# Determinants And Matrices Set-2

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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

${a^2} + {b^2} + {c^2} + 2abc = 1$

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