Q10. - Algebra 2

Question

Find each product, if possible.10 $\left[\begin{array}{cc}3& -1\\ 2& 5\end{array}\right].\left[\begin{array}{ccc}4& -1& -2\\ -3& 5& 4\end{array}\right]$

Step-by-Step Solution

Answer

The answer is $\left[\begin{array}{ccc}15& -8& -10\\ -7& 23& 16\end{array}\right]$

Step 1 - State the definition of a matrix.

A rectangular array of numbers arranged in rows and columns is called a matrix.

Step 2 - Find the product of the matrix.

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix.

 If two matrices say $A_{m\times n}$  and $B_{n\times r}$  are to multiplied. The multiplication is possible as number of columns in the first matrix that is n  is equal to the the number of rows in the second matrix n  and the resultant matrix ab  will be of dimension  .$m\times r$

$\begin{array}{d}{A}_{m\times n}.B_{n\times r}\\ m\times n.n\times r\end{array}$

Here n  and n  are the inner dimensions and m  and r  are the outer dimensions.

 Here the first matrix is:$\left[\begin{array}{cc}3& -1\\ 2& 5\end{array}\right]$

It is of dimension  $2\times 2$.

 And second matrix is:$\left[\begin{array}{ccc}4& -1& -2\\ -3& 5& 4\end{array}\right]$

And it of dimension  .$2\times 3$

Since, the number of columns in the first matrix is equal to the number of rows in the second matrix that is the inner dimensions are equal for the given matrices. So, the product of the matrices is possible.

The dimension of the resultant matrix will be $2\times 3$ .

 The product of the matrices is:

 First multiply columns by rows and then simplify,

$\begin{array}{c}\left[\begin{array}{cc}3& -1\\ 2& 5\end{array}\right].\left[\begin{array}{ccc}4& -1& -2\\ -3& 5& 4\end{array}\right]=\left[\begin{array}{ccc}(3\times 4)+(-1\times -3)& (3\times -1)+(-1\times 5)& (3\times -2)+(-1\times 4)\\ (2\times 4)+(5\times -3)& (2\times -1)+(5\times 5)& (2\times -2)+(5\times 4)\end{array}\right]\\ =\left[\begin{array}{ccc}12+3& -3-5& -6-4\\ 8-15& -2+25& -4+20\end{array}\right]\\ =\left[\begin{array}{ccc}15& -8& -10\\ -7& 23& 16\end{array}\right]\end{array}$

Step 3 - State the conclusion.

Therefore, the product the product is $\left[\begin{array}{ccc}15& -8& -10\\ -7& 23& 16\end{array}\right]$