Let A be a square matrix of order 3 \(\times\) 3, then | kA| is equal to
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Let A be a square matrix of order 3 \(\times\) 3, then | kA| is equal to
Answer
Let A be any 3×3 matrix
\(A=\left[\begin{array}{lll} {a} & {b} & {c} \\ {d} & {e} & {f} \\ {g} & {h} & {i} \end{array}\right]\)
\(\therefore|\mathrm{A}|=\left|\begin{array}{lll} {\mathrm{a}} & {\mathrm{b}} & {\mathrm{c}} \\ {\mathrm{d}} & {\mathrm{e}} & {\mathrm{f}} \\ {\mathrm{g}} & {\mathrm{h}} & {\mathrm{i}} \end{array}\right|\)
Clearly, \( \mathrm{k} \mathrm{A}=\left[\begin{array}{lll} {\mathrm{ka}} & {\mathrm{kb}} & {\mathrm{kc}} \\ {\mathrm{kd}} & {\mathrm{ke}} & {\mathrm{kf}} \\ {\mathrm{kg}} & {\mathrm{kh}} & {\mathrm{ki}} \end{array}\right]\)
\(\Rightarrow|\mathrm{k} \mathrm{A}|=\left|\begin{array}{lll} {\mathrm{ka}} & {\mathrm{kb}} & {\mathrm{kc}} \\ {\mathrm{kd}} & {\mathrm{ke}} & {\mathrm{kf}} \\ {\mathrm{kg}} & {\mathrm{kh}} & {\mathrm{ki}} \end{array}\right|\)
Taking out k from the 1st, 2nd and 3rd row, we get
\(|\mathrm{k} \mathrm{A}|=\mathrm{k}^{3}\left|\begin{array}{lll} {\mathrm{a}} & {\mathrm{b}} & {\mathrm{c}} \\ {\mathrm{d}} & {\mathrm{e}} & {\mathrm{f}} \\ {\mathrm{g}} & {\mathrm{h}} & {\mathrm{i}} \end{array}\right|\)
\(\therefore\) |kA| = k3|A|
Therefore, answer is option (C) k3|A|
Therefore the choice is: C
\(A=\left[\begin{array}{lll} {a} & {b} & {c} \\ {d} & {e} & {f} \\ {g} & {h} & {i} \end{array}\right]\)
\(\therefore|\mathrm{A}|=\left|\begin{array}{lll} {\mathrm{a}} & {\mathrm{b}} & {\mathrm{c}} \\ {\mathrm{d}} & {\mathrm{e}} & {\mathrm{f}} \\ {\mathrm{g}} & {\mathrm{h}} & {\mathrm{i}} \end{array}\right|\)
Clearly, \( \mathrm{k} \mathrm{A}=\left[\begin{array}{lll} {\mathrm{ka}} & {\mathrm{kb}} & {\mathrm{kc}} \\ {\mathrm{kd}} & {\mathrm{ke}} & {\mathrm{kf}} \\ {\mathrm{kg}} & {\mathrm{kh}} & {\mathrm{ki}} \end{array}\right]\)
\(\Rightarrow|\mathrm{k} \mathrm{A}|=\left|\begin{array}{lll} {\mathrm{ka}} & {\mathrm{kb}} & {\mathrm{kc}} \\ {\mathrm{kd}} & {\mathrm{ke}} & {\mathrm{kf}} \\ {\mathrm{kg}} & {\mathrm{kh}} & {\mathrm{ki}} \end{array}\right|\)
Taking out k from the 1st, 2nd and 3rd row, we get
\(|\mathrm{k} \mathrm{A}|=\mathrm{k}^{3}\left|\begin{array}{lll} {\mathrm{a}} & {\mathrm{b}} & {\mathrm{c}} \\ {\mathrm{d}} & {\mathrm{e}} & {\mathrm{f}} \\ {\mathrm{g}} & {\mathrm{h}} & {\mathrm{i}} \end{array}\right|\)
\(\therefore\) |kA| = k3|A|
Therefore, answer is option (C) k3|A|
Therefore the choice is: C
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