NCERT Solutions · Class 12 Mathematics Determinants

Consistency test by the matrix method: write the system as \(\displaystyle AX=B\). If \(\displaystyle |A|\neq 0\), then \(\displaystyle A^{-1}\) exists, \(\displaystyle X=A^{-1}B\) is the one and only solution, and the system is consistent.\[A=\left[\begin{array}{rr}1 & 2 \\ 2 & 3\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y\end{array}\right],\qquad B=\left[\begin{array}{r}2 \\ 3\end{array}\right] \]\[|A|=1(3)-2(2)=3-4=-1\neq 0 \]Since \(\displaystyle A\) is non-singular, the system has a unique solution.\[A^{-1}=\frac{1}{|A|}\,\mathrm{adj}\,A=\frac{1}{-1}\left[\begin{array}{rr}3 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{rr}-3 & 2 \\ 2 & -1\end{array}\right] \]\[X=A^{-1}B=\left[\begin{array}{rr}-3 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{r}2 \\ 3\end{array}\right]=\left[\begin{array}{r}-6+6 \\ 4-3\end{array}\right]=\left[\begin{array}{r}0 \\ 1\end{array}\right] \]The system is consistent, with the unique solution \(\displaystyle x=0,\ y=1\).

Write the system as \(\displaystyle AX=B\) and test \(\displaystyle |A|\): a non-zero determinant means \(\displaystyle A^{-1}\) exists, so \(\displaystyle X=A^{-1}B\) is the unique solution and the system is consistent.\[A=\left[\begin{array}{rr}2 & -1 \\ 1 & 1\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y\end{array}\right],\qquad B=\left[\begin{array}{r}5 \\ 4\end{array}\right] \]\[|A|=2(1)-(-1)(1)=2+1=3\neq 0 \]So \(\displaystyle A\) is non-singular and a unique solution exists.\[A^{-1}=\frac{1}{3}\left[\begin{array}{rr}1 & 1 \\ -1 & 2\end{array}\right] \]\[X=A^{-1}B=\frac{1}{3}\left[\begin{array}{rr}1 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{r}5 \\ 4\end{array}\right]=\frac{1}{3}\left[\begin{array}{r}9 \\ 3\end{array}\right]=\left[\begin{array}{r}3 \\ 1\end{array}\right] \]The system is consistent, with the unique solution \(\displaystyle x=3,\ y=1\).

When \(\displaystyle |A|=0\) the inverse does not exist, and the deciding test is the product \(\displaystyle (\mathrm{adj}\,A)B\): if \(\displaystyle (\mathrm{adj}\,A)B\neq O\) the system is inconsistent; if \(\displaystyle (\mathrm{adj}\,A)B=O\) it is consistent with infinitely many solutions (or no solution) and must be examined further.\[A=\left[\begin{array}{rr}1 & 3 \\ 2 & 6\end{array}\right],\qquad B=\left[\begin{array}{r}5 \\ 8\end{array}\right] \]\[|A|=1(6)-3(2)=6-6=0 \]So \(\displaystyle A\) is singular; form \(\displaystyle \mathrm{adj}\,A\) and multiply by \(\displaystyle B\).\[\mathrm{adj}\,A=\left[\begin{array}{rr}6 & -3 \\ -2 & 1\end{array}\right] \]\[(\mathrm{adj}\,A)B=\left[\begin{array}{rr}6 & -3 \\ -2 & 1\end{array}\right]\left[\begin{array}{r}5 \\ 8\end{array}\right]=\left[\begin{array}{r}30-24 \\ -10+8\end{array}\right]=\left[\begin{array}{r}6 \\ -2\end{array}\right]\neq O \](The same thing seen directly: doubling the first equation gives \(\displaystyle 2x+6y=10\), which contradicts \(\displaystyle 2x+6y=8\).)The system is inconsistent — it has no solution.

Write \(\displaystyle AX=B\) and evaluate \(\displaystyle |A|\); here \(\displaystyle a\) is a parameter, so the answer must be split on whether \(\displaystyle |A|\) vanishes.\[A=\left[\begin{array}{rrr}1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}1 \\ 2 \\ 4\end{array}\right] \]Take the common factor \(\displaystyle a\) out of the third row:\[|A|=a\left|\begin{array}{rrr}1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2\end{array}\right| = a\big[\,1(6-2)-1(4-2)+1(2-3)\,\big]=a(4-2-1)=a \]Case \(\displaystyle a\neq 0\): \(\displaystyle |A|=a\neq 0\), so \(\displaystyle A^{-1}\) exists and \(\displaystyle X=A^{-1}B\) is the unique solution — the system is consistent.Case \(\displaystyle a=0\): the third equation becomes \(\displaystyle 0\cdot x+0\cdot y+0\cdot z=4\), i.e. \(\displaystyle 0=4\), which is impossible — the system is inconsistent.Hence the system is consistent (unique solution) for every \(\displaystyle a\neq 0\), and inconsistent when \(\displaystyle a=0\).

Write the system as \(\displaystyle AX=B\) (a missing variable in an equation contributes a coefficient \(\displaystyle 0\)).\[A=\left[\begin{array}{rrr}3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right] \]Expand \(\displaystyle |A|\) along the first row:\[|A|=3(0-5)-(-1)(0+3)+(-2)(0-6)=-15+3+12=0 \]Since \(\displaystyle |A|=0\), the test is the sign of \(\displaystyle (\mathrm{adj}\,A)B\). The cofactors are\[A_{11}=-5,\ A_{12}=-3,\ A_{13}=-6,\quad A_{21}=10,\ A_{22}=6,\ A_{23}=12,\quad A_{31}=5,\ A_{32}=3,\ A_{33}=6 \]and \(\displaystyle \mathrm{adj}\,A\) is the transpose of the cofactor matrix:\[\mathrm{adj}\,A=\left[\begin{array}{rrr}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{array}\right] \]\[(\mathrm{adj}\,A)B=\left[\begin{array}{rrr}-5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6\end{array}\right]\left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right]=\left[\begin{array}{r}-10-10+15 \\ -6-6+9 \\ -12-12+18\end{array}\right]=\left[\begin{array}{r}-5 \\ -3 \\ -6\end{array}\right]\neq O \]The system is inconsistent — it has no solution.

