Question

Determine the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\) for the matrices shown in Exercise 13-18

16. \(A = \left( {\begin{array}{*{20}{c}}3&4\\{ - 6}&{10}\end{array}} \right)\)

Short answer

The dimension of \({\mathop{\rm Col}\nolimits} A\) is 2, and that of \({\mathop{\rm Nul}\nolimits} A\) is 0.

Step-by-step solution

State the condition for the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\)

Thedimension of \({\mathop{\rm Nul}\nolimits} A\) is thenumber of free variables in the equation \(A{\mathop{\rm x}\nolimits} = 0\), and the dimension of \({\mathop{\rm Col}\nolimits} A\)is thenumber of pivot columnsin \(A\).

Apply the row operation

Perform an elementary row operation to produce the row-reduced echelon form of the matrix.

At row 1, multiply row 1 by \(\frac{1}{3}\).

\(\left( {\begin{array}{*{20}{c}}1&{1.333}\\{ - 6}&{10}\end{array}} \right)\)

At row 2, multiply row 1 by 6 and add it to row 2.

\(\left( {\begin{array}{*{20}{c}}1&{1.333}\\0&{18}\end{array}} \right)\)

At row 2, multiply row 2 by \(\frac{1}{{18}}\).

\(\left( {\begin{array}{*{20}{c}}1&{1.333}\\0&1\end{array}} \right)\)

At row 1, multiply row 2 by \(1.3333\) and subtract it from row 1.

\(\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\)

Determine the dimensions of \({\mathop{\rm Nul}\nolimits} A\) and \({\mathop{\rm Col}\nolimits} A\)

\({\mathop{\rm Col}\nolimits} A\)

Matrix A has two pivot columns, so the dimension of \({\mathop{\rm Col}\nolimits} A\) is 2. The equation \(A{\mathop{\rm x}\nolimits} = 0\) has only a trivial solution since there are no columns without pivots.

Therefore, \({\mathop{\rm Nul}\nolimits} A = \left\{ 0 \right\}\) and the dimension of \({\mathop{\rm Nul}\nolimits} A\) is 0.

Thus, the dimension of \({\mathop{\rm Col}\nolimits} A\) is 2, and that of \({\mathop{\rm Nul}\nolimits} A\) is 0.