# prove that-`|~` b+c a a`~|` b c+a b =4abc `|__` c c a+b`__|`

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Show that `|[b+c,a,a],[b,a+c,b],[c,c,a+b]|=4abc ` where || indicates the determinant.

(1) There is a special case algorithm for the determinant of a 3x3 matrix. Rewrite the first two columns outside the original matrix on the right side.

The determinant is the sum of the products of the elements of the 3 diagonals read from upper left to lower right minus the sum of the products of the elements of the 3 diagonals as read from lower left to upper right.

`[[b+c,a,a,b+c,a],[b,a+c,b,b,a+c],[c,c,a+b,c,c]] ` :

The determinant is:

{(b+c)(a+c)(a+b)+abc+abc}-{c(a+c)a+cb(b+c)+(a+b)ba}

=`4abc+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2 `

`-{a^2b+a^2c+ab^2+b^2c+ac^2+bc^2} `

=4abc as required.

(2) The more general way of finding the determinant is by expansion by minors:

We choose an arbitrary row or column. Generally you would choose a row or column with as many zeros and ones as possible.

Here choose the first column. The determinant of the 3x3 matrix is the sum of the signed determinants of the product of the column entries and the determinant of the corresponding minor.

The determinant is:

`(b+c)|[a+c,b],[c,a+b]|-b|[a,a],[c,a+b]|+c|[a,a],[a+c,b]| `

Note that we have -b in the second term-- the sign is determined by `(-1)^(i+j) ` where i and j are the row and column numbers respectively.

The determinant of a 2x2 matrix is defined as the difference of the product of the elements in the main diagonal and the product of the other diagonal.

So the determinant is:

(b+c)((a+c)(a+b)-bc)-b(a(a+b)-ac)+c(ab-a(a+c))

=4abc .

**Sources:**

determinate:

`|~` b+c a a`~|`

b c+a b

` ``|__` c c a+b`__|`

**sol:**

determinate

`|~` b+c a a`~|`

b c+a b

`|__` c c a+b `__|`

`= (b+c)[(a+c)(a+b) -(b*c)] - a[(b(a+b)-(b*c))] +a[(b*c)-(c*(a+c))]`

= [bca+bca+bcc+bba+bbc+baa+caa+cca ]- a[ab+bb-bc]+a[bc-ac-cc]

= [bca+bca+bcc+bba+bbc+baa+caa+cca ]- aab-abb+abc+abc-aac-acc]

Upon cancellation we get

= 4abc

Please check the solution in the attachment for better understanding.

`|[b+c,a,a],[b,a+c,b],[c,c,a+b]|`

= `|[b,a,a],[b,a,b],[c,c,a]|+|[c,a,a],[b,c,b],[c,c,b]|`

so finding the det separately we get,

`|[b,a,a],[b,a,b],[c,c,a]| = b[(a^2)-bc] -a[(ab)-(bc)] +a[(bc)-(ac)]`

=`b(a^2)+abc+abc-(a^2)c-(a^2)b-(b^2)c`

= baa+abc+abc-aac-aba-bbc

and

`|[c,a,a],[b,c,b],[c,c,b]|=c[(a^2)-bc] -b[(ab)-(bc)] +c[(bc)-(ac)]`

= ccb+abc+abc-acc-abb-cbc

Now,

`|[b+c,a,a],[b,a+c,b],[c,c,a+b]|` = `|[b,a,a],[b,a,b],[c,c,a]|+|[c,a,a],[b,c,b],[c,c,b]|`

=[ baa+abc+abc-aac-aba-bbc] + [ccb+abc+abc-acc-abb-cbc]

=baa+abc+abc-aac-aba-bbc+ccb+abc+abc-acc-abb-cbc

=**4abc**