Question

The values of x, y, z and 'a' which satisfy the matrix equation $\begin{bmatrix}x+3& 2y+x \\ z-1& 4a-6\end{bmatrix}=\begin{bmatrix}0& -7 \\ 3& 2a\end{bmatrix},$ respectively, are:

• A. -3, -2, 4, 3
• B. 3, 2, 4, 3
• C. -2, -3, 3, 4
• D. 2, 3, 4, 3

Hint:

When equating matrices, always solve for the variables that appear alone in an equation first (like $x$ and $z$ here) to simplify the remaining substitutions.

Answer 1

The Correct Option is A

Concept: Two matrices are considered equal if and only if they have the same dimensions and their corresponding elements (entries in the same position) are equal. We can set up a system of linear equations by equating each corresponding element.

Step 1: Solve for $x$ and $z$.
From the element in the first row, first column ($a_{11}$): \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] From the element in the second row, first column ($a_{21}$): \[ z - 1 = 3 \quad \Rightarrow \quad z = 4 \]

Step 2: Solve for $y$.
From the element in the first row, second column ($a_{12}$): \[ 2y + x = -7 \] Substitute $x = -3$: \[ 2y + (-3) = -7 \] \[ 2y = -7 + 3 \] \[ 2y = -4 \quad \Rightarrow \quad y = -2 \]

Step 3: Solve for $a$.
From the element in the second row, second column ($a_{22}$): \[ 4a - 6 = 2a \] \[ 4a - 2a = 6 \] \[ 2a = 6 \quad \Rightarrow \quad a = 3 \] Final Result: The values are $x = -3$, $y = -2$, $z = 4$, and $a = 3$. This matches the sequence in option (1).