NCERT Solutions · Class 12 Mathematics Matrices

Both column matrices are of order \(\displaystyle 2\times1\). Multiply each by its scalar and add: \[x\left[\begin{array}{r}2 \\ 3\end{array}\right]+y\left[\begin{array}{r}-1 \\ 1\end{array}\right]=\left[\begin{array}{r}2x \\ 3x\end{array}\right]+\left[\begin{array}{r}-y \\ y\end{array}\right]=\left[\begin{array}{r}2x-y \\ 3x+y\end{array}\right]\]Equating corresponding entries with \(\displaystyle \left[\begin{array}{r}10 \\ 5\end{array}\right]\) turns the matrix equation into a pair of linear equations: \[2x-y=10\qquad\text{and}\qquad 3x+y=5\]Adding them eliminates \(\displaystyle y\): \[5x=15\ \Rightarrow\ x=3\] Substituting in \(\displaystyle 2x-y=10\): \[6-y=10\ \Rightarrow\ y=-4\] Check in the second equation: \(\displaystyle 3(3)+(-4)=5\).Final answer: \(\displaystyle x=3,\ y=-4\).

Compute both sides and equate corresponding entries; each position gives one equation.Left side: \(\displaystyle 3\left[\begin{array}{rr}x & y \\ z & w\end{array}\right]=\left[\begin{array}{rr}3x & 3y \\ 3z & 3w\end{array}\right]\).Right side: \(\displaystyle \left[\begin{array}{rr}x & 6 \\ -1 & 2w\end{array}\right]+\left[\begin{array}{rr}4 & x+y \\ z+w & 3\end{array}\right]=\left[\begin{array}{rr}x+4 & 6+x+y \\ -1+z+w & 2w+3\end{array}\right]\).The four equations are \[3x=x+4,\qquad 3y=6+x+y,\qquad 3z=-1+z+w,\qquad 3w=2w+3\]Solve them in an order that keeps each one in a single unknown. From the first, \[2x=4\ \Rightarrow\ x=2\] From the fourth, \[3w-2w=3\ \Rightarrow\ w=3\] Now the second, using \(\displaystyle x=2\): \[2y=6+x=8\ \Rightarrow\ y=4\] And the third, using \(\displaystyle w=3\): \[2z=-1+w=2\ \Rightarrow\ z=1\]Final answer: \(\displaystyle x=2,\ y=4,\ z=1,\ w=3\).

Multiply \(\displaystyle \mathrm{F}(x)\) by \(\displaystyle \mathrm{F}(y)\) using the row-by-column rule, then recognise the compound-angle formulas \[\cos(x+y)=\cos x\cos y-\sin x\sin y,\qquad \sin(x+y)=\sin x\cos y+\cos x\sin y.\]\[\mathrm{F}(x)\mathrm{F}(y)=\left[\begin{array}{rrr}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{rrr}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]\]Entry by entry: \[(1,1):\ \cos x\cos y-\sin x\sin y=\cos(x+y)\] \[(1,2):\ -\cos x\sin y-\sin x\cos y=-(\sin x\cos y+\cos x\sin y)=-\sin(x+y)\] \[(2,1):\ \sin x\cos y+\cos x\sin y=\sin(x+y)\] \[(2,2):\ -\sin x\sin y+\cos x\cos y=\cos(x+y)\] The third row and third column contain only the products of the zeros with the \(\displaystyle 1\)'s, giving \(\displaystyle 0,0,0\) and \(\displaystyle 0,0,1\). Hence \[\mathrm{F}(x)\mathrm{F}(y)=\left[\begin{array}{rrr}\cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1\end{array}\right]\]This is precisely \(\displaystyle \mathrm{F}\) evaluated at the angle \(\displaystyle x+y\). Therefore \(\displaystyle \mathrm{F}(x)\mathrm{F}(y)=\mathrm{F}(x+y)\), which is what was to be shown. (The sign in the \(\displaystyle (1,2)\) entry is the step to be careful with: both terms there are negative before they are collected.)

In a matrix polynomial, \(\displaystyle \mathrm{A}^{2}\) means the product \(\displaystyle \mathrm{AA}\), and the constant term \(\displaystyle 6\mathrm{I}\) means \(\displaystyle 6\) times the IDENTITY matrix of order \(\displaystyle 3\) (a scalar cannot be added to a matrix).First \(\displaystyle \mathrm{A}^{2}=\mathrm{AA}\). Row \(\displaystyle 1=(2,0,1)\) against the columns of \(\displaystyle \mathrm{A}\) gives \(\displaystyle 4+0+1=5\), \(\displaystyle 0+0-1=-1\), \(\displaystyle 2+0+0=2\); row \(\displaystyle 2=(2,1,3)\) gives \(\displaystyle 4+2+3=9\), \(\displaystyle 0+1-3=-2\), \(\displaystyle 2+3+0=5\); row \(\displaystyle 3=(1,-1,0)\) gives \(\displaystyle 2-2+0=0\), \(\displaystyle 0-1+0=-1\), \(\displaystyle 1-3+0=-2\). So \[\mathrm{A}^{2}=\left[\begin{array}{rrr}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]\]Next \[5\mathrm{A}=\left[\begin{array}{rrr}10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0\end{array}\right],\qquad 6\mathrm{I}=\left[\begin{array}{rrr}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{array}\right]\]Combining entrywise: \[\mathrm{A}^{2}-5\mathrm{A}+6\mathrm{I}=\left[\begin{array}{rrr}5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+6\end{array}\right]=\left[\begin{array}{rrr}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4\end{array}\right]\]Final answer: \(\displaystyle \mathrm{A}^{2}-5\mathrm{A}+6\mathrm{I}=\left[\begin{array}{rrr}1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4\end{array}\right]\) (not the zero matrix, so \(\displaystyle \mathrm{A}\) does not satisfy \(\displaystyle t^{2}-5t+6=0\)).

