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

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

& \Rightarrow 3A-4B+5X=C \\\ & \Rightarrow 5X=C-3A+4B \\\ \end{aligned}$$ Now, let us consider the matrix 3A. As, $$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{aligned} & \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{aligned}$$ As, $$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{aligned} & \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{aligned}$$ Now let us substitute the obtained values in $$5X=C-3A+4B$$. Then we get, $\begin{aligned} & \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{aligned}$ 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{aligned} & \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{aligned}$ 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{aligned} & \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{aligned}$$ 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.