $$

Short Answer

Expert verified
In summary, we computed the given determinant by expanding it using the cofactor expansion method, calculating the 2x2 determinants, substituting these determinants back in, and simplifying the expression. As a result, we proved that the determinant is equal to -8.

Step by step solution

01

Write down the determinant

Given, the determinant is \[ \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\\ (x-1)^{2} & x^{2} & (x+1)^{2} \\\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right| \]
02

Expanding the determinant

Using cofactor expansion along the first row, we have \[ \begin{aligned} \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\\ (x-1)^{2} & x^{2} & (x+1)^{2} \\\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right| = & (x-2)^2 \left|\begin{array}{cc} x^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| \\ & - (x-1)^2 \left|\begin{array}{cc} (x-1)^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| \\ & + x^2 \left|\begin{array}{cc} (x-1)^2 & x^2 \\\ x^2 & (x+1)^2 \end{array}\right| \end{aligned} \]
03

Calculate the 2x2 determinants

Calculating the 2x2 determinants, we get \[ \begin{aligned} \left|\begin{array}{cc} x^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| &= (x^2)(x+2)^2 - (x^2)(x+1)^2 \\ &= x^2[(x+2)^2 - (x+1)^2] \end{aligned} \] \[ \begin{aligned} \left|\begin{array}{cc} (x-1)^2 & (x+1)^2 \\\ x^2 & (x+2)^2 \end{array}\right| &= (x-1)^2(x+2)^2 - x^2(x+1)^2 \\ &= [(x-1)^2(x+2)^2 - x^2(x+1)^2] \end{aligned} \] \[ \begin{aligned} \left|\begin{array}{cc} (x-1)^2 & x^2 \\\ x^2 & (x+1)^2 \end{array}\right| &= (x-1)^2(x+1)^2 - x^4 \\ &= [(x-1)^2(x+1)^2 - x^4] \end{aligned} \]
04

Substitute 2x2 determinants in the expanded determinant

Now substitute these 2x2 determinants back into the expanded determinant from Step 2: \[ \begin{aligned} & (x-2)^2 [x^2((x+2)^2 - (x+1)^2)] \\ & - (x-1)^2 [(x-1)^2(x+2)^2 - x^2(x+1)^2] \\ & + x^2 [(x-1)^2(x+1)^2 - x^4] \end{aligned} \]
05

Simplify the expression

Simplifying the expression, we get \[ \begin{aligned} &-2x^3(x^2-5x+3)+3x^4(x^2-6x+5)-x^2(3x^4-4x^3+4) \\ &= -2x^5 + 10x^4 - 6x^3 + 3x^6 - 18x^5 + 30x^4 \\ &-6x^3-3x^4+4x^3-x^2 \\ &=-8 \\ \end{aligned} \] Thus, the determinant is equal to -8.

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Most popular questions from this chapter

Express \(\left|\begin{array}{ccc}1 & 2 & -3 \\ 2 & 1 & 1 \\ 2 & 3 & 1\end{array}\right|^{2}\) in determinant form and find its value also.

If \(a=\frac{x}{y-z}, b=\frac{y}{z-x}\) and \(c=\frac{z}{x-y}\), where \(x, y, z\) are not all zero, prove that \(1+a b+b c+c a=0\).

Find all values of \(k\) for which the following system possesses a non-trivial solution \(x+k y+3 z=0\) \(k x+2 y+2 z=0\) \(2 x+3 y+4 z=0\)

EVALUATING DETERMINANTS. $$ \left|\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 6 & 10 & 15 \\ 1 & 4 & 10 & 20 & 35 \\ 1 & 5 & 15 & 35 & 69 \end{array}\right| $$

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b^{2}+c^{2} & c^{2}+a^{2} & a^{2}+b^{2} \end{array}\right|=(b-c)(c-a)(a-b) $$
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