Question

Find the value of x, y and z if

1 \\\ 0 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 \\\ 1 \\\ 1 \\\ \end{matrix} \right]-3\left[ \begin{matrix} 1 \\\ -2 \\\ 3 \\\ \end{matrix}\text{ }\begin{matrix} 2 \\\ 3 \\\ 1 \\\ \end{matrix} \right] \right\\}\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right]=\left[ \begin{matrix} x+1 \\\ y-1 \\\ 2z \\\ \end{matrix} \right]$$

Answer 1

To find value of x, y and z we will compare matrices at LHS and RHS of above expression. To do so, first we will arrange the LHS by making it a single matrix. 5 and 3 can be multiplied inside using $k\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]=\left[ \begin{matrix} ka & kb \\\ kc & kd \\\ \end{matrix} \right]$ and then both matrices can be added or subtracted term wise to obtain a single matrix. Then, we will use matrix multiplication to finally form a last matrix at LHS then comparing to RHS given is result.

Complete step by step answer:
We will first open the matrix, we have the LHS of given expression as

1 \\\ 0 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 \\\ 1 \\\ 1 \\\ \end{matrix} \right]-3\left[ \begin{matrix} 1 \\\ -2 \\\ 3 \\\ \end{matrix}\text{ }\begin{matrix} 2 \\\ 3 \\\ 1 \\\ \end{matrix} \right] \right\\}\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right]$$ Now, matrix multiplication by scalar is done as given below: $$k\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]=\left[ \begin{matrix} ka & kb \\\ kc & kd \\\ \end{matrix} \right]$$ Where $\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]$ is a matrix and k is a scalar. That is, scalar is multiplied to each term of matrix. Applying this in above matrix where 5 is scalar of matrix $\left[ \begin{matrix} 1 \\\ 0 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 \\\ 1 \\\ 1 \\\ \end{matrix} \right]$ and +3 is scalar of matrix $\left[ \begin{matrix} 1 \\\ -2 \\\ 3 \\\ \end{matrix}\text{ }\begin{matrix} 2 \\\ 3 \\\ 1 \\\ \end{matrix} \right]$ we get: $$\begin{aligned} & \text{LHS}=\left\\{ 5\left[ \begin{matrix} 1 \\\ 0 \\\ 1 \\\ \end{matrix}\text{ }\begin{matrix} 0 \\\ 1 \\\ 1 \\\ \end{matrix} \right]-\left( 3 \right)\left[ \begin{matrix} 1 \\\ -2 \\\ 3 \\\ \end{matrix}\text{ }\begin{matrix} 2 \\\ 3 \\\ 1 \\\ \end{matrix} \right] \right\\}\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right] \\\ & \text{LHS}=\left\\{ \left[ \begin{matrix} 5 \\\ 0 \\\ 5 \\\ \end{matrix}\text{ }\begin{matrix} 0 \\\ 5 \\\ 5 \\\ \end{matrix} \right]-\left[ \begin{matrix} 3 \\\ -6 \\\ 9 \\\ \end{matrix}\text{ }\begin{matrix} 6 \\\ 9 \\\ 1 \\\ \end{matrix} \right] \right\\}\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right] \\\ \end{aligned}$$ Now, addition or subtraction is done term wise as given below: $$\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]+\left[ \begin{matrix} e & f \\\ g & h \\\ \end{matrix} \right]=\left[ \begin{matrix} a+e & b+f \\\ c+g & d+h \\\ \end{matrix} \right]$$ Where $\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\text{ and }\left[ \begin{matrix} e & f \\\ g & h \\\ \end{matrix} \right]$ are matrices of $2\times 2$ order. Applying this in above we get: $$\begin{aligned} & \text{LHS}=\left\\{ \left[ \begin{matrix} 5-3 \\\ 0+6 \\\ 5-9 \\\ \end{matrix}\text{ }\begin{matrix} 0-6 \\\ 5-9 \\\ 5-1 \\\ \end{matrix} \right] \right\\}\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right] \\\ & \text{LHS}=\left\\{ \left[ \begin{matrix} 2 \\\ 6 \\\ -4 \\\ \end{matrix}\text{ }\begin{matrix} -6 \\\ -4 \\\ 4 \\\ \end{matrix} \right] \right\\}\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right] \\\ \end{aligned}$$ So, finally we are left with multiplication of above obtained matrix. Matrix multiplication is done as below: $$\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\cdot \left[ \begin{matrix} e & f \\\ g & h \\\ \end{matrix} \right]=\left[ \begin{matrix} ae+bg & af+bh \\\ ce+dg & cf+dh \\\ \end{matrix} \right]$$ That is, multiplying the term wise of a row with corresponding column of the second matrix and adding them. Here $\left[ \begin{matrix} a & b \\\ c & d \\\ \end{matrix} \right]\text{ and }\left[ \begin{matrix} e & f \\\ g & h \\\ \end{matrix} \right]$ are matrices of $2\times 2$ order. Applying this matrix multiplication method to our question, we get: $$\begin{aligned} & \text{LHS}=\left[ \begin{matrix} 2 \\\ 6 \\\ -4 \\\ \end{matrix}\text{ }\begin{matrix} -6 \\\ -4 \\\ 4 \\\ \end{matrix} \right]\left[ \begin{matrix} 2 \\\ 1 \\\ \end{matrix} \right] \\\ & \text{LHS}=\left[ \begin{matrix} 2\times 2+(-6)\times 1 \\\ 6\times 2+\left( -4 \right)\times 1 \\\ -4\times 2+4\times 1 \\\ \end{matrix} \right] \\\ & \text{LHS}=\left[ \begin{matrix} -2 \\\ 8 \\\ -4 \\\ \end{matrix} \right] \\\ \end{aligned}$$ Comparing this obtained LHS by our given RHS we get: $$\left[ \begin{matrix} -2 \\\ 8 \\\ -4 \\\ \end{matrix} \right]=\left[ \begin{matrix} x+1 \\\ y-1 \\\ 2z \\\ \end{matrix} \right]=\left[ \begin{matrix} -2 \\\ 8 \\\ -4 \\\ \end{matrix} \right]$$ Comparing each term we get: $$\begin{aligned} & -2=x+1,8=y-1\text{ and }-4=2z \\\ & \Rightarrow x+1=-2 \\\ & x=-2-1 \\\ & x=-3 \\\ & \Rightarrow y-1=8 \\\ & y=8+1 \\\ & y=9 \\\ & \Rightarrow 2z=-4 \\\ & z=\dfrac{-4}{2}=-2 \\\ \end{aligned}$$ **Hence, value of x = -3, y = 9 and z = -2** **Note:** The key point to note here is that, the above given are matrices not determinants. Matrices are represented as $\left[ {} \right]$ and determinants as $\left| {} \right|$ If any scalar term is multiplied then it is multiplied to each and every term of a matrix, but if it is a determinant then that scalar is only multiplied to one row or column and not all. Here, we have a matrix, so we have multiplied 5 and 3 to each and every element inside.