Question:
Solve the matrix equation for \(x\) and \(y\).
\[ \begin{bmatrix} x^2 \\ y^2 \end{bmatrix} - 3\begin{bmatrix} x \\ 3y \end{bmatrix} = \begin{bmatrix} -2 \\ -20 \end{bmatrix} \]
Solve the matrix equation for \(x\) and \(y\).
\[ \begin{bmatrix} x^2 \\ y^2 \end{bmatrix} - 3\begin{bmatrix} x \\ 3y \end{bmatrix} = \begin{bmatrix} -2 \\ -20 \end{bmatrix} \]
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Subtract entry by entry to get two quadratic equations, then solve each by factorising.
Updated On: Oct 22, 2026
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Solution and Explanation
Step 1: Simplify the left side:
Multiply the matrix by 3 first. Then subtract entry by entry.
\[ \begin{bmatrix} x^2 - 3x \\ y^2 - 9y \end{bmatrix} = \begin{bmatrix} -2 \\ -20 \end{bmatrix} \]
Step 2: Equate corresponding entries:
Two matrices are equal only when matching entries are equal. So we get two separate equations.
\[ x^2 - 3x = -2, \qquad y^2 - 9y = -20 \]
Step 3: Solve for x:
Write \( x^2 - 3x + 2 = 0 \). Factorise it as \( (x-1)(x-2) = 0 \).
So \( x = 1 \) or \( x = 2 \).
Step 4: Solve for y:
Write \( y^2 - 9y + 20 = 0 \). Factorise it as \( (y-4)(y-5) = 0 \).
So \( y = 4 \) or \( y = 5 \).
Final Answer:
The two equations are independent, so any value of x goes with any value of y. \[ \boxed{x = 1 \text{ or } 2, \quad y = 4 \text{ or } 5} \]
Multiply the matrix by 3 first. Then subtract entry by entry.
\[ \begin{bmatrix} x^2 - 3x \\ y^2 - 9y \end{bmatrix} = \begin{bmatrix} -2 \\ -20 \end{bmatrix} \]
Step 2: Equate corresponding entries:
Two matrices are equal only when matching entries are equal. So we get two separate equations.
\[ x^2 - 3x = -2, \qquad y^2 - 9y = -20 \]
Step 3: Solve for x:
Write \( x^2 - 3x + 2 = 0 \). Factorise it as \( (x-1)(x-2) = 0 \).
So \( x = 1 \) or \( x = 2 \).
Step 4: Solve for y:
Write \( y^2 - 9y + 20 = 0 \). Factorise it as \( (y-4)(y-5) = 0 \).
So \( y = 4 \) or \( y = 5 \).
Final Answer:
The two equations are independent, so any value of x goes with any value of y. \[ \boxed{x = 1 \text{ or } 2, \quad y = 4 \text{ or } 5} \]
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