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Find minors and cofactors of the elements of the determinant $$\left| {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 6&0&4 \\ 1&5&{ - 7} \end{array}} \right|$$ Verify that a11A31 + a12A32 + a13A33 = 0.
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## QuestionMathsClass 12

Find minors and cofactors of the elements of the determinant $$\left| {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 6&0&4 \\ 1&5&{ - 7} \end{array}} \right|$$ Verify that a11A31 + a12A32 + a13A33 = 0.

We have, $$\mathrm{M}_{11}=\left|\begin{array}{cc}{0} & {4} \\{5} & {-7}\end{array}\right|=0-20=-20 ; \mathrm{A}_{11}=(-1)^{1+1}(-20)=-20$$
$$\mathrm{M}_{12}=\left|\begin{array}{cc}{6} & {4} \\{1} & {-7}\end{array}\right|=-42-4=-46 ; \quad \mathrm{A}_{12}=(-1)^{1+2}(-46)=46$$
$$\mathrm{M}_{13}=\left|\begin{array}{cc}{6} & {0} \\{1} & {5}\end{array}\right|=30-0=30 ; \quad \mathrm{A}_{13}=(-1)^{1+3}(30)=30$$
$$\mathrm{M}_{21}=\left|\begin{array}{cc}{-3} & {5} \\{5} & {-7}\end{array}\right|=21-25=-4 ; \quad \mathrm{A}_{21}=(-1)^{2+1}(-4)=4$$
$$\mathrm{M}_{22}=\left|\begin{array}{cc}{2} & {5} \\{1} & {-7}\end{array}\right|=-14-5=-19 ; \quad \mathrm{A}_{22}=(-1)^{2+2}(-19)=-19$$
$$\mathrm{M}_{23}=\left|\begin{array}{cc}{2} & {-3} \\{1} & {5}\end{array}\right|=10+3=13 ; \quad \mathrm{A}_{23}=(-1)^{2+3}(13)=-13$$
$$\mathbf{M}_{31}=\left|\begin{array}{cc}{-3} & {5} \\{0} & {4}\end{array}\right|=-12-0=-12 ; \quad \mathrm{A}_{31}=(-1)^{3+1}(-12)=-12$$
$$\mathrm{M}_{32}=\left|\begin{array}{cc}{2} & {5} \\{6} & {4}\end{array}\right|=8-30=-22 ;$$$$\quad \mathrm{A}_{32}=(-1)^{3+2}(-22)=22$$
and $$\mathrm{M}_{33}=\left|\begin{array}{cc}{2} & {-3} \\{6} & {0}\end{array}\right|=0+18=18;$$ $$A_{33}=(-1)^{3+3}(18)=18$$
Now a11 = 2, a12 = -3, a13 = 5; A31 = -12, A32 = 22, A33 = 18
So, a11 A31 + a12 A32 + a13 A33
= 2 (-12) + (-3) (22) + 5 (18) = -24 - 66 + 90 = 0