Chapter 16, Problem 7CT

To determine

To find: The value of a single matrix $B^2$ .

Answer to Problem 7CT

The value of single matrix $B^2=\left(\begin{array}{ccc}9& 8& -14\\ -35& 33& 19\\ 15& -18& 4\end{array}\right)$ .

Explanation of Solution

Given Information:

The given matrix is

  $A=\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)$ and $B=\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)$.

Calculation:

Consider the values:

  $\left(\begin{array}{ccc}2& 1& 3\\ 0& -1& 1\\ 4& -1& 0\end{array}\right)$for A, and $\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)$for B.

We know that $B^2$ is equivalent to $B\cdot B$ .

Substitute $\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)$for $B$ .

  $B^2=\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)$

Since the dimensions of the two matrices are equal, their product is defined and it will be a $3\times 3$ matrix.

The element in the $ath$ row and $bth$ column of the product matrix will be the sum of the products of the corresponding elements in the $ath$ row of the first matrix and $bth$ column of the second matrix.

Multiply the elements in the first row of the first matrix by the corresponding elements in the first column of the second matrix. Add the products and the result will be element in the first row, first column of the product matrix.

  $\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)=\left(\begin{array}{ccc}-2(-2)& 1(5)& 5(0)\\ & & \\ & & \end{array}\right)$ .

Now, multiply the elements in the first row of the first matrix by the elements in the second column of the second matrix. Add the products and put the result in the first row, second column of the product matrix.

Similarly, we can find the element in the first row third column of the product matrix using the elements in the first row of the first matrix and the elements in the third column of the second matrix.

  $\begin{array}{c}\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)\\ =\left(\begin{array}{ccc}-2(-2)+1(5)+5(0)& -2(1)+1(-5)+5(3)& -2(5)+1(1)+5(-1)\\ & & \\ & & \end{array}\right)\end{array}$

In order to find the elements in the second row of the product matrix, multiply the elements in the second row of the first matrix by the corresponding elements in each column of the second matrix. Similarly, we can find the elements in the third row of the product matrix.

  $\begin{array}{c}\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)\left(\begin{array}{ccc}-2& 1& 5\\ 5& -5& 1\\ 0& 3& -1\end{array}\right)\\ =\left(\begin{array}{ccc}-2(-2)+1(5)+5(0)& -2(1)+1(-5)+5(3)& -2(5)+1(1)+5(-1)\\ 5(-2)+(-5)(5)+1(0)& 5(1)+(-5)(-5)+1(3)& 5(5)+(-5)(1)+1(-1)\\ 0(-2)+3(5)+(-1)(0)& 0(1)+3(-5)+(-1)(3)& 0(5)+3(1)+(-1)(-1)\end{array}\right)\end{array}$

Simplify the product matrix.

  $\begin{array}{c}\left(\begin{array}{ccc}-2(-2)+1(5)+5(0)& -2(1)+1(-5)+5(3)& -2(5)+1(1)+5(-1)\\ 5(-2)+(-5)(5)+1(0)& 5(1)+(-5)(-5)+1(3)& 5(5)+(-5)(1)+1(-1)\\ 0(-2)+3(5)+(-1)(0)& 0(1)+3(-5)+(-1)(3)& 0(5)+3(1)+(-1)(-1)\end{array}\right)\\ =\left(\begin{array}{ccc}9& 8& -14\\ -35& 33& 19\\ 15& -18& 4\end{array}\right)\end{array}$

Therefore, the value of single matrix is $B^2=\left(\begin{array}{ccc}9& 8& -14\\ -35& 33& 19\\ 15& -18& 4\end{array}\right)$ .