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in Determinants by (15.2k points)

Using properties of determinants prove that:

\(\begin{bmatrix} {\text{x}} & a & 0 \\[0.3em] a & {\text{x}} & a \\[0.3em] a & a & {\text{x}} \end{bmatrix}\)

|(x,a,a)(a,x,a)(a,a,x)| = (x+2a)(x-a)2

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\(\begin{bmatrix} {\text{x}} & a & 0 \\[0.3em] a & {\text{x}} & a \\[0.3em] a & a & {\text{x}} \end{bmatrix}\)

= (x + 2a)(x - a)[x - ( - a) + ( - a - 0) + ( - a)] 

[expansion by first row] 

= (x + 2a)(x - a)(x + a - a - a) 

= (x + 2a)(x - a)2

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