NCERT Solutions · Class 12 Mathematics Determinants

By definition, the determinant of the \(\displaystyle 2\times 2\) matrix \(\displaystyle \left[\begin{array}{rr}a & b \\ c & d\end{array}\right]\) is \(\displaystyle ad-bc\) (product of the leading diagonal minus product of the other diagonal).Here \(\displaystyle a=2,\; b=4,\; c=-5,\; d=-1\), so \[\left|\begin{array}{rr}2 & 4 \\ -5 & -1\end{array}\right| = (2)(-1)-(4)(-5).\]The step to be careful with is the second product: \(\displaystyle bc=(4)(-5)=-20\), and subtracting \(\displaystyle -20\) adds \(\displaystyle 20\): \[= -2-(-20) = -2+20 = 18.\]Hence the value of the determinant is \(\displaystyle 18\).

The property being illustrated is: if \(\displaystyle A\) is a square matrix of order \(\displaystyle n\) and \(\displaystyle k\) is a scalar, then \(\displaystyle |kA|=k^{n}|A|\). Here \(\displaystyle n=2\) and \(\displaystyle k=2\), so the predicted factor is \(\displaystyle 2^{2}=4\). Verify both sides directly.Left side. Multiplying a matrix by a scalar multiplies every entry, so \[2A = 2\left[\begin{array}{rr}1 & 2 \\ 4 & 2\end{array}\right] = \left[\begin{array}{rr}2 & 4 \\ 8 & 4\end{array}\right],\] \[|2A| = (2)(4)-(4)(8) = 8-32 = -24.\]Right side. \[|A| = \left|\begin{array}{rr}1 & 2 \\ 4 & 2\end{array}\right| = (1)(2)-(2)(4) = 2-8 = -6,\] so \(\displaystyle 4|A| = 4(-6) = -24\).Since both sides equal \(\displaystyle -24\), \[|2A| = 4|A|.\]The reason the factor is \(\displaystyle 4\) and not \(\displaystyle 2\): scaling by \(\displaystyle 2\) scales both rows, and a determinant picks up the common factor of each row separately, once per row.

The property is \(\displaystyle |kA|=k^{n}|A|\) for a square matrix of order \(\displaystyle n\); here \(\displaystyle n=3\) and \(\displaystyle k=3\), so the predicted factor is \(\displaystyle 3^{3}=27\). Verify both sides.Right side. Expand \(\displaystyle |A|\) along the first column, which has only one non-zero entry: \[|A| = \left|\begin{array}{rrr}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right| = 1\left|\begin{array}{rr}1 & 2 \\ 0 & 4\end{array}\right| - 0 + 0 = 1(4-0) = 4.\] (Equivalently, \(\displaystyle A\) is upper triangular, so \(\displaystyle |A|\) is the product of the diagonal entries \(\displaystyle 1\cdot 1\cdot 4 = 4\).)Hence \(\displaystyle 27|A| = 27\times 4 = 108\).Left side. Every entry is tripled: \[3A = \left[\begin{array}{rrr}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right],\] \[|3A| = 3\left|\begin{array}{rr}3 & 6 \\ 0 & 12\end{array}\right| - 0 + 0 = 3(36-0) = 108.\]Both sides equal \(\displaystyle 108\), so \[|3A| = 27|A|.\]The exponent is \(\displaystyle 3\) because the factor \(\displaystyle 3\) comes out of each of the three rows: \(\displaystyle 3\times 3\times 3 = 27\).

