NCERT Solutions · Class 12 Mathematics Matrices

Recall the definitions: a square matrix \(\displaystyle \mathrm{M}\) is symmetric if \(\displaystyle \mathrm{M}^{\prime}=\mathrm{M}\), and skew symmetric if \(\displaystyle \mathrm{M}^{\prime}=-\mathrm{M}\). So the whole job is to compute \(\displaystyle (\mathrm{AB}-\mathrm{BA})^{\prime}\) and show it equals \(\displaystyle -(\mathrm{AB}-\mathrm{BA})\).Given: \(\displaystyle \mathrm{A}^{\prime}=\mathrm{A}\) and \(\displaystyle \mathrm{B}^{\prime}=\mathrm{B}\). Being symmetric, both are square of the same order, so \(\displaystyle \mathrm{AB}\), \(\displaystyle \mathrm{BA}\) and their difference are all defined and square.Use the two transpose rules: \(\displaystyle (\mathrm{X}-\mathrm{Y})^{\prime}=\mathrm{X}^{\prime}-\mathrm{Y}^{\prime}\), and the reversal law \(\displaystyle (\mathrm{XY})^{\prime}=\mathrm{Y}^{\prime}\mathrm{X}^{\prime}\) — the order of the factors REVERSES, which is the step that decides the sign here: \[(\mathrm{AB}-\mathrm{BA})^{\prime}=(\mathrm{AB})^{\prime}-(\mathrm{BA})^{\prime}=\mathrm{B}^{\prime}\mathrm{A}^{\prime}-\mathrm{A}^{\prime}\mathrm{B}^{\prime}.\]Now substitute \(\displaystyle \mathrm{A}^{\prime}=\mathrm{A}\) and \(\displaystyle \mathrm{B}^{\prime}=\mathrm{B}\): \[(\mathrm{AB}-\mathrm{BA})^{\prime}=\mathrm{BA}-\mathrm{AB}=-(\mathrm{AB}-\mathrm{BA}).\]The transpose is the negative of the matrix itself, so \(\displaystyle \mathrm{AB}-\mathrm{BA}\) is a skew symmetric matrix. Hence proved.

First check the product makes sense: if \(\displaystyle \mathrm{A}\) is \(\displaystyle n\times n\) and \(\displaystyle \mathrm{B}\) is \(\displaystyle n\times m\), then \(\displaystyle \mathrm{B}^{\prime}\) is \(\displaystyle m\times n\), so \(\displaystyle \mathrm{B}^{\prime}\mathrm{AB}\) is \(\displaystyle m\times m\) — square, so asking whether it is symmetric or skew symmetric is meaningful.Transpose it using the reversal law extended to three factors, \(\displaystyle (\mathrm{XYZ})^{\prime}=\mathrm{Z}^{\prime}\mathrm{Y}^{\prime}\mathrm{X}^{\prime}\), and the involution property \(\displaystyle (\mathrm{B}^{\prime})^{\prime}=\mathrm{B}\): \[(\mathrm{B}^{\prime}\mathrm{AB})^{\prime}=\mathrm{B}^{\prime}\mathrm{A}^{\prime}(\mathrm{B}^{\prime})^{\prime}=\mathrm{B}^{\prime}\mathrm{A}^{\prime}\mathrm{B}.\]Only now split into the two cases, because everything above is independent of what \(\displaystyle \mathrm{A}\) is.Case $\displaystyle 1$: \(\displaystyle \mathrm{A}\) symmetric, i.e. \(\displaystyle \mathrm{A}^{\prime}=\mathrm{A}\). Then \[(\mathrm{B}^{\prime}\mathrm{AB})^{\prime}=\mathrm{B}^{\prime}\mathrm{AB},\] so \(\displaystyle \mathrm{B}^{\prime}\mathrm{AB}\) is symmetric.Case $\displaystyle 2$: \(\displaystyle \mathrm{A}\) skew symmetric, i.e. \(\displaystyle \mathrm{A}^{\prime}=-\mathrm{A}\). Then \[(\mathrm{B}^{\prime}\mathrm{AB})^{\prime}=\mathrm{B}^{\prime}(-\mathrm{A})\mathrm{B}=-(\mathrm{B}^{\prime}\mathrm{AB}),\] so \(\displaystyle \mathrm{B}^{\prime}\mathrm{AB}\) is skew symmetric.Hence \(\displaystyle \mathrm{B}^{\prime}\mathrm{AB}\) is symmetric or skew symmetric according as \(\displaystyle \mathrm{A}\) is symmetric or skew symmetric.

