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3 years ago

When adding cells you must use a "+" symbol, you cannot use a ":" symbol. ☐ True ☐ False

Computers and Technology

1 answer:

enot [183]3 years ago

6 0

Answer:

False

Explanation:

We can use both. If the total number of cells are less we can use + and add like =Sum(A1 + A2 +A3) and if cells are in sequence, and in good number then we can use ":" like =Sum(A1:A7).

With + we can only add two cells at a time, and we need to use it again and again to add all cells, like:

If we want to add in between C1 to H1, then with plus we need to write =SUM(C1+D1+E1+F1+G1+H1).

However with : we only need to write =SUM(C1:H1), and that's it.

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What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent

Scrat [10]

Answer:

The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

$rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)$

Then satisfying this theorem the system is consistent and has one single solution.

Explanation:

1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

$rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)$

$rank(A)$

Then the system is consistent and has a unique solution.

$\left\{\begin{matrix}x-3y-2z=6 \\ 2x-4y-3z=8 \\ -3x+6y+8z=-5 \end{matrix}\right.$

2) Writing it as Linear system

$A=\begin{pmatrix}1 & -3 &-2 \\ 2& -4 &-3 \\ -3 &6 &8 \end{pmatrix}$ $B=\begin{pmatrix}6\\ 8\\ 5\end{pmatrix}$

$rank(A) =\left(\begin{matrix}7 & 0 & 0 \\0 & 7 & 0 \\0 & 0 & 7\end{matrix}\right)=3$

3) The Rank (A) is 3 found through Gauss elimination

$(A|B)=\begin{pmatrix}1 & -3 &-2 &6 \\ 2& -4 &-3 &8 \\ -3&6 &8 &-5 \end{pmatrix}$

$rank(A|B)=\left(\begin{matrix}1 & -3 & -2 \\0 & 2 & 1 \\0 & 0 & \frac{7}{2}\end{matrix}\right)=3$

4) The rank of (A|B) is also equal to 3, found through Gauss elimination:

So this linear system is consistent and has a unique solution.

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