# Use the graphing calculator to solve if possibleA=begin{bmatrix}1 & 0&5 1 & -5&70&3&-4 end{bmatrix}B=begin{bmatrix}3 & -5&3 2&3&14&1&-3end{bmatrix}C=begin{bmatrix}5 & 2&3 2& -1&0 end{bmatrix}D=begin{bmatrix}5 -34 end{bmatrix}Find the value in row 2 column 3 of AB-3B

Use the graphing calculator to solve if possible
$A=\left[\begin{array}{ccc}1& 0& 5\\ 1& -5& 7\\ 0& 3& -4\end{array}\right]\phantom{\rule{0ex}{0ex}}B=\left[\begin{array}{ccc}3& -5& 3\\ 2& 3& 1\\ 4& 1& -3\end{array}\right]\phantom{\rule{0ex}{0ex}}C=\left[\begin{array}{ccc}5& 2& 3\\ 2& -1& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}D=\left[\begin{array}{c}5\\ -3\\ 4\end{array}\right]$
Find the value in row 2 column 3 of $AB-3B$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

sweererlirumeX
Step 1
Given:
The given matrices are
$A=\left[\begin{array}{ccc}1& 0& 5\\ 1& -5& 7\\ 0& 3& -4\end{array}\right]$
$B=\left[\begin{array}{ccc}3& -5& 3\\ 2& 3& 1\\ 4& 1& -3\end{array}\right]$
$C=\left[\begin{array}{ccc}5& 2& 3\\ 2& -1& 0\end{array}\right]$
$D=\left[\begin{array}{c}5\\ -3\\ 4\end{array}\right]$
To find:
The value in row 2 column 3 of AB-3B.
Step 2
The matrices are $A=\left[\begin{array}{ccc}1& 0& 5\\ 1& -5& 7\\ 0& 3& -4\end{array}\right],B=\left[\begin{array}{ccc}3& -5& 3\\ 2& 3& 1\\ 4& 1& -3\end{array}\right]$
Now,
$AB-3B=\left[\begin{array}{ccc}1& 0& 5\\ 1& -5& 7\\ 0& 3& -4\end{array}\right]\left[\begin{array}{ccc}3& -5& 3\\ 2& 3& 1\\ 4& 1& -3\end{array}\right]-3\left[\begin{array}{ccc}3& -5& 3\\ 2& 3& 1\\ 4& 1& -3\end{array}\right]$
$AB-3B=\left[\begin{array}{ccc}23& 0& -12\\ 21& -13& -23\\ -10& 5& 15\end{array}\right]-\left[\begin{array}{ccc}9& -15& 9\\ 6& 9& 3\\ 12& 3& -9\end{array}\right]$
$AB-3B=\left[\begin{array}{ccc}14& 15& -21\\ 15& -22& -26\\ -22& 2& 24\end{array}\right]$
The value in row 2 column 3 of AB-3B is -26.
Jeffrey Jordon