Step 1: Solve for \( x \) and \( y \) in terms of \( z \).
From Equation 1:
\[
x = 5 - z.
\]
From Equation 2:
\[
y = 7 - z.
\]
Step 2: Substitute \( x \) and \( y \) into Equation 3.
Substitute \( x = 5 - z \) and \( y = 7 - z \) into Equation 3:
\[
(5 - z) + (7 - z) + z = 9.
\]
Simplify the equation:
\[
5 + 7 - z - z + z = 9 $\Rightarrow$ 12 - z = 9 $\Rightarrow$ z = 3.
\]
Step 3: Find \( x \) and \( y \).
Now that we know \( z = 3 \), substitute this value into the expressions for \( x \) and \( y \):
\[
x = 5 - 3 = 2,
\]
\[
y = 7 - 3 = 4.
\]
Conclusion:
The values of \( x \), \( y \), and \( z \) are:
\[
x = 2, y = 4, z = 3.
\]
Match List-I with List-II
List-I (Matrix) | List-II (Inverse of the Matrix) |
---|---|
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\) | (I) \(\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}\) |
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\) | (II) \(\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}\) |
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\) | (III) \(\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}\) |
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\) | (IV) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\) |