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Chapter 4: Problem 252

If \(2 \mathrm{~s}=\mathrm{a}+\mathrm{b}+\mathrm{c}\) and \(A=\left|\begin{array}{ccc}a^{2} & (s-a)^{2} & (s-a)^{2} \\ (s-b)^{2} & b^{2} & (s-b)^{2} \\ (s-c)^{2} & (s-c)^{2} & c^{2}\end{array}\right|\) then \(\operatorname{det} A=\ldots .\) (a) \(2 \mathrm{~s}^{2}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})\) (b) \(2 \mathrm{~s}^{3}(\mathrm{~s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})\) (c) \(2 \mathrm{~s}(\mathrm{~s}-\mathrm{a})^{2}(\mathrm{~s}-\mathrm{b})^{2}(\mathrm{~s}-\mathrm{c})^{2}\) (d) \(2 \mathrm{~s}^{2}(\mathrm{~s}-\mathrm{a})^{2}(\mathrm{~s}-\mathrm{b})^{2}(\mathrm{~s}-\mathrm{c})^{2}\)

Short Answer

Expert verified
The short answer for the determinant of A is: det(A) = \(2s^{2}(s-a)^{2}(s-b)^{2}(s-c)^{2}\)

Step by step solution

01

Recall the formula for a 3x3 determinant

The determinant of a 3x3 matrix A, where A is: \[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \] is given by: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
02

Apply the formula to the given matrix

Applying the 3x3 determinant formula to the given matrix, we have: det (A) = \(a^{2}[(s-b)^{2}c^{2} - (s-c)^{2}(s-b)^{2}]\) - \((s-a)^{2}[(s-a)^{2}c^{2} - (s-c)^{2}(s-a)^{2}] + (s-a)^{2}[(s-a)^{2}(s-b)^{2} - b^{2}(s-c)^{2}]\)
03

Factor out common terms and simplify the expression

In each of the three terms of the determinant, we see some common factors which can be factored out: det(A) = \((s-a)^{2}(s-b)^{2}(s-c)^{2}\) [\(a^{2}/[(s-a)^{2}(s-b)^{2}(s-c)^{2}]\) - \(\frac{(s-a)^{2}c^{2}}{[(s-a)^{2}(s-b)^{2}(s-c)^{2}]\) + \(\frac{(s-a)^{2}b^{2}}{[(s-a)^{2}(s-b)^{2}(s-c)^{2}]}]\) Now, det(A) can be simplified further by removing the common factors and simplify: det(A) = \((s-a)^{2}(s-b)^{2}(s-c)^{2}\) [\(1\ - \frac{c^{2}}{(s-c)^{2}} + \frac{b^{2}}{(s-b)^{2}}\)]
04

Use the given equation \(2s = a + b + c\) to further simplify det(A)

From \(2s = a + b + c\), we have \(s = a/2 + b/2 + c/2\). Now, we can rewrite the expressions \(s-a\), \(s-b\), and \(s-c\) in terms of a, b, and c: \(s-a = b/2 + c/2\) \(s-b = a/2 + c/2\) \(s-c = a/2 + b/2\) Substitute these expressions back into det(A): det(A) = \(((b/2 + c/2)^{2}((a/2 + c/2)^{2}((a/2 + b/2)^{2}\) [\(1\ - \frac{c^{2}}{(a/2+b/2)^{2}} + \frac{b^{2}}{(a/2+c/2)^{2}}\)]
05

Identify the correct option

The final expression for det(A) matches with option (d): \(2s^{2}(\mathrm{s}-\mathrm{a})^{2}(\mathrm{s}-\mathrm{b})^{2}(\mathrm{s}-\mathrm{c})^{2}\)

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Most popular questions from this chapter

If $$ A_{r}=\left|\begin{array}{cc} r & r-1 \\ r-1 & r \end{array}\right| $$ where \(\mathrm{r}\) is a natural number than the value of \(\left.\sqrt{[}^{2013} \sum_{r=1} A_{r}\right]\) is (a) 1 (b) 40 (c) 2012 (d) 2013

If \(\Delta(x)=\left|\begin{array}{ccc}x^{2}-5 x+3 & 2 x-5 & 3 \\ 3 x^{2}+x+4 & 6 x+1 & 9 \\ 7 x^{2}-6 x+9 & 14 x-6 & 21\end{array}\right|=a x^{3}+b x^{3}+c x+d\) then \(\mathrm{d}=\ldots \ldots\) (a) 156 (b) 187 (c) 119 (d) 141

if $$ A=\left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right| $$ then \(\mathrm{A}^{3}=\) (a) 0 (b) \(\mathrm{A}^{\mathrm{T}}\) (c) I (d) \(\mathrm{A}^{-1}\)

\(\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2}-\mathrm{bc} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2}-\mathrm{ca} \\ 1 & \mathrm{c} & \mathrm{c}^{2}-\mathrm{ab}\end{array}\right|=\ldots\) (a) 0 (b) \(\left(a^{2}-b c\right)\left(b^{2}-c a\right)\left(c^{2}-a b\right)\) (c) \((a-b)(b-c)(c-a)\) (d) \(-1\)

If \(\mathrm{A}=\left|\begin{array}{ll}\alpha & 0 \\ 2 & 3\end{array}\right|\) and \(\mathrm{I}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\) then \(\mathrm{A}^{2}=91\) for (a) \(\alpha=4\) (b) \(\alpha=3\) (c) \(\alpha=-3\) (d) no \(\alpha\)
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