Build the powers one at a time: \(\displaystyle \mathrm{A}^{2}=\mathrm{AA}\), then \(\displaystyle \mathrm{A}^{3}=\mathrm{A}^{2}\mathrm{A}\). The \(\displaystyle 0\) on the right-hand side is the null matrix of order \(\displaystyle 3\), and \(\displaystyle 2\mathrm{I}\) is twice the identity matrix.\[\mathrm{A}^{2}=\left[\begin{array}{rrr}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{rrr}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]=\left[\begin{array}{rrr}1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9\end{array}\right]=\left[\begin{array}{rrr}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\]\[\mathrm{A}^{3}=\mathrm{A}^{2}\mathrm{A}=\left[\begin{array}{rrr}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{rrr}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]=\left[\begin{array}{rrr}5+0+16 & 0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0 & 16+0+39\end{array}\right]=\left[\begin{array}{rrr}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]\]Now assemble the four terms: \[6\mathrm{A}^{2}=\left[\begin{array}{rrr}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right],\quad 7\mathrm{A}=\left[\begin{array}{rrr}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{array}\right],\quad 2\mathrm{I}=\left[\begin{array}{rrr}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]\]Adding entrywise, \(\displaystyle \mathrm{A}^{3}-6\mathrm{A}^{2}+7\mathrm{A}+2\mathrm{I}\) has entries \[(1,1):\ 21-30+7+2=0,\qquad (1,3):\ 34-48+14+0=0\] \[(2,1):\ 12-12+0+0=0,\qquad (2,2):\ 8-24+14+2=0,\qquad (2,3):\ 23-30+7+0=0\] \[(3,1):\ 34-48+14+0=0,\qquad (3,3):\ 55-78+21+2=0\] and the remaining positions \(\displaystyle (1,2),(3,2)\) are \(\displaystyle 0-0+0+0=0\). Hence \[\mathrm{A}^{3}-6\mathrm{A}^{2}+7\mathrm{A}+2\mathrm{I}=\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=0\] which is the required result.

Compute \(\displaystyle \mathrm{A}^{2}\) and \(\displaystyle k\mathrm{A}-2\mathrm{I}\) separately, then equate corresponding entries; \(\displaystyle k\) must satisfy all four equations simultaneously, or no such \(\displaystyle k\) exists.\[\mathrm{A}^{2}=\left[\begin{array}{rr}3 & -2 \\ 4 & -2\end{array}\right]\left[\begin{array}{rr}3 & -2 \\ 4 & -2\end{array}\right]=\left[\begin{array}{rr}9-8 & -6+4 \\ 12-8 & -8+4\end{array}\right]=\left[\begin{array}{rr}1 & -2 \\ 4 & -4\end{array}\right]\]\[k\mathrm{A}-2\mathrm{I}=\left[\begin{array}{rr}3k & -2k \\ 4k & -2k\end{array}\right]-\left[\begin{array}{rr}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{rr}3k-2 & -2k \\ 4k & -2k-2\end{array}\right]\]Equating entrywise: \[3k-2=1\ \Rightarrow\ k=1;\qquad -2k=-2\ \Rightarrow\ k=1;\qquad 4k=4\ \Rightarrow\ k=1;\qquad -2k-2=-4\ \Rightarrow\ k=1\]All four positions give the same value, so the equation \(\displaystyle \mathrm{A}^{2}=k\mathrm{A}-2\mathrm{I}\) is genuinely satisfied.Final answer: \(\displaystyle k=1\).

