Question

Find x1 and x2.

Answer 1

Answer:

$x_1 = 7\\x_2 = 6$

Step-by-step explanation:

To find x₁ and x₂ :

$\left[\begin{array}{ccc}-4&1\\5&4\\\end{array}\right] \times \left[\begin{array}{ccc}x_1\\x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\\\end{array}\right]$

Step 1

Multiply first 2 x 2 matrix with 2 x 1 vector, we get

$\left[\begin{array}{ccc}-4x_1&+ x_2\\5x_1&+ 4x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]$

Step 2

Add the 2 x 1 matrices on LHS, we get

$\left[\begin{array}{ccc}-4x_1&+x_2&+11\\5x_1&+4x_2&-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]$

Step 3,

we get

$-4x_1 + x_2 + 11 = -11$

and

$5x_1 + 4x_2-19=40$

Step 4,

Simplify, we get

$-4x_1+x_2=-22----(1)\\ 5x_1+4x_2=59----(2)$

Step 5,

multiply eqn(1) by 4

we get

$16x_1+4x_2=-88$

Step 6,

eqn (2) - eqn(3)

we get

$21x_1 = 147\\x_1 =\frac{147}{21} \\x_1= 7$

substituting in eqn (1), we get

$(-4 \times 7) + x_2 = -22$

so, we get

$x_2 = 6$

Therefore

$x_1 = 7\\x_2 = 6$