Find the Product of Two Matrices

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Pre-Calculus › Find the Product of Two Matrices

Questions 1 - 10

We consider the matrices and that we assume of the same size .

$A=\begin{bmatrix} 34&0 &\cdots &0\ 34 &0 &\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 34&0 &0 &0 \end{bmatrix},B=\begin{bmatrix} 0&2 &\cdots &0\ 0 &2&\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 0&2 &0 &0 \end{bmatrix}$

Find the product .

$AB=\begin{bmatrix} 0&68 &\cdots &0\ 0&68&\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 0&68&0 &0 \end{bmatrix}$

$AB=\begin{bmatrix} 0&0 &\cdots &0\ 0 &0 &\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 0&0 &0 &0 \end{bmatrix}$

$AB=\begin{bmatrix} 68&0 &\cdots &0\ 68 &0 &\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 68&0 &0 &0 \end{bmatrix}$

$AB=\begin{bmatrix} 0&0 &\cdots &0\ 0 &0 &\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 68&68 &\cdots&68 \end{bmatrix}$

$AB=\begin{bmatrix} 0&68 &\cdots &0\ 0 &68 &\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 0&0 &0 &0 \end{bmatrix}$

Explanation

Note that multiplying every row of by the first column of gives .

Mutiplying every row of by the second column of gives .

Now the remaining columns are columns of zeros, and therefore this product gives zero in every row-column product.

Knowing these three aspects we get the resulting matrix.

$AB=\begin{bmatrix} 0&68 &\cdots &0\ 0&68&\cdots &0 \\vdots &\vdots &\vdots &\vdots \ 0&68&0 &0 \end{bmatrix}$

$A=\begin{bmatrix} 3\ -10\ 21 \end{bmatrix}$

$B=\begin{bmatrix} 5 & 0 & -1 \ 2 & 1 & 3 \end{bmatrix}$

Find .

No Solution

$\begin{bmatrix} -6\ 59 \end{bmatrix}$

$\begin{bmatrix} 15 & 0 & -21 \ 6 & -10 & 63 \end{bmatrix}$

$\begin{bmatrix} 105 & 0 & -3\ 42 & -10 & 9 \end{bmatrix}$

Explanation

The dimensions of A and B are as follows: A= 3x1, B= 2x3

In order to be able to multiply matrices, the inner numbers need to be the same. In this case, they are 1 and 2. As such, we cannot find their product.

The answer is No Solution.

We consider the two matrices and defined below:

$A=\begin{bmatrix} 1\2 \\3 \4 \end{bmatrix}$ , $B=\begin{bmatrix} 1 &0 &1 \end{bmatrix}$

What is the matrix ?

$\begin{bmatrix} 1 & 0 &1 \ 2& 0 & 2\ 3 & 0 & 3\4 & 0 &4 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 &1 \ 2& 0 & 2\ 3 & 0 & 3\0 & 0 &0 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 &0 \ 2& 0 & 0\ 3 & 0 & 0\4 & 0 &0\end{bmatrix}$

We can't find the product

$\begin{bmatrix} 1 & 1 &1 \ 2& 0 & 2\ 3 & 0 & 3\4 & 0 &4 \end{bmatrix}$

Explanation

The first matrix is (4x1) and the second matrix is (1x3). We can perform the matrix multiplication in this case. The resulting matrix is (4x3).

The first entry in the formed matrix is on the first row and the first column.

It is coming from the product of the first row of A and the first column of B.

This gives $AB(1,1)=1\cdot1=1$ .We continue in this fashion.

The entry (4,3) is coming from the 4th row of A and the 3rd column of B.

This gives $AB(4,3)=4\cdot1=4$ . To obtain the whole matrix we need to remember that any entry on AB say(i,j) is coming from the product of the rom i from A and the column j of B.

After doing all these calculations we obtain:

$\begin{bmatrix} 1 & 0 &1 \ 2& 0 & 2\ 3 & 0 & 3\4 & 0 &4 \end{bmatrix}$

Let

$A=\begin{bmatrix} 1 & 1& 1&1 \ -1& -1 & -1&-1 \1 &1 &1 &1 \ -1&-1 &-1 &-1 \end{bmatrix}$ and

$B=\begin{bmatrix} -1\ 1\ -1 \1 \end{bmatrix}$

What is the matrix ?

