Answer:
Statement c:
a₁₁ + ( b₁₁+ c₁₁ ) = 0
Explanation:
The options are:
a: a₁₁ = 0 and b₁₁ = 0
b: a₁₁ - ( b₁₁ + c₁₁ ) = 0
c. a₁₁ + ( b₁₁+ c₁₁ ) = 0
d. a₁₁x( b₁₁ + c₁₁) = 0
We know that A, B, and C are matrices of the same size, so we can write:
![A = \left[\begin{array}{ccc}a_(11)&a_(21)&...\\a_(21)&a_(22)&...\\...&...&...\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/high-school/a5jsgsnrmzdnbe51jn49jdgyi4vzqfmlw7.png)
![B = \left[\begin{array}{ccc}b_(11)&b_(21)&...\\b_(21)&b_(22)&...\\...&...&...\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/high-school/kooftbkqf9xr3p71yo6ss0uenkvxhwnf7n.png)
![C = \left[\begin{array}{ccc}c_(11)&c_(21)&...\\c_(21)&c_(22)&...\\...&...&...\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/high-school/wicdf73wdp7bcv3ozy60xz4lgercn9f6v3.png)
And here we have the sum of matrices, remember that:
![A + B = \left[\begin{array}{ccc}a_(11)&a_(21)&...\\a_(21)&a_(22)&...\\...&...&...\end{array}\right] + \left[\begin{array}{ccc}b_(11)&b_(21)&...\\b_(21)&b_(22)&...\\...&...&...\end{array}\right] = \left[\begin{array}{ccc}a_(11) + b_(11)&a_(21) + b_(12)&...\\a_(21) + b_(21)&a_(22) + a_(22)&...\\...&...&...\end{array}\right]]()
And we know that (A + B) + C = 0 (a matrix full of zeros)
then:
![\left[\begin{array}{ccc}(a_(11) + b_(11)) + c_(11) &(a_(21) + b_(12)) + c_(12)&...\\(a_(21) + b_(21)) + c_(21)&(a_(22) + b_(22)) + c_(22)&...\\...&...&...\end{array}\right] = \left[\begin{array}{ccc}0&0&...\\0&...&...\\...&...&...\end{array}\right]](https://img.qammunity.org/2022/formulas/mathematics/high-school/8q95r3x8sgvr9n0kheod8gmoop9py45elq.png)
Then:
(a₁₁ + b₁₁ )+ c₁₁ = 0
These are real numbers, so we can rewrite this as:
a₁₁ + ( b₁₁+ c₁₁ ) = 0
Then statement c is the correct one.