Determinants And Matrices Set-2

Example -4

Let \(f,g,h\) be differentiable functions of x and \(\Delta \) and be such that

\[\Delta  = \left| {\;\begin{gathered}f&g&h\\{(xf)'}&{(xg)'}&{(xh)'}\\{({x^2}f)''}&{({x^2}g)''}&{({x^2}h)''}\end{gathered}\;} \right|\]

Find \(\Delta '\).

Solution: We have

\(\Delta  = \left| {\;\begin{gathered}f&g&h\\{f + xf'}&{g + xg'}&{h + xh'}\\{{x^2}f'' + 4xf' + 2f}&{{x^2}g'' + 4xg' + 2g}&{{x^2}h'' + 4xh' + 2h}\end{gathered}\;} \right|\)

Applying \({R_2} \to {R_2} - {R_1},\;{R_3} \to {R_3} - 4{R_2} + 2{R_1},\) we have

\[\begin{align} &\Delta  = {x^3}\left| {\;\begin{array}{*{20}{c}}f&g&h\\{f'}&{g'}&{h'}\\{f''}&{g''}&{h''}\end{array}\;} \right| = {x^3}{\Delta _1}\\ \Rightarrow  \quad &\Delta ' = 3{x^2}{\Delta _1} + {x^3}\Delta _1^\prime \end{align} \]

where

\[\begin{align}\Delta _1^\prime &= \left| {\;\begin{gathered}{f'}&{g'}&{h'}\\{f'}&{g'}&{h'}\\{f''}&{g''}&{h''}\end{gathered}\;} \right| + \left| {\;\begin{gathered}f&g&h\\{f''}&{g''}&{h''}\\{f''}&{g''}&{h''}\end{gathered}\;} \right| + \left| {\;\begin{gathered}f&g&h\\{f'}&{g'}&{h'}\\{f'''}&{g'''}&{h'''}\end{gathered}\;} \right|\\ \\&= 0 + 0 + {\Delta _2}\\\\\Rightarrow  \quad  \Delta^\prime &= 3{x^2}{\Delta _1} + {x^3}{\Delta _2}\end{align}\]

Example -5

Let a, b, c be real numbers such that \({a^2} + {b^2} + {c^2} = 1\) . Let

\[f(x,y) = \left| {\;\begin{gathered}{ax - by - c}&{bx + ay}&{cx + a}\\{bx + ay}&{ - ax + by - c}&{cy + b}\\{cx + a}&{cy + b}&{ - ax - by + c}\end{gathered}\;} \right|\]

what does \(f(x,y) = 0\) represent?

Solution: We have to move towards generating \({a^2} + {b^2} + {c^2}\) in the determinant, since we know its value.

By \({R_1} \;\to a{R_1},\,{R_2} \to b{R_2},\;{R_3} \to c{R_3} \; \text{and then} \;{R_1} \to {R_1} + \,{R_2} + {R_3},\) then we have

\[\begin{align}f(x,y) &= \frac{1}{{abc}}\left| {\;\begin{gathered}x&y&1\\{{b^2}x + aby}&{ - abx + {b^2}y - bc}&{bcy + {b^2}}\\{{c^2}x + ac}&{{c^2}y + bc}&{ - acx - bcy + {c^2}}\end{gathered}\;} \right|\\ &  = \frac{1}{a}\left| {\;\begin{gathered}x&y&1\\{bx + ay}&{ - ax + by - c}&{cy + b}\\{cx + a}&{cy + b}&{ - ax - by + c}\end{gathered}\;} \right|\end{align}\]

Now, using \(R{ _2} \to {R_2} - b{R_1},\;{R_3} \to {R_3} - c{R_1},\) we have

\[f(x,y) = \frac{1}{a}\left| {\;\begin{array}{*{20}{c}}x&y&1\\{ay}&{ - ax - c}&{cy}\\a&b&{ - ax - by}\end{array}\;} \right|\]

Finally, \({C_3} \to \,{C_3} + y{C_2} - x{C_1}\) gives

\[f(x,y) = \frac{1}{a}\left| {\;\begin{array}{*{20}{c}}x&y&{{x^2} + {y^2}}\\{ay}&{ - ax - c}&0\\a&b&0\end{array}\;} \right| = ({x^2} + {y^2})(x + by + c)\]

Thus, \(f(x,y) = 0\) represents the point (0, 0) on the straight line

\[x + by + c = 0\]

Example -6

Consider the system of equations \(x = cy + bz,\;y = az + cx,\;z = bx + ay,\) where \(x,y,z\) are not all zero. Find the value of \({a^2} + {b^2} + {c^2} + 2abc\).

Solution: The given system is homogenous:

\[\begin{align} - x + cy + bz &= 0\\cx - y\,\, + az &= 0\\bx + ay - z &= 0\end{align}\]

For this system to have a non-trivial solution, the corresponding determinant must be zero:

\[\left| {\;\begin{align}{ - 1}\quad c \quad b\\c\quad{ - 1}\quad a\\b \quad a \quad { - 1}\end{align}\;}\right| = 0\]

expanding this yields

\[{a^2} + {b^2} + {c^2} + 2abc = 1\]