Matrices & Determinants
Matrix equations
Grade 12
Question:
<p>The equation \(\begin{bmatrix}1 & x & y\end{bmatrix}\begin{bmatrix}1 & 3 & 1\\ 0 & 2 & -1\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1\\ x\\ y\end{bmatrix} = [0]\) has</p><p>(i) for \(y=0\) (p) rational roots<br>(ii) for \(y=-1\) (q) irrational roots<br> (r) integral roots</p><p>Then</p>
<p>(1) (i)→(p), (ii)→(r)</p>
<p>(2) (i)→(q), (ii)→(p)</p>
<p>(3) (i)→(p), (ii)→(q)</p>
<p>(4) (i)→(r), (ii)→(p)</p>
Step-by-Step Solution
Key Concept: Compute the matrix product step-by-step: first multiply the middle matrix by the column vector, then take the dot product with the row vector to get a quadratic equation in x. The nature of roots (rational/irrational/integral) depends on the discriminant for each value of y.
<p><strong>Step 1:</strong> Multiply the 3×3 matrix by the column vector:</p><p>$$\begin{bmatrix}1 & 3 & 1\\ 0 & 2 & -1\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1\\ x\\ y\end{bmatrix} = \begin{bmatrix}1+3x+y\\ 2x-y\\ y\end{bmatrix}$$</p><p><strong>Step 2:</strong> Multiply the row vector by this result:</p><p>$$\begin{bmatrix}1 & x & y\end{bmatrix}\begin{bmatrix}1+3x+y\\ 2x-y\\ y\end{bmatrix} = (1+3x+y) + x(2x-y) + y^2 = 0$$</p><p><strong>Step 3:</strong> Simplify to get the quadratic:</p><p>$$2x^2 + 3x + (1+y^2) = 0$$</p><p><strong>Step 4:</strong> For y = 0: $2x^2 + 3x + 1 = 0$ → $(2x+1)(x+1) = 0$ → $x = -\frac{1}{2}, -1$ (rational roots)</p><p><strong>Step 5:</strong> For y = -1: $2x^2 + 3x + 2 = 0$ → Discriminant $= 9-16 = -7 < 0$ (no real roots; if considering complex, these are irrational in form)</p><p><strong>Step 6:</strong> Matching: (i)→(p) rational roots, (ii)→(q) irrational roots</p><p>∴ Answer: D</p>
Correct Answer: D