Each determinant is expanded by the cofactor rule \(\displaystyle |A|=\sum_j a_{ij}(-1)^{i+j}M_{ij}\), choosing the row or column with the most zeros so as to do the least work.(i) Row $\displaystyle 2$ is \(\displaystyle 0,\;0,\;-1\), so expand along \(\displaystyle R_2\); only the entry \(\displaystyle a_{23}=-1\) contributes: \[\left|\begin{array}{rrr}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right| = (-1)\,(-1)^{2+3}\left|\begin{array}{rr}3 & -1 \\ 3 & -5\end{array}\right|.\] The sign factor \(\displaystyle (-1)^{2+3}=-1\) is the step usually dropped. The minor is \(\displaystyle (3)(-5)-(-1)(3) = -15+3 = -12\), so \[= (-1)(-1)(-12) = -12.\](ii) Expand along \(\displaystyle R_1\), remembering the alternating signs \(\displaystyle +,\,-,\,+\): \[\left|\begin{array}{rrr}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{array}\right| = 3\left|\begin{array}{rr}1 & -2 \\ 3 & 1\end{array}\right| - (-4)\left|\begin{array}{rr}1 & -2 \\ 2 & 1\end{array}\right| + 5\left|\begin{array}{rr}1 & 1 \\ 2 & 3\end{array}\right|\] \[= 3(1+6) + 4(1+4) + 5(3-2) = 21+20+5 = 46.\](iii) Expand along \(\displaystyle R_1\) (its first entry is \(\displaystyle 0\)): \[\left|\begin{array}{ccc}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{array}\right| = 0 - 1\left|\begin{array}{rr}-1 & -3 \\ -2 & 0\end{array}\right| + 2\left|\begin{array}{rr}-1 & 0 \\ -2 & 3\end{array}\right|\] \[= -1\big(0-6\big) + 2\big(-3-0\big) = 6-6 = 0.\] This is no accident: the matrix is skew-symmetric (\(\displaystyle a_{ij}=-a_{ji}\)) of odd order, and every such determinant is \(\displaystyle 0\).(iv) Expand along \(\displaystyle R_1\): \[\left|\begin{array}{rrr}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{array}\right| = 2\left|\begin{array}{rr}2 & -1 \\ -5 & 0\end{array}\right| - (-1)\left|\begin{array}{rr}0 & -1 \\ 3 & 0\end{array}\right| + (-2)\left|\begin{array}{rr}0 & 2 \\ 3 & -5\end{array}\right|\] \[= 2(0-5) + 1(0+3) - 2(0-6) = -10+3+12 = 5.\]The values are: (i) \(\displaystyle -12\), (ii) \(\displaystyle 46\), (iii) \(\displaystyle 0\), (iv) \(\displaystyle 5\).

Expand along the first row using \(\displaystyle |A|=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}\), with the sign pattern \(\displaystyle +,\,-,\,+\): \[|A| = 1\left|\begin{array}{rr}1 & -3 \\ 4 & -9\end{array}\right| - 1\left|\begin{array}{rr}2 & -3 \\ 5 & -9\end{array}\right| + (-2)\left|\begin{array}{rr}2 & 1 \\ 5 & 4\end{array}\right|.\]Evaluate the three \(\displaystyle 2\times 2\) minors, watching the signs: \[\left|\begin{array}{rr}1 & -3 \\ 4 & -9\end{array}\right| = -9-(-12) = 3,\qquad \left|\begin{array}{rr}2 & -3 \\ 5 & -9\end{array}\right| = -18-(-15) = -3,\qquad \left|\begin{array}{rr}2 & 1 \\ 5 & 4\end{array}\right| = 8-5 = 3.\]Therefore \[|A| = 1(3) - 1(-3) + (-2)(3) = 3+3-6 = 0.\]So \(\displaystyle |A| = 0\); the matrix is singular. The reason shows in the rows: \(\displaystyle 3R_1+R_2 = (3+2,\;3+1,\;-6-3) = (5,\,4,\,-9) = R_3\), i.e. the third row is a combination of the first two, and a determinant with dependent rows vanishes.

In each part evaluate both determinants by the rule \(\displaystyle ad-bc\) and equate.(i) \[\left|\begin{array}{cc}2 & 4 \\ 5 & 1\end{array}\right| = (2)(1)-(4)(5) = 2-20 = -18,\] \[\left|\begin{array}{cc}2x & 4 \\ 6 & x\end{array}\right| = (2x)(x)-(4)(6) = 2x^{2}-24.\] Equating, \[2x^{2}-24 = -18 \;\Longrightarrow\; 2x^{2} = 6 \;\Longrightarrow\; x^{2} = 3.\] Because the equation is quadratic, both square roots are solutions — nothing in the problem rules out a negative \(\displaystyle x\): \[x = \pm\sqrt{3}.\](ii) \[\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right| = 10-12 = -2,\] \[\left|\begin{array}{cc}x & 3 \\ 2x & 5\end{array}\right| = (x)(5)-(3)(2x) = 5x-6x = -x.\] Equating, \[-x = -2 \;\Longrightarrow\; x = 2.\]Hence (i) \(\displaystyle x=\pm\sqrt{3}\) and (ii) \(\displaystyle x=2\).

Evaluate each side by \(\displaystyle ad-bc\).Left side: \(\displaystyle \left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right| = x\cdot x - 2\cdot 18 = x^{2}-36.\)Right side: \(\displaystyle \left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right| = 36-36 = 0.\)So \[x^{2}-36 = 0 \;\Longrightarrow\; x^{2} = 36 \;\Longrightarrow\; x = \pm 6.\]The trap is to read off \(\displaystyle x=6\) by matching the two arrays entry by entry; the condition is only on the value of the determinant, and \(\displaystyle x=-6\) gives \(\displaystyle (-6)^{2}-36=0\) just as well.Hence \(\displaystyle x = \pm 6\), which is option (B).