math_matrix_doubt_questionsFrequently asked Questions: Matrix and Determinants
Matrix and Determinants
Question: Show that$|\begin{array}{ccc}a+b& a& b\\ a& a+c& c\\ b& c& b+c\end{array}|=4abc\phantom{\rule{0.66em}{0ex}}$ Answer$\begin{array}{l}LHS=|\begin{array}{ccc}a+b& a& b\\ a& a+c& c\\ b& c& b+c\end{array}|\\ \\ R1\to R1-R2-R3\\ \\ =|\begin{array}{ccc}0& -2c& -2c\\ a& a+c& c\\ b& c& b+c\end{array}|\\ \\ C2\to C2-C3\\ \phantom{\rule{0.44em}{0ex}}\\ =|\begin{array}{ccc}0& 0& -2c\\ a& a& c\\ b& -b& b+c\end{array}|\\ =\left(-2c\right)\left(-ab-ab\right)\\ =4abc=RHS\\ \end{array}$
Question: Show that$|\begin{array}{ccc}1& lo{g}_{x}y& lo{g}_{x}z\\ lo{g}_{y}x& 1& lo{g}_{y}z\\ lo{g}_{z}x& lo{g}_{y}y& 1\end{array}|=0\phantom{\rule{0.66em}{0ex}}$ Answer$\begin{array}{l}LHS=|\begin{array}{ccc}1& lo{g}_{x}y& lo{g}_{x}z\\ lo{g}_{y}x& 1& lo{g}_{y}z\\ lo{g}_{z}x& lo{g}_{y}y& 1\end{array}|\\ \\ \\ =|\begin{array}{ccc}lo{g}_{x}x& lo{g}_{x}y& lo{g}_{x}z\\ lo{g}_{y}x& lo{g}_{y}y& lo{g}_{y}z\\ lo{g}_{z}x& lo{g}_{z}y& lo{g}_{z}z\end{array}|\\ \\ \phantom{\rule{0.44em}{0ex}}\\ =|\begin{array}{ccc}\frac{logx}{logx}& \frac{logy}{logx}& \frac{logz}{logx}\\ \frac{logx}{logy}& \frac{logy}{logy}& \frac{logz}{logy}\\ \frac{logx}{logz}& \frac{logy}{logz}& \frac{logz}{logx}\end{array}|\\ =\frac{1}{logx\phantom{\rule{0.22em}{0ex}}logy\phantom{\rule{0.22em}{0ex}}logz}|\begin{array}{ccc}logx& logy& logz\\ logx& logy& logz\\ logx& logy& logz\end{array}|\\ \\ =\frac{1}{logx\phantom{\rule{0.22em}{0ex}}logy\phantom{\rule{0.22em}{0ex}}logz}×0=0=RHS\\ \end{array}$
Question: If$|\begin{array}{ccc}4+x& 4-x& 4-x\\ 4-x& 4+x& 4-x\\ 4-x& 4-x& 4+x\end{array}|=0\phantom{\rule{0.44em}{0ex}}$ Find x Answer$\begin{array}{l}|\begin{array}{ccc}4+x& 4-x& 4-x\\ 4-x& 4+x& 4-x\\ 4-x& 4-x& 4+x\end{array}|=0\\ \\ C2\to C2-C3\\ ⟹|\begin{array}{ccc}4+x& 0& 4-x\\ 4-x& 2x& 4-x\\ 4-x& -2x& 4+x\end{array}|=0\phantom{\rule{0.22em}{0ex}}\\ \\ C1\to C1+C2\\ \\ ⟹|\begin{array}{ccc}4+x& 0& 4-x\\ 4+x& 2x& 4-x\\ 4-3x& -2x& 4+x\end{array}|=0\\ \\ R1\to R1-R2\\ \\ ⟹|\begin{array}{ccc}0& -2x& 0\\ 4+x& 2x& 4-x\\ 4-3x& -2x& 4+x\end{array}|=0\\ \\ \\ ⟹2x\left[\left(4+x\right)\left(4+x\right)-\left(4-x\right)\left(4-3x\right)\right]\phantom{\rule{0.22em}{0ex}}=0\\ \\ ⟹x=0\phantom{\rule{0.22em}{0ex}}Answer\\ \\ or\\ \\ 16+8x+{x}^{2}-16-3{x}^{2}+16x=0\\ ⟹2{x}^{2}-24x=0\\ ⟹2x\left(x-12\right)=0\\ ⟹x=12\phantom{\rule{0.22em}{0ex}}Answer\\ \end{array}$