Product cannot be found.

$A=\begin{bmatrix} 1 & 1& 1&1 \ -1& 1 & -1&-1 \1 &1 &0 &1 \ -1&-1 &-1 &-1 \end{bmatrix}$

$A=\begin{bmatrix} 0& 1& 1&0 \ -1& -1 & -1&-1 \0 &1 &1 &0 \ -1&-1 &-1 &-1 \end{bmatrix}$

$A=\begin{bmatrix} -1 & 1& 1&-1 \ -1& -1 & -1&-1 \1 &1 &1 &1 \ -1&-1 &-1 &-1 \end{bmatrix}$

$A=\begin{bmatrix} 1 & 1& 1&1 \ -1& -1 & -1&-1 \1 &1 &1 &1 \ 1&1 &1 &1 \end{bmatrix}$

Explanation

We note first that A is 4x4 , B is 4x1.

To be able to do BA the number of columns of B must equal the number of rows

of A.

Since the number of columns of B is 1 and the number of rows of A is 4, we do not have equality and therefore we can't have the product BA.

We consider the matrix equality:

$\begin{bmatrix} x & 1\ 1&x\end{bmatrix}=\begin{bmatrix} x^{2}&1 \ 1&3\end{bmatrix}$

Find the that makes the matrix equality possible.

There is no that satisfies the above equality.

$x=\frac{3}{2}$

Explanation

To have the above equality we need to have $x^{2}=x$ and .

means that , or . Trying all different values of , we see that no can satisfy both matrices.

Therefore there is no that satisfies the above equality.

Multiply $\small \begin{bmatrix} 2 &-1 &4 \ 1& 0& 2\3 &1 &-3 \end{bmatrix}*\begin{bmatrix} 10\-4 \1 \end{bmatrix}$

$\small \begin{bmatrix} 28\ 12\ 23\end{bmatrix}$

$\small \begin{bmatrix} 28 & 12& 23\end{bmatrix}$

$\small \begin{bmatrix} 20\ 12\ 37\end{bmatrix}$

$\small \begin{bmatrix} 20 &12 & 37 \end{bmatrix}$

$\small \begin{bmatrix} 12\ 12 \ 23 \end{bmatrix}$

Explanation

To find the product, line up the rows of the left matrix individually with the one column in the right matrix:

$\small \begin{bmatrix} 2(10)+-1(-4)+4(1)\1(10)+0(-4)+2(1) \ 3(10)+1(-4)+-3(1)\end{bmatrix}$

$\small =\begin{bmatrix} 28\ 12\ 23\end{bmatrix}$

$A=\begin{bmatrix} 14 & -5 & -13 \end{bmatrix}$

$B=\begin{bmatrix} 4 \ -3\ 2 \end{bmatrix}$

Find .

$\begin{bmatrix} 45 \end{bmatrix}$

$\begin{bmatrix} 56\ 15\ 26 \end{bmatrix}$

$\begin{bmatrix} -9 \end{bmatrix}$

No Solution

Explanation

The dimensions of A and B are as follows: A=1x3, B= 3x1.

Because the two inner numbers are the same, we can find the product.

The two outer numbers will tell us the dimensions of the product: 1x1.

$\begin{bmatrix} a & b & c \end{bmatrix} \cdot \begin{bmatrix} d \ e\ f\end{bmatrix}= \begin{bmatrix} ad + be +cf \end{bmatrix}$

Therefore, plugging in our values for this problem we get the following:

Let be the matrix defined by:

$A=\begin{bmatrix} 1 & 1\ 1& 1 \end{bmatrix}$

The value of $A^{n}$ ( the nth power of ) is:

$\begin{bmatrix} 2^{n-1} &2^{n-1}\ 2^{n-1}& 2^{n-1} \end{bmatrix}$

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

$\begin{bmatrix} n+1&n+1 \ n+1& n+1\end{bmatrix}$

$\begin{bmatrix} n & n+2\ n+2&n \end{bmatrix}$

$\begin{bmatrix} n &n+1 \ & n+1\end{bmatrix}$

Explanation

We will use an induction proof to show this result.

