solve the matrix equation for ax=bx+c for x={x1}
{ x2}
where a= { 6 0} b={-1 -2} c= {1]
[15 1} { 4 -2} [2}
please show work if you can
To solve the matrix equation ax = bx + c for x = {x1, x2}, where a = {{6, 0}, {15, 1}}, b = {{-1, -2}, {4, -2}}, and c = {1, 2}, we need to perform the following steps:
Step 1: Start by isolating x in the equation.
ax - bx = c
(x(a - b) = c
x = c/(a - b)
Step 2: Calculate the matrix subtraction (a - b).
(a - b) = {{6, 0}, {15, 1}} - {{-1, -2}, {4, -2}}
(a - b) = {{6 - (-1), 0 - (-2)}, {15 - 4, 1 - (-2)}}
(a - b) = {{7, 2}, {11, 3}}
Step 3: Calculate the inverse of matrix (a - b)^-1.
(a - b)^-1 = (1/((7 * 3) - (2 * 11))) * {{3, -2}, {-11, 7}}
(a - b)^-1 = (1/(21 - 22)) * {{3, -2}, {-11, 7}}
(a - b)^-1 = (1/(-1)) * {{3, -2}, {-11, 7}}
(a - b)^-1 = {{-3, 2}, {11, -7}}
Step 4: Calculate the product of (a - b)^-1 and c.
(a - b)^-1 * c = {{-3, 2}, {11, -7}} * {1, 2}
(a - b)^-1 * c = {{(-3 * 1) + (2 * 2)},{(11 * 1) + (-7 * 2)}}
(a - b)^-1 * c = {{-3 + 4},{11 - 14}}
(a - b)^-1 * c = {{1},{-3}}
Therefore, the solution to the matrix equation ax = bx + c for x = {x1, x2} is x = {1, -3}.