Question

Consider a 4x4 matrix A with rows $\overrightarrow{{\mathbf{V}}_{\mathbf{1}}}\mathbf{,}\overrightarrow{{\mathbf{V}}_{\mathbf{2}}}\mathbf{,}\overrightarrow{{\mathbf{V}}_{\mathbf{3}}}\mathbf{,}\overrightarrow{{\mathbf{V}}_{\mathbf{4}}}$. If det(A) = 8, find the determinants in Exercises 11 through 16.

16. role="math" localid="1659506283449" $\mathit{d}\mathit{e}\mathit{t}\mathbf{\left[}\begin{array}{c}\mathbf{6}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}\mathbf{2}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{4}}\\ {\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{2}}\\ {\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{3}}\\ \mathbf{3}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{1}}\mathbf{+}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}}_{\mathbf{4}}\end{array}\mathbf{\right]}$

Short answer

Therefore, the determinant of given determinant matrix is given by,

$det\left[\begin{array}{c}6{\stackrel{\rightharpoonup }{v}}_1+2{\stackrel{\rightharpoonup }{v}}_4\\ {\stackrel{\rightharpoonup }{v}}_1\\ {\stackrel{\rightharpoonup }{v}}_3\\ 3{\stackrel{\rightharpoonup }{v}}_1+{\stackrel{\rightharpoonup }{v}}_4\end{array}\right]=0$.

Step-by-step solution

Definition

A determinant is a unique number associatedwith a square matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

Given

Given determinant,

$det\left[\begin{array}{c}6{\stackrel{\rightharpoonup }{v}}_1+2{\stackrel{\rightharpoonup }{v}}_4\\ {\stackrel{\rightharpoonup }{v}}_2\\ {\stackrel{\rightharpoonup }{v}}_3\\ 3{\stackrel{\rightharpoonup }{v}}_1+{\stackrel{\rightharpoonup }{v}}_4\end{array}\right]$

To find determinant

To determine,

$$\begin{gathered} det\left[\begin{array}{c}6{\stackrel{\rightharpoonup }{v}}_1+2{\stackrel{\rightharpoonup }{v}}_4\\ {\stackrel{\rightharpoonup }{v}}_1\\ {\stackrel{\rightharpoonup }{v}}_3\\ 3{\stackrel{\rightharpoonup }{v}}_1+{\stackrel{\rightharpoonup }{v}}_4\end{array}\right]=\frac{1}{2}det\left[\begin{array}{c}3{\stackrel{\rightharpoonup }{v}}_1+{\stackrel{\rightharpoonup }{v}}_4\\ {\stackrel{\rightharpoonup }{v}}_1\\ {\stackrel{\rightharpoonup }{v}}_3\\ 3{\stackrel{\rightharpoonup }{v}}_1+{\stackrel{\rightharpoonup }{v}}_4\end{array}\right] \\ det\left[\begin{array}{c}6{\stackrel{\rightharpoonup }{v}}_1+{\stackrel{\rightharpoonup }{v}}_4\\ {\stackrel{\rightharpoonup }{v}}_1\\ {\stackrel{\rightharpoonup }{v}}_3\\ 3{\stackrel{\rightharpoonup }{v}}_1+{\stackrel{\rightharpoonup }{v}}_4\end{array}\right]=0 \end{gathered}$$