NCERT Solutions: Class 12 Maths Chapter 3 - Matrices

The matrix has 3 rows and 4 columns, so its order is $3\times4$ and it has $3\cdot4=12$ elements. Reading row $i$, column $j$: $a_{13}=19$, $a_{21}=35$, $a_{33}=-5$, $a_{24}=12$ and $a_{23}=\dfrac{5}{2}$.

An $m\times n$ matrix has $mn$ elements. Factor pairs of 24 give the eight listed orders. Since 13 is prime, its only positive factor pairs are $(1,13)$ and $(13,1)$.

The order $m\times n$ must satisfy $mn=18$ or $mn=5$. Listing positive factor pairs gives the stated orders; 5 is prime.

For a $2\times2$ matrix, put $i,j\in\{1,2\}$. Substituting each pair $(1,1),(1,2),(2,1),(2,2)$ in the three formulae gives the three matrices in the answer.

Use $i=1,2,3$ and $j=1,2,3,4$. Substitution in $\dfrac12|-3i+j|$ gives rows $(1,\frac12,0,\frac12)$, $(\frac52,2,\frac32,1)$, $(4,\frac72,3,\frac52)$. Substitution in $2i-j$ gives rows $(1,0,-1,-2)$, $(3,2,1,0)$, $(5,4,3,2)$.

Equate corresponding elements. (i) $y=4$, $z=3$, $x=1$. (ii) $x+y=6$, $z=0$, $xy=8$, so $x,y$ are 2 and 4 in either order. (iii) From $(x+y+z)-(x+z)=4$, get $y=4$; then $y+z=7$ gives $z=3$ and $x+z=5$ gives $x=2$.

Equating corresponding elements gives $a-b=-1$, $2a+c=5$, $2a-b=0$, $3c+d=13$. From the first and third equations, $a=1$ and $b=2$. Then $c=5-2a=3$ and $d=13-3c=4$.

1. i. $m \lt n$
2. ii. $m \gt n$
3. iii. $m=n$
4. iv. None of these

A matrix is square exactly when its number of rows equals its number of columns, so $m=n$.

1. i. $x=-\dfrac{1}{3},\ y=7$
2. ii. Not possible to find
3. iii. $y=7,\ x=-\dfrac{2}{3}$
4. iv. $x=-\dfrac{1}{3},\ y=-\dfrac{2}{3}$

Equality would require $3x+7=0$, so $x=-\dfrac73$, and also $2-3x=4$, so $x=-\dfrac23$. These conditions contradict each other. Hence no such $x,y$ exist.

1. i. $27$
2. ii. $18$
3. iii. $81$
4. iv. $512$

A $3\times3$ matrix has 9 entries. Each entry has 2 choices, 0 or 1, so the total number is $2^9=512$.

Add and subtract corresponding entries for (i), (ii), and (iii). For products, $AB=\begin{bmatrix}2\cdot1+4(-2) & 2\cdot3+4\cdot5 \\ 3\cdot1+2(-2) & 3\cdot3+2\cdot5\end{bmatrix}=\begin{bmatrix}-6&26\\-1&19\end{bmatrix}$ and $BA=\begin{bmatrix}11&10\\11&2\end{bmatrix}$.

Add corresponding entries. In (ii), simplify entries such as $a^2+b^2+2ab=(a+b)^2$ and $a^2+c^2-2ac=(a-c)^2$. In (iv), use $\sin^2x+\cos^2x=1$.

Use row-by-column multiplication. For example, in (iv), the first row gives $(2,3,4)\cdot(1,0,3)=14$, $(2,3,4)\cdot(-3,2,0)=0$, and $(2,3,4)\cdot(5,4,5)=42$. The same rule gives all listed products.

Compute corresponding entries: $A+B$ and $B-C$ are as listed. Then adding $A$ to $B-C$ gives $\begin{bmatrix}0&0&-3\\9&-1&5\\2&1&1\end{bmatrix}$, and subtracting $C$ from $A+B$ gives the same matrix, so the identity is verified.