We first note the above result holds for n=1. This means $A^{1}=\begin{bmatrix} 1 &1 \ 1 &1 \end{bmatrix}$

We suppose that $A^n=\begin{bmatrix} 2^{n-1}& 2^{n-1} \ 2^{n-1}& 2^{n-1} \end{bmatrix}$ and we need to show that: $A^{n+1}=\begin{bmatrix} 2^{n} &2^{n} \2^{n}&2^{n} \end{bmatrix}$

By definition $A^{n+1}=A^{n}A$ . By inductive hypothesis, we have:

$A^{n}=\begin{bmatrix} 2^{n-1}&2^{n-1}\ 2^{n-1}&2^{n-1}\end{bmatrix}$

Therefore, $A^{n+1}=\begin{bmatrix} 2^{n-1}&2^{n-1} \ 2^{n-1} &2^{n} \end{bmatrix}\begin{bmatrix} 1 & 1\ 1&1 \end{bmatrix}$

$A^{n+1}=\begin{bmatrix} 2^{n} &2^{n} \ 2^{n} &2^{n} \end{bmatrix}$

This shows that the result is true for n+1. By the principle of mathematical induction we have the result.

We consider the matrices and below. We suppose that and are of the same size

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

What is the product ?

$\begin{bmatrix} 2n &2n &\cdots &2n\ 2n &2n &\cdots &2n \\vdots &\vdots &\vdots &\vdots \ 2n &2n &2n &2n \end{bmatrix}$

$\begin{bmatrix} 2n+1 &2n+1 &\cdots &2n+1\ 2n+1 &2n+1 &\cdots &2n+1 \\vdots &\vdots &\vdots &\vdots \ 2n+1 &2n+1 &2n+1 &2n+1 \end{bmatrix}$

$\begin{bmatrix} n+1 &n+1 &\cdots &n+1\ n+1 &n+1 &\cdots &n+1 \\vdots &\vdots &\vdots &\vdots \ n+1 &n+1 &n+1 &n+1 \end{bmatrix}$

$\begin{bmatrix} n &n &\cdots &n\ n &n &\cdots &n \\vdots &\vdots &\vdots &\vdots \ n &n &n &n \end{bmatrix}$

$\begin{bmatrix} 2n-1 &2n-1 &\cdots &2n-1\ 2n-1 &2n -1&\cdots &2n-1 \\vdots &\vdots &\vdots &\vdots \ 2n-1 &2n -1&\cdots &2n-1 \end{bmatrix}$

Explanation

Note that every entry of the product matrix is the sum of ( times) .

$AB=\begin{bmatrix} 1 &1 &\cdots &1 \ 1 &1 &\cdots &1 \\vdots &\vdots &\vdots &\vdots \ 1 &1 &\cdots &1 \end{bmatrix} \times \begin{bmatrix} 2 &2 &\cdots &2 \ 2 &2 &\cdots &2 \\vdots &\vdots &\vdots &\vdots \ 2&2 &\cdots &2 \end{bmatrix}$

$= \begin{bmatrix} 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2 \ 1(2)+1(2)+... 2 & 1(2)+1(2)+... 2\\vdots& \vdots\end{bmatrix}=\begin{bmatrix} 2n & 2n &2n ... \2n & 2n &2n ... \\vdots &\vdots\end{bmatrix}$

This gives as every entry of the product of the two matrices.

$A=\begin{bmatrix} -3 & 0 \ 8 & 5 \end{bmatrix}$

$B=\begin{bmatrix} 2 & 5 \ 10 & 6 \end{bmatrix}$

Find .

$\begin{bmatrix} -6 & -15\ 66 & 70 \end{bmatrix}$

$\begin{bmatrix} 34 & 25 \ 18 & 30 \end{bmatrix}$

$\begin{bmatrix} -6 & 0 \ 80 & 30 \end{bmatrix}$

No Solution

Explanation

The dimensions of both A and B are 2x2. Therefore, the matrix that results from their product will have the same dimensions.

$\begin{bmatrix} a & b \ c & d \end{bmatrix} \cdot \begin{bmatrix} e & f \ g & h \end{bmatrix}=\begin{bmatrix} ae+bg & af+bh \ ce+dg & cf+dh \end{bmatrix}$

Thus plugging in our values for this particular problem we get the following:

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