Question

If the matrices A, B and C are given as \[A=\left[ \begin{matrix}
   1 \\
   3 \\
   -6 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   -5 \\
   0 \\
\end{matrix} \right]\], \[B=\left[ \begin{matrix}
   -1 \\
   4 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   2 \\
   5 \\
\end{matrix} \right]\] and \[C=\left[ \begin{matrix}
   2 \\
   -1 \\
   -3 \\
\end{matrix}\text{ }\begin{matrix}
   4 \\
   -4 \\
   6 \\
\end{matrix} \right]\]. Find the matrix X such that \[3A-4B+5X=C\].

Answer 1

Hint: We start solving this problem by first finding the values of 3A, 4B by multiplying the matrices A and B with 3 and 4. Then we substitute the values of 3A, 4B and C in the given condition \[3A-4B+5X=C\] and solve it to find the value of 5X. then we divide the obtained value with 5 to find the matrix X.

Complete step by step answer:
We are given that matrix \[A=\left[ \begin{matrix}
   1 \\
   3 \\
   -6 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   -5 \\
   0 \\
\end{matrix} \right]\], matrix \[B=\left[ \begin{matrix}
   -1 \\
   4 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   2 \\
   5 \\
\end{matrix} \right]\] and matrix \[C=\left[ \begin{matrix}
   2 \\
   -1 \\
   -3 \\
\end{matrix}\text{ }\begin{matrix}
   4 \\
   -4 \\
   6 \\
\end{matrix} \right]\].
We are given that \[3A-4B+5X=C\]. We can write it as
\[\begin{align}
  & \Rightarrow 3A-4B+5X=C \\
 & \Rightarrow 5X=C-3A+4B \\
\end{align}\]
Now, let us consider the matrix 3A.
As, \[A=\left[ \begin{matrix}
   1 \\
   3 \\
   -6 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   -5 \\
   0 \\
\end{matrix} \right]\], multiplying it with 3 we get,
\[\begin{align}
  & \Rightarrow 3A=3\left[ \begin{matrix}
   1 \\
   3 \\
   -6 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   -5 \\
   0 \\
\end{matrix} \right] \\
 & \Rightarrow 3A=\left[ \begin{matrix}
   3 \\
   9 \\
   -18 \\
\end{matrix}\text{ }\begin{matrix}
   -6 \\
   -15 \\
   0 \\
\end{matrix} \right].................\left( 1 \right) \\
\end{align}\]
As, \[B=\left[ \begin{matrix}
   -1 \\
   4 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   2 \\
   5 \\
\end{matrix} \right]\], multiplying it with 4 we get,
\[\begin{align}
  & \Rightarrow 4B=4\left[ \begin{matrix}
   -1 \\
   4 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   2 \\
   5 \\
\end{matrix} \right] \\
 & \Rightarrow 4B=\left[ \begin{matrix}
   -4 \\
   16 \\
   4 \\
\end{matrix}\text{ }\begin{matrix}
   -8 \\
   8 \\
   20 \\
\end{matrix} \right].................\left( 2 \right) \\
\end{align}\]
Now let us substitute the obtained values in \[5X=C-3A+4B\]. Then we get,
$\begin{align}
  & \Rightarrow 5X=\left[ \begin{matrix}
   2 \\
   -1 \\
   -3 \\
\end{matrix}\text{ }\begin{matrix}
   4 \\
   -4 \\
   6 \\
\end{matrix} \right]-\left[ \begin{matrix}
   3 \\
   9 \\
   -18 \\
\end{matrix}\text{ }\begin{matrix}
   -6 \\
   -15 \\
   0 \\
\end{matrix} \right]+\left[ \begin{matrix}
   -4 \\
   16 \\
   4 \\
\end{matrix}\text{ }\begin{matrix}
   -8 \\
   8 \\
   20 \\
\end{matrix} \right] \\
 & \Rightarrow 5X=\left[ \begin{matrix}
   -1 \\
   -10 \\
   -21 \\
\end{matrix}\text{ }\begin{matrix}
   10 \\
   11 \\
   6 \\
\end{matrix} \right]+\left[ \begin{matrix}
   -4 \\
   16 \\
   4 \\
\end{matrix}\text{ }\begin{matrix}
   -8 \\
   8 \\
   20 \\
\end{matrix} \right] \\
 & \Rightarrow 5X=\left[ \begin{matrix}
   -5 \\
   6 \\
   -17 \\
\end{matrix}\text{ }\begin{matrix}
   2 \\
   19 \\
   26 \\
\end{matrix} \right] \\
\end{align}$
Now, dividing it with 5 we can find the value of matrix X.
$\Rightarrow X=\dfrac{1}{5}\left[ \begin{matrix}
   -5 \\
   6 \\
   -17 \\
\end{matrix}\text{ }\begin{matrix}
   2 \\
   19 \\
   26 \\
\end{matrix} \right]$
Now, let us take the 5, that is outside the matrix in the above value of X, inside the matrix. Then we get,
$\begin{align}
  & \Rightarrow X=\left[ \begin{matrix}
   \dfrac{-5}{5} \\
   \dfrac{6}{5} \\
   \dfrac{-17}{5} \\
\end{matrix}\text{ }\begin{matrix}
   \dfrac{2}{5} \\
   \dfrac{19}{5} \\
   \dfrac{26}{5} \\
\end{matrix} \right] \\
 & \\
 & \Rightarrow X=\left[ \begin{matrix}
   -1 \\
   \dfrac{6}{5} \\
   \dfrac{-17}{5} \\
\end{matrix}\text{ }\begin{matrix}
   \dfrac{2}{5} \\
   \dfrac{19}{5} \\
   \dfrac{26}{5} \\
\end{matrix} \right] \\
\end{align}$
So, we get the value of matrix X as $\left[ \begin{matrix}
   -1 \\
   \dfrac{6}{5} \\
   \dfrac{-17}{5} \\
\end{matrix}\text{ }\begin{matrix}
   \dfrac{2}{5} \\
   \dfrac{19}{5} \\
   \dfrac{26}{5} \\
\end{matrix} \right]$.

Hence answer is $\left[ \begin{matrix}
   -1 \\
   \dfrac{6}{5} \\
   \dfrac{-17}{5} \\
\end{matrix}\text{ }\begin{matrix}
   \dfrac{2}{5} \\
   \dfrac{19}{5} \\
   \dfrac{26}{5} \\
\end{matrix} \right]$.

Note: The common mistake one does while solving this problem is when multiplying a matrix with a number one might multiply only the first element in the matrix. For example, when finding the value of 3A, one might make a mistake while multiplying 3 with A as,
\[\begin{align}
  & \Rightarrow 3A=3\left[ \begin{matrix}
   1 \\
   3 \\
   -6 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   -5 \\
   0 \\
\end{matrix} \right] \\
 & \Rightarrow 3A=\left[ \begin{matrix}
   3 \\
   3 \\
   -6 \\
\end{matrix}\text{ }\begin{matrix}
   -2 \\
   -5 \\
   0 \\
\end{matrix} \right] \\
\end{align}\]
But it is wrong. When a matrix is multiplied by a number, we need to multiply all the elements in the given matrix not only the first one.