Questionwithout expanding show that$\begin{array}{l}|\begin{array}{ccc}b+c& bc& {b}^{2}{c}^{2}\\ c+a& ca& {c}^{2}{a}^{2}\\ a+b& ab& {a}^{2}{b}^{2}\end{array}|=0\\ \end{array}$ Answer$\begin{array}{l}\\ \\ LHS=|\begin{array}{ccc}b+c& bc& {b}^{2}{c}^{2}\\ c+a& ca& {c}^{2}{a}^{2}\\ a+b& ab& {a}^{2}{b}^{2}\end{array}|\\ \\ aR1,bR2,cR3\\ \\ =\frac{1}{abc}|\begin{array}{ccc}a\left(b+c\right)& abc& a{b}^{2}{c}^{2}\\ b\left(c+a\right)& bca& b{c}^{2}{a}^{2}\\ c\left(a+b\right)& cab& c{a}^{2}{b}^{2}\end{array}|\\ \\ take\phantom{\rule{0.22em}{0ex}}abc\phantom{\rule{0.22em}{0ex}}common\phantom{\rule{0.22em}{0ex}}from\phantom{\rule{0.22em}{0ex}}C3\\ \\ =\frac{abc}{abc}|\begin{array}{ccc}ab+ca& abc& bc\\ bc+ab& bca& ca\\ ca+bc& cab& ab\end{array}|\\ \\ {C}_{1}\to {C}_{1}+{C}_{3}\\ \\ =|\begin{array}{ccc}ab+bc+ca& abc& bc\\ ab+bc+ca& abc& ca\\ ab+bc+ca& abc& ab\end{array}|\\ \phantom{\rule{0.22em}{0ex}}\end{array}$as all the values in ${C}_{1}$ is same and all the values in ${C}_{2}$is same so $\begin{array}{l}|\begin{array}{ccc}ab+bc+ca& abc& bc\\ ab+bc+ca& abc& ca\\ ab+bc+ca& abc& ab\end{array}|=0\phantom{\rule{0.22em}{0ex}}\\ \\ So\phantom{\rule{0.22em}{0ex}}|\begin{array}{ccc}b+c& bc& {b}^{2}{c}^{2}\\ c+a& ca& {c}^{2}{a}^{2}\\ a+b& ab& {a}^{2}{b}^{2}\end{array}|=0\end{array}$
Questionwithout expanding show that$\begin{array}{l}|\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}|=0\\ \end{array}$ Answer$\begin{array}{l}LHS=|\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}|\\ \\ c{R}_{1},\phantom{\rule{0.22em}{0ex}}-b{R}_{2},\phantom{\rule{0.22em}{0ex}}a{R}_{3}\\ \\ =-\frac{1}{abc}|\begin{array}{ccc}0& ca& bc\\ ba& 0& -bc\\ -ab& -ca& 0\end{array}|\\ \\ {R}_{1}\to {R}_{1}+{R}_{2}+{R}_{3}\\ \\ =-\frac{1}{abc}|\begin{array}{ccc}0& 0& 0\\ ba& 0& -bc\\ -ab& -ca& 0\end{array}|\\ \end{array}$As all the values of ${R}_{1}$ is 0 (zero), so $-\frac{1}{abc}|\begin{array}{ccc}0& 0& 0\\ ba& 0& -bc\\ -ab& -ca& 0\end{array}|$=0 So $\begin{array}{l}|\begin{array}{ccc}0& a& b\\ -a& 0& c\\ -b& -c& 0\end{array}|=0\\ \end{array}$