Write down \(\displaystyle \mathrm{A}^{\prime}\) by turning the rows of \(\displaystyle \mathrm{A}\) into columns: \[\mathrm{A}=\left[\begin{array}{ccr}0 & 2y & z \\ x & y & -z \\ x & -y & z\end{array}\right],\qquad \mathrm{A}^{\prime}=\left[\begin{array}{rrr}0 & x & x \\ 2y & y & -y \\ z & -z & z\end{array}\right].\]Now form \(\displaystyle \mathrm{A}^{\prime}\mathrm{A}\), taking (row of \(\displaystyle \mathrm{A}^{\prime}\))\(\displaystyle \cdot\)(column of \(\displaystyle \mathrm{A}\)). Note \(\displaystyle (\mathrm{A}^{\prime}\mathrm{A})^{\prime}=\mathrm{A}^{\prime}\mathrm{A}\), so the product is automatically symmetric and only six entries need checking.Diagonal entries — these are the sums of squares of the COLUMNS of \(\displaystyle \mathrm{A}\): \[(\mathrm{A}^{\prime}\mathrm{A})_{11}=0^{2}+x^{2}+x^{2}=2x^{2},\] \[(\mathrm{A}^{\prime}\mathrm{A})_{22}=(2y)^{2}+y^{2}+(-y)^{2}=4y^{2}+y^{2}+y^{2}=6y^{2},\] \[(\mathrm{A}^{\prime}\mathrm{A})_{33}=z^{2}+(-z)^{2}+z^{2}=3z^{2}.\]Off-diagonal entries: \[(\mathrm{A}^{\prime}\mathrm{A})_{12}=0\cdot 2y+x\cdot y+x\cdot(-y)=0,\] \[(\mathrm{A}^{\prime}\mathrm{A})_{13}=0\cdot z+x\cdot(-z)+x\cdot z=0,\] \[(\mathrm{A}^{\prime}\mathrm{A})_{23}=2y\cdot z+y\cdot(-z)+(-y)\cdot z=2yz-yz-yz=0.\]So the off-diagonal conditions hold for ALL \(\displaystyle x,y,z\), and \[\mathrm{A}^{\prime}\mathrm{A}=\left[\begin{array}{rrr}2x^{2} & 0 & 0 \\ 0 & 6y^{2} & 0 \\ 0 & 0 & 3z^{2}\end{array}\right].\]Equating this to \(\displaystyle \mathrm{I}\) and comparing corresponding elements: \[2x^{2}=1,\qquad 6y^{2}=1,\qquad 3z^{2}=1.\]Each equation gives two roots, since squaring loses the sign: \[x=\pm\frac{1}{\sqrt{2}},\qquad y=\pm\frac{1}{\sqrt{6}},\qquad z=\pm\frac{1}{\sqrt{3}}.\]The three signs are independent of one another, so any of the eight sign combinations works.

Matrix multiplication is associative, so bracket the product in whichever order is easier. Multiplying the \(\displaystyle 1\times 3\) row into the \(\displaystyle 3\times 3\) matrix first keeps everything one row wide; the result is \(\displaystyle 1\times 1\), so \(\displaystyle \mathrm{O}\) here means the \(\displaystyle 1\times 1\) zero matrix \(\displaystyle [0]\).Step $\displaystyle 1$: \[\left[\begin{array}{lll}1 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}1+4+1 & 2+0+0 & 0+2+2\end{array}\right]=\left[\begin{array}{lll}6 & 2 & 4\end{array}\right].\]Step $\displaystyle 2$: \[\left[\begin{array}{lll}6 & 2 & 4\end{array}\right]\left[\begin{array}{l}0 \\ 2 \\ x\end{array}\right]=\left[6\cdot 0+2\cdot 2+4\cdot x\right]=\left[4+4x\right].\]Set this equal to \(\displaystyle [0]\) and compare the single element: \[4+4x=0\ \Longrightarrow\ x=-1.\]Hence \(\displaystyle x=-1\).

