Sum or Difference of Two Matrices

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Pre-Calculus › Sum or Difference of Two Matrices

Questions 1 - 10

Add:

$\begin{bmatrix} 11& 9\ 7& -5 \end{bmatrix}+$ $\begin{bmatrix} 6& 9\ 10& 13 \end{bmatrix}$

$\begin{bmatrix} 17& 18\ 17& 8 \end{bmatrix}$

$\begin{bmatrix} 17& 19\ 16& 15 \end{bmatrix}$

$\begin{bmatrix} 35\ 25 \end{bmatrix}$

$\begin{bmatrix} 36\ 31 \end{bmatrix}$

Explanation

In order to add matrices, they have to be of the same dimension. In this case, they are both 2x2.

The formula to add matrices is as follows:

$\begin{bmatrix} a& b\ c& d \end{bmatrix}+\begin{bmatrix} e& f\ g& h \end{bmatrix}=\begin{bmatrix} a+e& b+f\ c+g& d+h \end{bmatrix}$

Plugging in our values to solve we get:

$\begin{bmatrix} 11& 9\ 7& -5 \end{bmatrix}+\begin{bmatrix} 6& 9\ 10& 13 \end{bmatrix}=\begin{bmatrix} 11+6& 9+9\ 7+10& -5+13 \end{bmatrix}=\begin{bmatrix} 17& 18\ 17& 8 \end{bmatrix}$

We consider the following matrices:

$A=\begin{bmatrix} 1 & 2 &2 \ 1&2 &2 \1 &2 &2 \end{bmatrix},B=\begin{bmatrix} 1 &2 & 2 &2 \1 &2 &2 & 2\1 &2 &2 &2 \end{bmatrix}$

Find .

We can't find .

$\begin{bmatrix} 0 &0 & 0\ 0&0 &0 \0 &0 &0 \end{bmatrix}$

$\begin{bmatrix} 0 & 0 &2 \0 & 0 & 2\0 &0 &2 \end{bmatrix}$

$\begin{bmatrix} 0 &0 &0 &0 \ 0&0 &0 &0 \ 0&0 &0 &0 \ 0 &0 &0 &0 \end{bmatrix}$

$\begin{bmatrix} 0 &0 &0 &2 \0 & 0 &0 &2 \ 0&0 &0 &2 \ 0 &0 &0 &2 \end{bmatrix}$

Explanation

To be able to subtract matrices, the matrices must have the same size.

A is 3x3 and B is 3x4. Therefore we can't perform this operation.

We consider the following matrices:

$A=\begin{bmatrix} 1 & 2 &2 \ 1&2 &2 \1 &2 &2 \end{bmatrix},B=\begin{bmatrix} 1 &2 & 2 &2 \1 &2 &2 & 2\1 &2 &2 &2 \end{bmatrix}$

Find .

We can't find .

$\begin{bmatrix} 0 &0 & 0\ 0&0 &0 \0 &0 &0 \end{bmatrix}$

$\begin{bmatrix} 0 & 0 &2 \0 & 0 & 2\0 &0 &2 \end{bmatrix}$

$\begin{bmatrix} 0 &0 &0 &0 \ 0&0 &0 &0 \ 0&0 &0 &0 \ 0 &0 &0 &0 \end{bmatrix}$

$\begin{bmatrix} 0 &0 &0 &2 \0 & 0 &0 &2 \ 0&0 &0 &2 \ 0 &0 &0 &2 \end{bmatrix}$

Explanation

To be able to subtract matrices, the matrices must have the same size.

A is 3x3 and B is 3x4. Therefore we can't perform this operation.

Add:

$\begin{bmatrix} 11& 9\ 7& -5 \end{bmatrix}+$ $\begin{bmatrix} 6& 9\ 10& 13 \end{bmatrix}$

$\begin{bmatrix} 17& 18\ 17& 8 \end{bmatrix}$

$\begin{bmatrix} 17& 19\ 16& 15 \end{bmatrix}$

$\begin{bmatrix} 35\ 25 \end{bmatrix}$

$\begin{bmatrix} 36\ 31 \end{bmatrix}$

Explanation

In order to add matrices, they have to be of the same dimension. In this case, they are both 2x2.

The formula to add matrices is as follows:

$\begin{bmatrix} a& b\ c& d \end{bmatrix}+\begin{bmatrix} e& f\ g& h \end{bmatrix}=\begin{bmatrix} a+e& b+f\ c+g& d+h \end{bmatrix}$

Plugging in our values to solve we get:

$\begin{bmatrix} 11& 9\ 7& -5 \end{bmatrix}+\begin{bmatrix} 6& 9\ 10& 13 \end{bmatrix}=\begin{bmatrix} 11+6& 9+9\ 7+10& -5+13 \end{bmatrix}=\begin{bmatrix} 17& 18\ 17& 8 \end{bmatrix}$

Let

$A=\begin{pmatrix} 5&1 \ 3&-7 \end{pmatrix}$ $B=\begin{pmatrix} 4&2 \ 9& 0 \end{pmatrix}$

Determine the sum .

$\begin{pmatrix} 9&3 \ 12&-7 \end{pmatrix}$

DNE

$\begin{pmatrix} 1&-1 \ -6&7 \end{pmatrix}$

$\begin{pmatrix} 4&2 \ 9&-7 \end{pmatrix}$

Explanation

Since the dimensions of the two matrices are equal the sum of the two matrices exists.

To find the sum, add each component entry from the first matrix to the same component entry of the second matrix.

