Question

Solve this question.

Answer 1

`\displaystyle{X = \left[\begin{array}{ccc}1&1\\1&1\end{array}\right] }`

Explanation:

Solve the matrices like normal equation, you can add 2X both sides so we have:

`\displaystyle{\left[\begin{array}{ccc}2&3\\3&2\end{array}\right] = \left[\begin{array}{ccc}0&1\\1&0\end{array}\right] + 2X}`

Now, subtract the matrices:

`\displaystyle{\left[\begin{array}{ccc}2&3\\3&2\end{array}\right] -\left[\begin{array}{ccc}0&1\\1&0\end{array}\right] = 2X}`

Follow the matrices subtraction laws:

`\displaystyle{\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] -\left[\begin{array}{ccc}e&f\\g&h\end{array}\right] = \left[\begin{array}{ccc}a-e&b-f\\c-g&d-h\end{array}\right] }`

Therefore:

`\displaystyle{\left[\begin{array}{ccc}2-0&3-1\\3-1&2-0\end{array}\right] = 2X}\\\\\displaystyle{\left[\begin{array}{ccc}2&2\\2&2\end{array}\right] = 2X}`

Divide both sides by 2, leaves us with:

`\displaystyle{(1)/(2)\left[\begin{array}{ccc}2&2\\2&2\end{array}\right] = X}`

Expand 1/2 inside the matrix, multiplying whole elements. Therefore:

`\displaystyle{\left[\begin{array}{ccc}1&1\\1&1\end{array}\right] = X}`

Hence,

`\displaystyle{X = \left[\begin{array}{ccc}1&1\\1&1\end{array}\right] }`