Question 43: Verify that \(\det A = \det B + \det C\), where \(A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{11}}}&{{u_1} + {v_1}}\\{{a_{21}}}&{{a_{22}}}&{{u_2} + {v_2}}\\{{a_{31}}}&{{a_{32}}}&{{u_3} + {v_3}}\end{array}} \right]\), \(B = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\), \(C = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{v_1}}\\{{a_{21}}}&{{a_{22}}}&{{v_2}}\\{{a_{31}}}&{{a_{32}}}&{{v_3}}\end{array}} \right]\). Note, however, that \(A\) is not the same as \(B + C\).

Theorem 1states that the determinant of an\(n \times n\) matrix can be computed by a cofactor expansion across any row or down any column. The expansion across the \(i{\mathop{\rm th}\nolimits} \) row using the cofactor in \({C_{ij}} = {\left( { - 1} \right)^{i + j}}\det {A_{ij}}\)gives \(\det A = {a_{i1}}{C_{i1}} + {a_{i2}}{C_{i2}} + ... + {a_{in}}{C_{in}}\).

The cofactor expansion down the \(j{\mathop{\rm th}\nolimits} \) column gives\(\det A = {a_{1j}}{C_{1j}} + {a_{2j}}{C_{2j}} + ... + {a_{nj}}{C_{nj}}\)

Use cofactor expansion across the third column to compute \(\det A\) as shown below:

\(\begin{aligned}{}\det A &= \left( {{u_1} + {v_1}} \right)\det {A_{13}} - \left( {{u_2} + {v_2}} \right)\det {A_{23}} + \left( {{u_3} + {v_3}} \right)\det {A_{33}}\\ &= {u_1}\det {A_{13}} - {u_2}\det {A_{23}} + {u_3}\det {A_{33}} + {v_1}\det {A_{13}} - {v_2}\det {A_{23}} + {v_3}\det {A_{33}}\\ &= \det B + \det C\end{aligned}\)

Thus, it is verified that \(\det A = \det B + \det C\).