Solveeit Logo
Login

Question

Mathematics Question on Matrices

Find X,if Y=[32\14]Y=\begin{bmatrix}3&2\\\1&4\end{bmatrix}and 2X+Y=[10−32]2X+Y=\begin{bmatrix}1&0\\\\-3&2\end{bmatrix}

Answer

The correct answer is [−1−1−2−1]\begin{bmatrix}-1&-1\\\\-2&-1\end{bmatrix}
2X+Y=[10−32]2X+Y=\begin{bmatrix}1&0\\\\-3&2\end{bmatrix}
  ⟹  2X+[32\14]=[10−32]\implies 2X+\begin{bmatrix}3&2\\\1&4\end{bmatrix}=\begin{bmatrix}1&0\\\\-3&2\end{bmatrix}
  ⟹  2X=[10−32]−[32\14]=[1−30−2−3−12−4]\implies 2X=\begin{bmatrix}1&0\\\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\\1&4\end{bmatrix}=\begin{bmatrix}1-3&0-2\\\\-3-1&2-4\end{bmatrix}
  ⟹  2X=[−2−2−4−2]\implies 2X=\begin{bmatrix}-2&-2\\\\-4&-2\end{bmatrix}
∴X=12[−2−2−4−2]=[−1−1−2−1]\therefore X=\frac{1}{2}\begin{bmatrix}-2&-2\\\\-4&-2\end{bmatrix}=\begin{bmatrix}-1&-1\\\\-2&-1\end{bmatrix}