Q9. - Algebra 2

Question

Use matrices$A$, B, and C to find the following.

 

$A=\left[\begin{array}{cc}2& 3\\ 5& 6\end{array}\right],B=\left[\begin{array}{cc}-1& 7\\ 0& 4\end{array}\right],C=\left[\begin{array}{cc}9& -4\\ -6& 5\end{array}\right]$

 

9. $3B-2C$

Step-by-Step Solution

Answer

The final matrix after performing the indicated matrix operations is

$3B-2C=\left[\begin{array}{cc}-21& 29\\ 12& 2\end{array}\right]$ .

Step 1 - Substitute the matrices

Substitute $\left[\begin{array}{cc}-1& 7\\ 0& 4\end{array}\right]$ for B and $\left[\begin{array}{cc}9& -4\\ -6& 5\end{array}\right]$ for C into $3B-2C$.

 

$3B-2C=3\left[\begin{array}{cc}-1& 7\\ 0& 4\end{array}\right]-2\left[\begin{array}{cc}9& -4\\ -6& 5\end{array}\right]$

Step 2 - Perform the scalar multiplication

Scalar multiplication - The product of a scalar k and an $m\times n$ matrix is an $m\times n$ matrix in which each element equals k times the corresponding elements of the original matrix.

Multiply each element of the first matrix $\left[\begin{array}{cc}-1& 7\\ 0& 4\end{array}\right]$ by 3 and multiply each element of the second matrix $\left[\begin{array}{cc}9& -4\\ -6& 5\end{array}\right]$ by 2.

$\begin{array}{c}3B-2C=3\left[\begin{array}{cc}-1& 7\\ 0& 4\end{array}\right]-2\left[\begin{array}{cc}9& -4\\ -6& 5\end{array}\right]\\ =\left[\begin{array}{cc}3\left(-1\right)& 3\left(7\right)\\ 3\left(0\right)& 3\left(4\right)\end{array}\right]-\left[\begin{array}{cc}2\left(9\right)& 2\left(-4\right)\\ 2\left(-6\right)& 2\left(5\right)\end{array}\right]\\ =\left[\begin{array}{cc}-3& 21\\ 0& 12\end{array}\right]-\left[\begin{array}{cc}18& -8\\ -12& 10\end{array}\right]\end{array}$

Step 3 - Perform the matrix subtraction

Matrix subtraction - If A and B are two matrices of order $m\times n$, then the subtraction of the two matrices will also yield a matrix of order $m\times n$ wherein each element is the difference of the corresponding elements.

$\begin{array}{c}3B-2C=\left[\begin{array}{cc}-3& 21\\ 0& 12\end{array}\right]-\left[\begin{array}{cc}18& -8\\ -12& 10\end{array}\right]\\ =\left[\begin{array}{cc}-3-18& 21-\left(-8\right)\\ 0-\left(-12\right)& 12-10\end{array}\right]\\ =\left[\begin{array}{cc}-21& 29\\ 12& 2\end{array}\right]\end{array}$