(i) \(\left[\begin{array}{ll} 4 & 3 \\ x & 5 \end{array}\right]\) = \(\left[\begin{array}{ll} y & 3 \\ 1 & 5 \end{array}\right]\)
Equating corresponding elements of the two matrices, we get
4 = y, 3 = z, x = 1
i.e.,
x = 1, y = 4, z = 3.
(ii) \(\left[\begin{array}{ll} x+y & 2 \\ 5+z & xy \end{array}\right]\) = \(\left[\begin{array}{ll} 6 & 5 \\ 2 & 8 \end{array}\right]\)
Comparing the corresponding elements, we get
x + y = 6, 5 + z = 5, xy = 8
∴ y = 6 – x. Putting value of y in xy = 8, we get
x(6 – x) = 8 or x² – 6x + 8 = 0
∴ (x-4)(x-2) = 0 ∴ x = 4, 2
y = 2, 4
Also, 5 + z = 5 ∴ z = 0
∴ x = 4, y = 2, z = 0
or x = 2, y = 4, z = 0
(iii) \(\left[\begin{array}{c} x+y+z \\ x+z \\ y+z \end{array}\right]\) = \(\left[\begin{array}{c} 9 \\ 5 \\ 7 \end{array}\right]\)
Equating the corresponding elements, we get
x + y + z = 9 … (1)
x + z = 5 … (2)
y + z = 7 … (3)
Adding (2) and (3),
or (x + y + z) + z = 12
or 9 + z = 12
∴ z = 12 – 9
= 3.
From (2),
x + z = 5
or x + 3 = 5
∴ x = 2.
From (3),
y + z = 7
y + 3 = 7
∴ y = 4.
Thus, x = 2, y = 4, z = 3.
Alternatively : Equation (1) – (2)
y = 4
(1)-(3)
x = 2 [∴ z = 3 [from(1)]]
⇒ x = 2, y = 4, z = 3.
