Question:
If the rank of the matrix
\[
A=
\begin{bmatrix}
1 & 2 & 1 & -1 \\
-1 & 2 & 3 & 5 \\
0 & 1 & k & k
\end{bmatrix}
\]
is \(2\) and \(k\) is a real number, then \(k\) is a root of the following quadratic equation:
If the rank of the matrix
\[
A=
\begin{bmatrix}
1 & 2 & 1 & -1 \\
-1 & 2 & 3 & 5 \\
0 & 1 & k & k
\end{bmatrix}
\]
is \(2\) and \(k\) is a real number, then \(k\) is a root of the following quadratic equation:
Show Hint
If the rank of a \(3 \times 4\) matrix is \(2\), then every \(3 \times 3\) minor must be equal to zero.
Updated On: Jun 24, 2026
- \(x^2+3x+2=0\)
- \(x^2+x-2=0\)
- \(x^2+x-6=0\)
- \(x^2-x-6=0\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use the condition for rank \(2\).
Since the matrix has rank \(2\), all \(3 \times 3\) minors must be zero.
Consider the minor formed by the first three columns: \[ \begin{vmatrix} 1 & 2 & 1 \\ -1 & 2 & 3 \\ 0 & 1 & k \end{vmatrix} =0 \]
Step 2: Evaluate the determinant.
Expanding along the first row: \[ 1(2k-3)-2(-k-0)+1(-1-0)=0 \] \[ 2k-3+2k-1=0 \] \[ 4k-4=0 \] \[ k=1 \]
Step 3: Check which quadratic equation has \(k=1\) as a root.
For option (2): \[ x^2+x-2=0 \] Substitute \(x=1\): \[ 1^2+1-2=0 \] \[ 1+1-2=0 \] \[ 0=0 \] Hence, \(x=1\) is a root of \[ x^2+x-2=0 \]
Step 4: Final conclusion.
Therefore, \(k\) is a root of \[ \boxed{x^2+x-2=0} \]
Since the matrix has rank \(2\), all \(3 \times 3\) minors must be zero.
Consider the minor formed by the first three columns: \[ \begin{vmatrix} 1 & 2 & 1 \\ -1 & 2 & 3 \\ 0 & 1 & k \end{vmatrix} =0 \]
Step 2: Evaluate the determinant.
Expanding along the first row: \[ 1(2k-3)-2(-k-0)+1(-1-0)=0 \] \[ 2k-3+2k-1=0 \] \[ 4k-4=0 \] \[ k=1 \]
Step 3: Check which quadratic equation has \(k=1\) as a root.
For option (2): \[ x^2+x-2=0 \] Substitute \(x=1\): \[ 1^2+1-2=0 \] \[ 1+1-2=0 \] \[ 0=0 \] Hence, \(x=1\) is a root of \[ x^2+x-2=0 \]
Step 4: Final conclusion.
Therefore, \(k\) is a root of \[ \boxed{x^2+x-2=0} \]
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