Exercise – 4.4
1. Write Minors and Cofactors of the elements of the following determinants:
Solution:
-
$\left| {\begin{array}{*{20}{r}}2& -4\\0& 3\end{array}} \right|$
Let $P = \left| {\begin{array}{*{20}{r}}2& -4\\0& 3\end{array}} \right|$
Minor of the element $a_{ij}$ is $M_{ij}$. Here,
$M_{11} = 3$, $M_{12} = 0$, $M_{21} = -4$, $M_{22} = 2$
For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
$\therefore P_{11} = 3$, $P_{12} = 0$, $P_{21} = 4$, $P_{22} = 3$ -
$\left| {\begin{array}{*{20}{r}}a& c\\b& d\end{array}} \right|$
Let $P = \left| {\begin{array}{*{20}{r}}a& c\\b& d\end{array}} \right|$, Minor of the element $a_{ij}$ is $M_{ij}$. Here,
$M_{11} = d$, $M_{12} = b$, $M_{21} = c$, $M_{22} = a$
For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
$\therefore P_{11} = d$, $P_{12} = -b$, $P_{21} = -c$, $P_{22} = a$
2. Write Minors and Cofactors of the elements of the following determinants:
Solution:
-
$\left| {\begin{array}{*{20}{r}}1& 0& 0\\0& 1& 0\\0& 0& 1\end{array}} \right|$
Let $P = \left| {\begin{array}{*{20}{r}}1& 0& 0\\0& 1& 0\\0& 0& 1\end{array}} \right|$, we have
$M_{11} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 1$, $M_{12} = \left| {\begin{array}{*{20}{r}}0& 0\\0& 1\end{array}} \right| = 0$, $M_{13} = \left| {\begin{array}{*{20}{r}}0& 1\\0& 1\end{array}} \right| = 0$
$M_{21} = \left| {\begin{array}{*{20}{r}}0& 0\\0& 1\end{array}} \right| = 0$, $M_{22} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 1$, $M_{23} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 0$
$M_{31} = \left| {\begin{array}{*{20}{r}}0& 0\\0& 1\end{array}} \right| = 0$, $M_{32} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 0$, $M_{33} = \left| {\begin{array}{*{20}{r}}1& 1\\0& 1\end{array}} \right| = 1$
For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
$P_{11} = 1$, $P_{12} = 0$, $P_{13} = 0$, $P_{21} = 0$, $P_{22} = 1$, $P_{23} = 0$, $P_{31} = 0$, $P_{32} = 0$, $P_{33} = 1$ -
$\left| {\begin{array}{*{20}{r}}1& 0& 4\\3& 5& -1\\0& 1& 2\end{array}} \right|$
Let $P = \left| {\begin{array}{*{20}{r}}1& 0& 4\\3& 5& -1\\0& 1& 2\end{array}} \right|$, we have
$M_{11} = \left| {\begin{array}{*{20}{r}}5& -1\\1& 2\end{array}} \right| = 10 + 1 = 11$, $M_{12} = \left| {\begin{array}{*{20}{r}}3& -1\\0& 2\end{array}} \right| = 6$, $M_{13} = \left| {\begin{array}{*{20}{r}}3& 5\\0& 1\end{array}} \right| = 3$
$M_{21} = \left| {\begin{array}{*{20}{r}}0& 4\\1& 2\end{array}} \right| = -4$, $M_{22} = \left| {\begin{array}{*{20}{r}}1& 4\\0& 2\end{array}} \right| = 2$, $M_{23} = \left| {\begin{array}{*{20}{r}}1& 0\\0& 1\end{array}} \right| = 1$
$M_{31} = \left| {\begin{array}{*{20}{r}}0& 4\\5& -1\end{array}} \right| = -20$, $M_{32} = \left| {\begin{array}{*{20}{r}}1& 4\\3& -1\end{array}} \right| = -13$, $M_{33} = \left| {\begin{array}{*{20}{r}}1& 0\\3& 5\end{array}} \right| = 5$
For cofactors, we know that $P_{ij} = (-1)^{i+j}M_{ij}$
$P_{11} = 11$, $P_{12} = -6$, $P_{13} = 3$, $P_{21} = 4$, $P_{22} = 2$, $P_{23} = -1$, $P_{31} = -20$, $P_{32} = 13$, $P_{33} = 5$
3. Using Cofactors of elements of the second row, evaluate $\Delta = \left| {\begin{array}{*{20}{r}}5& 3& 8\\2& 0& 1\\1& 2& 3\end{array}} \right|$.
Solution:
$\Delta = \left| {\begin{array}{*{20}{r}}5& 3& 8\\2& 0& 1\\1& 2& 3\end{array}} \right|$
Cofactors of elements of the second row are
$A_{21} = (-1)^{2+1}\left| {\begin{array}{*{20}{r}}3& 8\\2& 3\end{array}} \right| = -(9 – 16) = 7$
$A_{22} = (-1)^{2+2}\left| {\begin{array}{*{20}{r}}5& 8\\1& 3\end{array}} \right| = 15 – 8 = 7$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}5& 3\\1& 2\end{array}} \right| = -(10 – 3) = -7$
Now, $\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2 \times 7 + 0 \times 7 + 1 \times (-7) = 14 + 0 – 7 = 7$
4. Using Cofactors of elements of the third column, evaluate $\Delta = \left| {\begin{array}{*{20}{r}}1& x& yz\\1& y& zx\\1& z& xy\end{array}} \right|$.
Solution:
Let $\Delta = \left| {\begin{array}{*{20}{r}}1& x& yz\\1& y& zx\\1& z& xy\end{array}} \right|$
Cofactors of elements of the third column are
$A_{13} = (-1)^{1+3}\left| {\begin{array}{*{20}{r}}1& y\\1& z\end{array}} \right| = z – y$
$A_{23} = (-1)^{2+3}\left| {\begin{array}{*{20}{r}}1& x\\1& z\end{array}} \right| = -(z – x)$
$A_{33} = (-1)^{3+3}\left| {\begin{array}{*{20}{r}}1& x\\1& y\end{array}} \right| = y – x = -(x – y)$
Now, $\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$
$\Delta = -yz(y – z) – zx(z – x) – xy(x – y)$
$= zy(z – y) + zx(x – z) + xy(y – x)$
$= yz^2 – y^2z + zx^2 – z^2x + xy^2 – x^2y$
$= x^2z – x^2y + xy^2 – xz^2 + yz^2 – y^2z$
$= x^2(z – y) + x(y^2 – z^2) + yz(z – y)$
$= (z – y)[x^2 – x(y + z) + yz] = (z – y)[x(x – y) – z(x – y)]$
$= (z – y)(x – y)(x – z) = (x – y)(y – z)(z – x)$
5. If $\Delta = \left| {\begin{array}{*{20}{r}}a_{11}& a_{12}& a_{13}\\a_{21}& a_{22}& a_{23}\\a_{31}& a_{32}& a_{33}\end{array}} \right|$ and $A_{ij}$ is the Cofactor of $a_{ij}$, then the value of $\Delta$ is given by:
Solution:
(D) $a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$
We know that the value of the determinant is given by:
$a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}$ (where $A_{ij}$ is the Cofactor of $a_{ij}$)