Write \(\displaystyle t=\tan\dfrac{\alpha}{2}\) throughout, and use the half-angle (Weierstrass) identities \[\cos\alpha=\frac{1-t^{2}}{1+t^{2}},\qquad \sin\alpha=\frac{2t}{1+t^{2}}.\]The two sides to compare are \[\mathrm{I}+\mathrm{A}=\left[\begin{array}{rr}1 & -t \\ t & 1\end{array}\right],\qquad \mathrm{I}-\mathrm{A}=\left[\begin{array}{rr}1 & t \\ -t & 1\end{array}\right].\]Multiply out the right-hand side (keeping the order of the factors, since matrix multiplication is not commutative): \[(\mathrm{I}-\mathrm{A})\left[\begin{array}{rr}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]=\left[\begin{array}{rr}\cos\alpha+t\sin\alpha & -\sin\alpha+t\cos\alpha \\ -t\cos\alpha+\sin\alpha & t\sin\alpha+\cos\alpha\end{array}\right]\]Now substitute the half-angle expressions. For the \(\displaystyle (1,1)\) entry: \[\cos\alpha+t\sin\alpha=\frac{1-t^{2}}{1+t^{2}}+t\cdot\frac{2t}{1+t^{2}}=\frac{1-t^{2}+2t^{2}}{1+t^{2}}=\frac{1+t^{2}}{1+t^{2}}=1\] The \(\displaystyle (2,2)\) entry is the same expression, so it is also \(\displaystyle 1\). For the \(\displaystyle (1,2)\) entry: \[-\sin\alpha+t\cos\alpha=\frac{-2t+t(1-t^{2})}{1+t^{2}}=\frac{-t-t^{3}}{1+t^{2}}=\frac{-t(1+t^{2})}{1+t^{2}}=-t\] The \(\displaystyle (2,1)\) entry, \(\displaystyle \sin\alpha-t\cos\alpha\), is the negative of that, hence \(\displaystyle +t\).Therefore \[(\mathrm{I}-\mathrm{A})\left[\begin{array}{rr}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{array}\right]=\left[\begin{array}{rr}1 & -t \\ t & 1\end{array}\right]=\mathrm{I}+\mathrm{A},\] which is the required identity. (It holds wherever \(\displaystyle \tan\dfrac{\alpha}{2}\) is defined, i.e. \(\displaystyle \alpha\neq(2n+1)\pi\).)

Convert dozens to units first: \(\displaystyle 1\) dozen \(\displaystyle =12\) books, so the stock is \(\displaystyle 10\times12=120\) chemistry, \(\displaystyle 8\times12=96\) physics and \(\displaystyle 10\times12=120\) economics books.Put the quantities in a \(\displaystyle 1\times3\) row matrix and the corresponding selling prices in a \(\displaystyle 3\times1\) column matrix, keeping the subjects in the same order in both. Their product is a \(\displaystyle 1\times1\) matrix whose single entry is the total receipt: \[\left[\begin{array}{rrr}120 & 96 & 120\end{array}\right]\left[\begin{array}{r}80 \\ 60 \\ 40\end{array}\right]=\left[\begin{array}{r}120(80)+96(60)+120(40)\end{array}\right]=\left[\begin{array}{r}9600+5760+4800\end{array}\right]=\left[\begin{array}{r}20160\end{array}\right]\]Final answer: the bookshop receives ₹$\displaystyle 20,160$ from selling all the books.

Use the two order rules: a product \(\displaystyle \mathrm{MN}\) exists only if the number of columns of \(\displaystyle \mathrm{M}\) equals the number of rows of \(\displaystyle \mathrm{N}\), and a SUM exists only if the two matrices have the same order.Here \(\displaystyle \mathrm{P}\) is \(\displaystyle p\times k\), \(\displaystyle \mathrm{Y}\) is \(\displaystyle 3\times k\) and \(\displaystyle \mathrm{W}\) is \(\displaystyle n\times3\).For \(\displaystyle \mathrm{PY}\) to exist: columns of \(\displaystyle \mathrm{P}\) \(\displaystyle =\) rows of \(\displaystyle \mathrm{Y}\), i.e. \[k=3,\qquad\text{and then }\mathrm{PY}\text{ is }p\times k=p\times3.\]For \(\displaystyle \mathrm{WY}\) to exist: columns of \(\displaystyle \mathrm{W}\) \(\displaystyle =\) rows of \(\displaystyle \mathrm{Y}\), i.e. \(\displaystyle 3=3\), which holds automatically, and \[\mathrm{WY}\text{ is }n\times k=n\times3.\]For the SUM \(\displaystyle \mathrm{PY}+\mathrm{WY}\) the two orders must agree: \[p\times3=n\times3\ \Rightarrow\ p=n.\]So both conditions are needed: \(\displaystyle k=3\) and \(\displaystyle p=n\).Final answer: (A) \(\displaystyle k=3,\ p=n\).

A scalar multiple does not change the order of a matrix, and a difference of two matrices exists only when both have the same order — in which case the result has that same order.\(\displaystyle \mathrm{X}\) is \(\displaystyle 2\times n\), so \(\displaystyle 7\mathrm{X}\) is \(\displaystyle 2\times n\). \(\displaystyle \mathrm{Z}\) is \(\displaystyle 2\times p\), so \(\displaystyle 5\mathrm{Z}\) is \(\displaystyle 2\times p\). With the given condition \(\displaystyle n=p\) the two orders coincide, the subtraction is defined, and \[\text{order of }(7\mathrm{X}-5\mathrm{Z})=2\times n.\]Final answer: (B) \(\displaystyle 2\times n\).