QUESTION

# Match the following elements of $\left| \begin{matrix} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \\\end{matrix} \right|$ with their cofactors and choose the correct answer  $\text{Element} \;\;\;\;\;\; \text{Cofactor}$\;\;\;-1 \;\;\;\;\;\; \;\;\;\;\; \;\;\;\;\; a) -2$ \;\;\;\;\;1 \;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\; b)\;32$\;\;\;\;\;3 \;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\; c)\; 4$ \;\;\;\;\;6 \;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\; d)\; 6$ $\;\;\;\;\;\; \;\;\;\;\;\; \;\;\;\;\; \;\;\;\;\;\; e) -6$ (A) I - b, II – d, III – a, IV – c (B) I - b, II – d, III – c, IV - a (C) I - d, II – b, III – a, IV - c (D) I - d, II – a, III – b, IV - c

Hint: To solve the question, we have to apply the formula for cofactor of 3x3 matrix to calculate the cofactors of the given elements. To solve further apply the formula for determinant of 2x2 matrix, which will ease the procedure of solving. Solve independently for each element to avoid mistakes.

The given matrix $\left| \begin{matrix} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \\ \end{matrix} \right|$
We know that the formula for cofactor of 3x3 matrix $\left| \begin{matrix} {{a}_{1,1}} & {{a}_{1,2}} & {{a}_{1,3}} \\ {{a}_{2,1}} & {{a}_{2,2}} & {{a}_{2,3}} \\ {{a}_{3,1}} & {{a}_{3,2}} & {{a}_{3,3}} \\ \end{matrix} \right|$ is given by
${{\Delta }_{m,n}}={{\left( -1 \right)}^{m+n}}\left| {{M}_{m,n}} \right|$
Where matrix ${{M}_{m,n}}$ is formed after deleting ${{m}^{th}}$ row and ${{n}^{th}}$ column.
By comparing the given matrix with the general 3x3 matrix we get
${{a}_{1,1}}=1,{{a}_{1,2}}=-1,{{a}_{3,1}}=3,{{a}_{3,3}}=6$
For ${{a}_{1,1}}=1$
${{\Delta }_{1,1}}={{\left( -1 \right)}^{1+1}}\left| {{M}_{1,1}} \right|$
${{\Delta }_{1,1}}={{\left( -1 \right)}^{2}}\left| \begin{matrix} 4 & 2 \\ -4 & 6 \\ \end{matrix} \right|$
${{\Delta }_{1,1}}=(1)\left| \begin{matrix} 4 & 2 \\ -4 & 6 \\ \end{matrix} \right|$
We know the determinant of 2x2 matrix $\left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right|$ is equal to ad – bc. Thus, by substituting the values, we get
\begin{align} & {{\Delta }_{1,1}}=\left( 4\times 6-(-4)2 \right) \\ & =24-(-8) \\ & =24+8 \\ & =32 \\ \end{align}
Thus, the cofactor of element 1 is equal to 32.
For ${{a}_{1,2}}=-1$
${{\Delta }_{1,2}}={{\left( -1 \right)}^{1+2}}\left| {{M}_{1,2}} \right|$
${{\Delta }_{1,2}}={{\left( -1 \right)}^{3}}\left| \begin{matrix} 0 & 2 \\ 3 & 6 \\ \end{matrix} \right|$
${{\Delta }_{1,2}}=(-1)\left| \begin{matrix} 0 & 2 \\ 3 & 6 \\ \end{matrix} \right|$
By applying the determinant of 2x2 matrix formula, we get
\begin{align} & {{\Delta }_{1,2}}=-\left( 0\times 6-(3)2 \right) \\ & =-\left( 0-6 \right) \\ & =-(-6) \\ & =6 \\ \end{align}
Thus, the cofactor of element -1 is equal to 6.
For ${{a}_{3,1}}=3$
${{\Delta }_{3,1}}={{\left( -1 \right)}^{3+1}}\left| {{M}_{3,1}} \right|$
${{\Delta }_{3,1}}={{\left( -1 \right)}^{4}}\left| \begin{matrix} -1 & 0 \\ 4 & 2 \\ \end{matrix} \right|$
${{\Delta }_{3,1}}=(1)\left| \begin{matrix} -1 & 0 \\ 4 & 2 \\ \end{matrix} \right|$
By applying the determinant of 2x2 matrix formula, we get
\begin{align} & {{\Delta }_{3,1}}=\left( (-1)\times 2-(4)0 \right) \\ & =-2-(0) \\ & =-2 \\ \end{align}
Thus, the cofactor of element 3 is equal to -2.
For ${{a}_{3,3}}=6$
${{\Delta }_{3,3}}={{\left( -1 \right)}^{3+3}}\left| {{M}_{3,3}} \right|$
${{\Delta }_{3,3}}={{\left( -1 \right)}^{6}}\left| \begin{matrix} 1 & -1 \\ 0 & 4 \\ \end{matrix} \right|$
${{\Delta }_{3,3}}=(1)\left| \begin{matrix} 1 & -1 \\ 0 & 4 \\ \end{matrix} \right|$
By applying the determinant of 2x2 matrix formula, we get
\begin{align} & {{\Delta }_{3,3}}=\left( 1\times 4-(-1)0 \right) \\ & =4-(0) \\ & =4 \\ \end{align}
Thus, the cofactor of element 6 is equal to 4.
Thus, the match the following is I - b, II – d, III – a, IV – c
Hence, option (A) is the right choice.

Note: The possibility of mistake can be, not applying the formula for cofactor of 3x3 matrix and the formula for determinant of 2x2 matrix, which will ease the procedure of solving. The alternative way of solving is option elimination method, we have to calculate cofactor of element I and eliminate the other options and find the cofactor of element 3 or 6 to find the right choice. Thus, by only finding cofactors of two elements, we can find the right choice.