Here \(\displaystyle \mathrm{A}^{2}\) means \(\displaystyle \mathrm{A}\cdot\mathrm{A}\) (matrix product, not entrywise squares) and \(\displaystyle \mathrm{I}\) is the \(\displaystyle 2\times 2\) identity.Compute \(\displaystyle \mathrm{A}^{2}\): \[\mathrm{A}^{2}=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{rr}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{rr}8 & 5 \\ -5 & 3\end{array}\right].\]Compute the other two terms by scalar multiplication: \[5\mathrm{A}=\left[\begin{array}{rr}15 & 5 \\ -5 & 10\end{array}\right],\qquad 7\mathrm{I}=\left[\begin{array}{rr}7 & 0 \\ 0 & 7\end{array}\right].\]Add and subtract entrywise: \[\mathrm{A}^{2}-5\mathrm{A}+7\mathrm{I}=\left[\begin{array}{rr}8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7\end{array}\right]=\left[\begin{array}{rr}0 & 0 \\ 0 & 0\end{array}\right]=0.\]Hence \(\displaystyle \mathrm{A}^{2}-5\mathrm{A}+7\mathrm{I}=0\).

The product of a \(\displaystyle 1\times 3\), a \(\displaystyle 3\times 3\) and a \(\displaystyle 3\times 1\) matrix is \(\displaystyle 1\times 1\), so \(\displaystyle \mathrm{O}\) is the \(\displaystyle 1\times 1\) zero matrix \(\displaystyle [0]\). Using associativity, multiply the first two factors first.Step $\displaystyle 1$: \[\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]=\left[\begin{array}{lll}x-0-2 & 0-10-0 & 2x-5-3\end{array}\right]=\left[\begin{array}{lll}x-2 & -10 & 2x-8\end{array}\right].\]Step $\displaystyle 2$ — this is where the \(\displaystyle x\) in the row and the \(\displaystyle x\) in the column meet, producing the quadratic: \[\left[\begin{array}{lll}x-2 & -10 & 2x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=\left[(x-2)x+(-10)(4)+(2x-8)(1)\right].\]Simplify the single entry: \[x^{2}-2x-40+2x-8=x^{2}-48.\] The \(\displaystyle -2x\) and \(\displaystyle +2x\) cancel, so the equation is purely quadratic.Set it equal to \(\displaystyle 0\): \[x^{2}-48=0\ \Longrightarrow\ x^{2}=48\ \Longrightarrow\ x=\pm\sqrt{48}=\pm 4\sqrt{3}.\]Hence \(\displaystyle x=4\sqrt{3}\) or \(\displaystyle x=-4\sqrt{3}\).

First fix the order of \(\displaystyle \mathrm{X}\). If \(\displaystyle \mathrm{X}\) is \(\displaystyle m\times n\), then for \(\displaystyle \mathrm{X}\) times a \(\displaystyle 2\times 3\) matrix to be defined we need \(\displaystyle n=2\), and the product is then \(\displaystyle m\times 3\); since the right-hand side is \(\displaystyle 2\times 3\), \(\displaystyle m=2\). So \(\displaystyle \mathrm{X}\) is \(\displaystyle 2\times 2\). Put \[\mathrm{X}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right].\]Multiply out: \[\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d\end{array}\right]=\left[\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right].\]Equality of matrices means equality of corresponding elements. The two rows give two independent systems.Row $\displaystyle 1$: \[a+4b=-7,\qquad 2a+5b=-8.\] From the first, \(\displaystyle a=-7-4b\). Substituting into the second: \[2(-7-4b)+5b=-8\ \Rightarrow\ -14-3b=-8\ \Rightarrow\ b=-2,\qquad a=-7-4(-2)=1.\] The third equation must be checked, not assumed: \(\displaystyle 3a+6b=3(1)+6(-2)=-9\) \(\displaystyle \checkmark\).Row $\displaystyle 2$: \[c+4d=2,\qquad 2c+5d=4.\] From the first, \(\displaystyle c=2-4d\); substituting, \(\displaystyle 2(2-4d)+5d=4\Rightarrow 4-3d=4\Rightarrow d=0\), so \(\displaystyle c=2\). Check: \(\displaystyle 3c+6d=6\) \(\displaystyle \checkmark\).Hence \[\mathrm{X}=\left[\begin{array}{rr}1 & -2 \\ 2 & 0\end{array}\right].\]