$A+B=\begin{pmatrix} 5&1 \ 3&-7 \end{pmatrix}+\begin{pmatrix} 4&2 \ 9& 0 \end{pmatrix}}=\begin{pmatrix} 9&3 \ 12&-7 \end{pmatrix}$

Let

$A=\begin{pmatrix} 5&1 \ 3&-7 \end{pmatrix}$ $B=\begin{pmatrix} 4&2 \ 9& 0 \end{pmatrix}$

Determine the sum .

$\begin{pmatrix} 9&3 \ 12&-7 \end{pmatrix}$

DNE

$\begin{pmatrix} 1&-1 \ -6&7 \end{pmatrix}$

$\begin{pmatrix} 4&2 \ 9&-7 \end{pmatrix}$

Explanation

Since the dimensions of the two matrices are equal the sum of the two matrices exists.

To find the sum, add each component entry from the first matrix to the same component entry of the second matrix.

$A+B=\begin{pmatrix} 5&1 \ 3&-7 \end{pmatrix}+\begin{pmatrix} 4&2 \ 9& 0 \end{pmatrix}}=\begin{pmatrix} 9&3 \ 12&-7 \end{pmatrix}$

If $\begin{bmatrix} a&b\ c&d \end{bmatrix} + \begin{bmatrix} 3&14\ 12&4 \end{bmatrix} = \begin{bmatrix} 19&2\ 4&5 \end{bmatrix}$ , what is ?

Explanation

You can treat matrices just like you treat other members of an equation. Therefore, you can subtract the matrix

$\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

from both sides of the equation. This gives you:

$\begin{bmatrix} a&b\ c&d \end{bmatrix}=\begin{bmatrix} 19&2\ 4&5 \end{bmatrix}-\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

Now, matrix subtraction is simple. You merely subtract each element, matching the correlative spaces with each other:

$\begin{bmatrix} a&b\ c&d \end{bmatrix} = \begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix}$

Then, you simplify:

$\begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix} = \begin{bmatrix} 16&-12\ -8&1 \end{bmatrix}$

Therefore,

$\begin{bmatrix} 7& 16& -3\ -13& 5&8 \ 4&-2 &21 \end{bmatrix}+\begin{bmatrix} -11& 13& 4\ 6& 5& -18\ 16& 3& -9 \end{bmatrix}=$

$\begin{bmatrix} -4& 29& 1\ -7& 10& -10\ 20& 1& 12 \end{bmatrix}$

$\begin{bmatrix} -4& 29& 1\ -9& 10& -11\ 21& 11& 12 \end{bmatrix}$

$\begin{bmatrix} -4& 29& 1\ -7& 8& -10\ 20& -3& 12 \end{bmatrix}$

$\begin{bmatrix} -4& -29& -1\ -7& -5& -1\ 20& 8& 12 \end{bmatrix}$

$\begin{bmatrix} 4& 29& -1\ -17& 10& 10\ 20& -1& -12 \end{bmatrix}$

Explanation

To find the sum of two matrices, we simply add each entry from one matrix to the corresponding entry of the other matrix, and the result becomes the entry in the same location of the matrix for their sum:

$\begin{bmatrix} 7& 16& -3\ -13& 5&8 \ 4&-2 &21 \end{bmatrix}+\begin{bmatrix} -11& 13& 4\ 6& 5& -18\ 16& 3& -9 \end{bmatrix}=\begin{bmatrix} -4& 29& 1\ -7& 10& -10\ 20& 1& 12 \end{bmatrix}$

$\begin{bmatrix} 7& 16& -3\ -13& 5&8 \ 4&-2 &21 \end{bmatrix}+\begin{bmatrix} -11& 13& 4\ 6& 5& -18\ 16& 3& -9 \end{bmatrix}=$

$\begin{bmatrix} -4& 29& 1\ -7& 10& -10\ 20& 1& 12 \end{bmatrix}$

$\begin{bmatrix} -4& 29& 1\ -9& 10& -11\ 21& 11& 12 \end{bmatrix}$

$\begin{bmatrix} -4& 29& 1\ -7& 8& -10\ 20& -3& 12 \end{bmatrix}$

$\begin{bmatrix} -4& -29& -1\ -7& -5& -1\ 20& 8& 12 \end{bmatrix}$

$\begin{bmatrix} 4& 29& -1\ -17& 10& 10\ 20& -1& -12 \end{bmatrix}$

Explanation

$\begin{bmatrix} 7& 16& -3\ -13& 5&8 \ 4&-2 &21 \end{bmatrix}+\begin{bmatrix} -11& 13& 4\ 6& 5& -18\ 16& 3& -9 \end{bmatrix}=\begin{bmatrix} -4& 29& 1\ -7& 10& -10\ 20& 1& 12 \end{bmatrix}$

If $\begin{bmatrix} a&b\ c&d \end{bmatrix} + \begin{bmatrix} 3&14\ 12&4 \end{bmatrix} = \begin{bmatrix} 19&2\ 4&5 \end{bmatrix}$ , what is ?

Explanation

You can treat matrices just like you treat other members of an equation. Therefore, you can subtract the matrix

$\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

from both sides of the equation. This gives you:

$\begin{bmatrix} a&b\ c&d \end{bmatrix}=\begin{bmatrix} 19&2\ 4&5 \end{bmatrix}-\begin{bmatrix} 3&14\ 12&4 \end{bmatrix}$

Now, matrix subtraction is simple. You merely subtract each element, matching the correlative spaces with each other:

$\begin{bmatrix} a&b\ c&d \end{bmatrix} = \begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix}$

Then, you simplify:

$\begin{bmatrix} 19-3&2-14\ 4-12&5-4 \end{bmatrix} = \begin{bmatrix} 16&-12\ -8&1 \end{bmatrix}$

Therefore,

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