Write the system as \(\displaystyle AX=B\) and test \(\displaystyle |A|\).\[A=\left[\begin{array}{rrr}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y \\ z\end{array}\right],\qquad B=\left[\begin{array}{r}5 \\ 2 \\ -1\end{array}\right] \]Expanding along the first row:\[|A|=5\,(18+10)-(-1)(12-25)+4\,(-4-15)=5(28)+(-13)+4(-19) \] \[|A|=140-13-76=51\neq 0 \]Since \(\displaystyle |A|\neq 0\), \(\displaystyle A^{-1}\) exists and \(\displaystyle X=A^{-1}B\) is the unique solution.The system is consistent, and it has exactly one solution.

Matrix method: write \(\displaystyle AX=B\), check \(\displaystyle |A|\neq 0\), then \(\displaystyle X=A^{-1}B\) with \(\displaystyle A^{-1}=\dfrac{1}{|A|}\,\mathrm{adj}\,A\).\[A=\left[\begin{array}{rr}5 & 2 \\ 7 & 3\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y\end{array}\right],\qquad B=\left[\begin{array}{r}4 \\ 5\end{array}\right] \]\[|A|=5(3)-2(7)=15-14=1\neq 0 \]For a \(\displaystyle 2\times 2\) matrix, \(\displaystyle \mathrm{adj}\,A\) swaps the diagonal entries and changes the sign of the other two:\[A^{-1}=\frac{1}{1}\left[\begin{array}{rr}3 & -2 \\ -7 & 5\end{array}\right] \]\[X=A^{-1}B=\left[\begin{array}{rr}3 & -2 \\ -7 & 5\end{array}\right]\left[\begin{array}{r}4 \\ 5\end{array}\right]=\left[\begin{array}{r}12-10 \\ -28+25\end{array}\right]=\left[\begin{array}{r}2 \\ -3\end{array}\right] \]\[x=2,\qquad y=-3 \]

Write \(\displaystyle AX=B\) and solve by \(\displaystyle X=A^{-1}B\).\[A=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y\end{array}\right],\qquad B=\left[\begin{array}{r}-2 \\ 3\end{array}\right] \]\[|A|=2(4)-(-1)(3)=8+3=11\neq 0 \]\[A^{-1}=\frac{1}{11}\left[\begin{array}{rr}4 & 1 \\ -3 & 2\end{array}\right] \]\[X=A^{-1}B=\frac{1}{11}\left[\begin{array}{rr}4 & 1 \\ -3 & 2\end{array}\right]\left[\begin{array}{r}-2 \\ 3\end{array}\right]=\frac{1}{11}\left[\begin{array}{r}-8+3 \\ 6+6\end{array}\right]=\frac{1}{11}\left[\begin{array}{r}-5 \\ 12\end{array}\right] \]\[x=-\frac{5}{11},\qquad y=\frac{12}{11} \]

Write \(\displaystyle AX=B\) and solve by \(\displaystyle X=A^{-1}B\). Watch the sign of the determinant here — it is negative, and that sign must be carried into \(\displaystyle A^{-1}\).\[A=\left[\begin{array}{rr}4 & -3 \\ 3 & -5\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y\end{array}\right],\qquad B=\left[\begin{array}{r}3 \\ 7\end{array}\right] \]\[|A|=4(-5)-(-3)(3)=-20+9=-11\neq 0 \]\[\mathrm{adj}\,A=\left[\begin{array}{rr}-5 & 3 \\ -3 & 4\end{array}\right],\qquad A^{-1}=\frac{1}{-11}\left[\begin{array}{rr}-5 & 3 \\ -3 & 4\end{array}\right] \]\[X=A^{-1}B=-\frac{1}{11}\left[\begin{array}{rr}-5 & 3 \\ -3 & 4\end{array}\right]\left[\begin{array}{r}3 \\ 7\end{array}\right]=-\frac{1}{11}\left[\begin{array}{r}-15+21 \\ -9+28\end{array}\right]=-\frac{1}{11}\left[\begin{array}{r}6 \\ 19\end{array}\right] \]\[x=-\frac{6}{11},\qquad y=-\frac{19}{11} \]

Write \(\displaystyle AX=B\) and solve by \(\displaystyle X=A^{-1}B\).\[A=\left[\begin{array}{rr}5 & 2 \\ 3 & 2\end{array}\right],\qquad X=\left[\begin{array}{r}x \\ y\end{array}\right],\qquad B=\left[\begin{array}{r}3 \\ 5\end{array}\right] \]\[|A|=5(2)-2(3)=10-6=4\neq 0 \]\[A^{-1}=\frac{1}{4}\left[\begin{array}{rr}2 & -2 \\ -3 & 5\end{array}\right] \]\[X=A^{-1}B=\frac{1}{4}\left[\begin{array}{rr}2 & -2 \\ -3 & 5\end{array}\right]\left[\begin{array}{r}3 \\ 5\end{array}\right]=\frac{1}{4}\left[\begin{array}{r}6-10 \\ -9+25\end{array}\right]=\frac{1}{4}\left[\begin{array}{r}-4 \\ 16\end{array}\right] \]\[x=-1,\qquad y=4 \]