Take \(\displaystyle \mathrm{A}=\left[\begin{array}{rr}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]\) and square it: \[\mathrm{A}^{2}=\left[\begin{array}{rr}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]\left[\begin{array}{rr}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]=\left[\begin{array}{cc}\alpha^{2}+\beta\gamma & \alpha\beta-\beta\alpha \\ \gamma\alpha-\alpha\gamma & \gamma\beta+\alpha^{2}\end{array}\right]=\left[\begin{array}{cc}\alpha^{2}+\beta\gamma & 0 \\ 0 & \alpha^{2}+\beta\gamma\end{array}\right].\] The off-diagonal entries vanish automatically because the entries are scalars and commute.Set \(\displaystyle \mathrm{A}^{2}=\mathrm{I}\) and compare corresponding elements: \[\alpha^{2}+\beta\gamma=1\ \Longrightarrow\ 1-\alpha^{2}-\beta\gamma=0.\]Hence the correct option is (C) \(\displaystyle 1-\alpha^{2}-\beta\gamma=0\).

Apply both definitions at once. \(\displaystyle \mathrm{A}\) symmetric gives \(\displaystyle \mathrm{A}^{\prime}=\mathrm{A}\); \(\displaystyle \mathrm{A}\) skew symmetric gives \(\displaystyle \mathrm{A}^{\prime}=-\mathrm{A}\). Therefore \[\mathrm{A}=\mathrm{A}^{\prime}=-\mathrm{A}\ \Longrightarrow\ 2\mathrm{A}=\mathrm{O}\ \Longrightarrow\ \mathrm{A}=\mathrm{O}.\] (Equivalently, elementwise: \(\displaystyle a_{ij}=a_{ji}\) and \(\displaystyle a_{ij}=-a_{ji}\) give \(\displaystyle 2a_{ij}=0\), so \(\displaystyle a_{ij}=0\) for every \(\displaystyle i,j\).)So \(\displaystyle \mathrm{A}\) must be the zero matrix. Options (A) and (C) are true of the zero square matrix but are weaker statements that do not pin \(\displaystyle \mathrm{A}\) down; the conclusion forced by the hypothesis is that \(\displaystyle \mathrm{A}\) is a zero matrix.Hence the correct option is (B) \(\displaystyle \mathrm{A}\) is a zero matrix.

Since \(\displaystyle \mathrm{I}\) commutes with every matrix (\(\displaystyle \mathrm{IA}=\mathrm{AI}=\mathrm{A}\)), the binomial expansion is valid for \(\displaystyle \mathrm{I}+\mathrm{A}\): \[(\mathrm{I}+\mathrm{A})^{3}=\mathrm{I}^{3}+3\mathrm{I}^{2}\mathrm{A}+3\mathrm{IA}^{2}+\mathrm{A}^{3}=\mathrm{I}+3\mathrm{A}+3\mathrm{A}^{2}+\mathrm{A}^{3}.\]Now use the given idempotency \(\displaystyle \mathrm{A}^{2}=\mathrm{A}\), and reduce the cube through it — this is the step to get right: \[\mathrm{A}^{3}=\mathrm{A}^{2}\cdot\mathrm{A}=\mathrm{A}\cdot\mathrm{A}=\mathrm{A}^{2}=\mathrm{A}.\]Substituting \(\displaystyle \mathrm{A}^{2}=\mathrm{A}\) and \(\displaystyle \mathrm{A}^{3}=\mathrm{A}\): \[(\mathrm{I}+\mathrm{A})^{3}=\mathrm{I}+3\mathrm{A}+3\mathrm{A}+\mathrm{A}=\mathrm{I}+7\mathrm{A}.\]Therefore \[(\mathrm{I}+\mathrm{A})^{3}-7\mathrm{A}=\mathrm{I}+7\mathrm{A}-7\mathrm{A}=\mathrm{I}.\]Hence the correct option is (C) \(\displaystyle \mathrm{I}\).