Question-8

Let \(A\) be a real \(3\times3\) matrix:

\[ A=\begin{bmatrix}C_{1} & C_{2} & C_{3}\end{bmatrix} \]

Find the determinant of all the following matrices in terms of \(|A|\):

\[ \begin{aligned} A_{1} & =\begin{bmatrix}C_{1} & C_{2}+5C_{3} & C_{3}\end{bmatrix}\\ A_{2} & =\begin{bmatrix}C_{1}+C_{2}+C_{3} & C_{2} & C_{3}\end{bmatrix}\\ A_{3} & =\begin{bmatrix}C_{1} & C_{2}+5C_{3} & 0\end{bmatrix}\\ A_{4} & =\begin{bmatrix}C_{1}+C_{2} & C_{2}+C_{3} & C_{3}+C_{1}\end{bmatrix} \end{aligned} \]

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Since \(A_{3}\) has a zero column, \(|A_{3}|=0\).

• To get \(A_{1}\), we perform \(C_{2}\rightarrow C_{2}+5C_{3}\).

• To get \(A_{2}\), we perform \(C_{1}\rightarrow C_{1}+C_{2}+C_{3}\)

Both these operations do not change the determinant. So \(|A_{1}|=|A_{2}|=|A|\). \(A_{4}\) is slightly tricky:

\[ \begin{aligned} \begin{vmatrix}C_{1}+C_{2} & C_{2}+C_{3} & C_{3}+C_{1}\end{vmatrix} & =\begin{vmatrix}2(C_{1}+C_{2}+C_{3}) & C_{2}+C_{3} & C_{3}+C_{1}\end{vmatrix}\\ & =2\begin{vmatrix}C_{1}+C_{2}+C_{3} & C_{2}+C_{3} & C_{3}+C_{1}\end{vmatrix}\\ & =2\begin{vmatrix}C_{1} & C_{2}+C_{3} & C_{3}+C_{1}\end{vmatrix}\\ & =2\begin{vmatrix}C_{1} & C_{2}+C_{3} & C_{3}\end{vmatrix}\\ & =2\begin{vmatrix}C_{1} & C_{2} & C_{3}\end{vmatrix}\\ & =2|A| \end{aligned} \]