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Consider the following statements:
$1.$ The matrix $\left(\begin{array}{lll}1 & 2 & 1 \\ a & 2 a & 1 \\ b & 2 b & 1\end{array}\right)$ is singular.
$2.$ The matrix $\left(\begin{array}{lll}c & 2 c & 1 \\ a & 2 a & 1 \\ b & 2 b & 1\end{array}\right)$ is non-singular.
Which of the above statements is/are correct?
Options:
$1.$ The matrix $\left(\begin{array}{lll}1 & 2 & 1 \\ a & 2 a & 1 \\ b & 2 b & 1\end{array}\right)$ is singular.
$2.$ The matrix $\left(\begin{array}{lll}c & 2 c & 1 \\ a & 2 a & 1 \\ b & 2 b & 1\end{array}\right)$ is non-singular.
Which of the above statements is/are correct?
Solution:
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Verified Answer
The correct answer is:
1 only
$1.$ $\left(\begin{array}{ccc}1 & 2 & 1 \\ a & 2 a & 1 \\ b & 2 b & 1\end{array}\right)=2\left(\begin{array}{lll}1 & 1 & 1 \\ a & a & 1 \\ b & b & 1\end{array}\right)=2\left|\begin{array}{lll}1 & 1 & 1 \\ a & a & 1 \\ b & b & 1\end{array}\right|=0$
Hence matrix is singular.
$2.$ $\left|\begin{array}{lll}\mathrm{c} & 2 \mathrm{c} & 1 \\ \mathrm{a} & 2 \mathrm{a} & 1 \\ \mathrm{~b} & 2 \mathrm{~b} & 1\end{array}\right|=2\left|\begin{array}{lll}\mathrm{c} & \mathrm{c} & 1 \\ \mathrm{a} & \mathrm{a} & 1 \\ \mathrm{~b} & \mathrm{~b} & 1\end{array}\right|=2\left|\begin{array}{lll}\mathrm{c} & \mathrm{c} & 1 \\ \mathrm{a} & \mathrm{a} & 1 \\ \mathrm{~b} & \mathrm{~b} & 1\end{array}\right|=0$
Hence matrix is singular.
Hence matrix is singular.
$2.$ $\left|\begin{array}{lll}\mathrm{c} & 2 \mathrm{c} & 1 \\ \mathrm{a} & 2 \mathrm{a} & 1 \\ \mathrm{~b} & 2 \mathrm{~b} & 1\end{array}\right|=2\left|\begin{array}{lll}\mathrm{c} & \mathrm{c} & 1 \\ \mathrm{a} & \mathrm{a} & 1 \\ \mathrm{~b} & \mathrm{~b} & 1\end{array}\right|=2\left|\begin{array}{lll}\mathrm{c} & \mathrm{c} & 1 \\ \mathrm{a} & \mathrm{a} & 1 \\ \mathrm{~b} & \mathrm{~b} & 1\end{array}\right|=0$
Hence matrix is singular.
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