Solve for x and y, when [(3,-4)(1,2)] [(x,y)] = [(3,11)]

Question

Solve for x and y, when

\(\begin{bmatrix} 3& -4 \\[0.3em] 1&2 \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} x \\[0.3em] y \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 3 \\[0.3em] 11 \\[0.3em] \end{bmatrix}.\) 

[(3,-4)(1,2)]

[(x,y)]

[(3,11)]

Answer 1

Given : \(\begin{bmatrix} 3& -4 \\[0.3em] 1&2 \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} x \\[0.3em] y \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 3 \\[0.3em] 11 \\[0.3em] \end{bmatrix}.\)

To find : x and y

Formula used :

Where c_ij = a_i1b_1j + a_i2b_2j + a_i3b_3j + ……………… + a_inb_nj

If A is a matrix of order a x b and B is a matrix of order c x d ,then matrix AB exists and is of order a x d ,

if and only if b = c

The resulting matrix obtained on multiplying \(\begin{bmatrix} 3& -4 \\[0.3em] 1& 2 \\[0.3em] \end{bmatrix}\) and \(\begin{bmatrix} x \\[0.3em] y \\[0.3em] \end{bmatrix}\) is of order 2 × 1

Equating similar terms,

3x – 4y = 3 equation 1

x + 2y = 11 equation 2

equation 1 + 2(equation 2) and solving the above equations,

5x = 25

x = \(\frac{25}{5}=5\)

x = 5 , substituting x = 2 in equation 2,

5 + 2y = 11 

2y =11 – 5 = 6 

2y = 6

y = \(\frac{6}{2}=3\) 

x = 5 and y = 3