Question

Which equation can be used to solve

Answer 1

The Answer is last option

`\left[\begin{array}{ccc}x_(1)\\x_(2)\\\end{array}\right]=\left[\begin{array}{ccc}0.5&-3\\0&1\\\end{array}\right]\left[\begin{array}{ccc}2\\-3\\\end{array}\right]`

Explanation:

The Answer is last option

`\left[\begin{array}{ccc}x_(1)\\x_(2)\\\end{array}\right]=\left[\begin{array}{ccc}0.5&-3\\0&1\\\end{array}\right]\left[\begin{array}{ccc}2\\-3\\\end{array}\right]`

Write The Given Matrix in

AX= B

Now Pre multiplying by
`A^(-1)` we get

`A^(-1)AX=A^(-1)B\\ \\We\ know\\A^(-1)A = I\ and\ IX = X\\\\\therefore X = A^(-1)B\\\\\therefore \left[\begin{array}{ccc}x_(1)\\x_(2)\\\end{array}\right]=\left[\begin{array}{ccc}0.5&-3\\0&1\\\end{array}\right]\left[\begin{array}{ccc}2\\-3\\\end{array}\right]`

Here,

`A^(-1)=\left[\begin{array}{ccc}0.5&-3\\0&1\end{array}\right]\\\\X=\left[\begin{array}{ccc}x_(1)\\ x_(2)\end{array}\right] \\\\B=\left[\begin{array}{ccc}2\\-3\end{array}\right]`