$3A=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}$ and $5B=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}$. Hence $3A-5B=O$.

The $(1,1)$ and $(2,2)$ entries become $\cos^2\theta+\sin^2\theta=1$. The off-diagonal entries become $\cos\theta\sin\theta-\sin\theta\cos\theta=0$ and $-\sin\theta\cos\theta+\sin\theta\cos\theta=0$.

(i) Add the two equations: $2X=\begin{bmatrix}10&0\\2&8\end{bmatrix}$, then subtract to get $2Y=\begin{bmatrix}4&0\\2&2\end{bmatrix}$. (ii) From $2X+3Y=A$ and $3X+2Y=B$, $5X=3B-2A$ and $5Y=3A-2B$, giving the listed matrices.

$2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}$, so $X=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}$.

The left side is $\begin{bmatrix}2+y&6\\1&2x+2\end{bmatrix}$. Equating entries gives $2+y=5$ and $2x+2=8$, so $y=3$ and $x=3$.

$2\begin{bmatrix}x&z\\y&t\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix}-\begin{bmatrix}3&-3\\0&6\end{bmatrix}=\begin{bmatrix}6&18\\12&12\end{bmatrix}$. Dividing by 2 gives $\begin{bmatrix}x&z\\y&t\end{bmatrix}=\begin{bmatrix}3&9\\6&6\end{bmatrix}$.

Equating entries gives $2x-y=10$ and $3x+y=5$. Adding gives $5x=15$, so $x=3$; then $y=5-3x=-4$.

Equating entries: $3x=x+4$, so $x=2$; $3y=6+x+y$, so $2y=8$ and $y=4$; $3w=2w+3$, so $w=3$; $3z=-1+z+w=z+2$, so $z=1$.

Multiplying the two matrices, the top-left entry is $\cos x\cos y-\sin x\sin y=\cos(x+y)$, the top-middle entry is $-(\cos x\sin y+\sin x\cos y)=-\sin(x+y)$, the second-left entry is $\sin(x+y)$, and the second-middle entry is $\cos(x+y)$. The third row and column remain $(0,0,1)$, so the product is $F(x+y)$.

Compute each side by row-by-column multiplication. Since the corresponding products differ in each part, matrix multiplication is not commutative for these examples.

First $A^2=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}$. Therefore $A^2-5A+6I=\begin{bmatrix}5&-1&2\\9&-2&5\\0&-1&-2\end{bmatrix}-\begin{bmatrix}10&0&5\\10&5&15\\5&-5&0\end{bmatrix}+\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=\begin{bmatrix}1&-1&-3\\-1&-1&-10\\-5&4&4\end{bmatrix}$.

Compute $A^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}$ and $A^3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}$. Substituting, $A^3-6A^2+7A+2I$ has entries $21-30+7+2=0$, $34-48+14=0$, and similarly all other entries reduce to 0. Hence the expression is the zero matrix.

$A^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$. Also $kA-2I=\begin{bmatrix}3k-2&-2k\\4k&-2k-2\end{bmatrix}$. Equating entries gives $k=1$, and then every entry matches.

Let $t=\tan\dfrac{\alpha}{2}$. Then $\cos\alpha=\dfrac{1-t^2}{1+t^2}$ and $\sin\alpha=\dfrac{2t}{1+t^2}$. Multiplying $(I-A)$ by the given rotation matrix gives $\begin{bmatrix}1&t\\-t&1\end{bmatrix}\begin{bmatrix}\frac{1-t^2}{1+t^2}&-\frac{2t}{1+t^2}\\\frac{2t}{1+t^2}&\frac{1-t^2}{1+t^2}\end{bmatrix}=\begin{bmatrix}1&-t\\t&1\end{bmatrix}=I+A$.

Let the investments be $x$ and $y$. Then $x+y=30000$. For ₹1800 interest, $0.05x+0.07y=1800$; solving gives $x=y=15000$. For ₹2000 interest, $0.05x+0.07y=2000$ with $x+y=30000$; solving gives $y=25000$ and $x=5000$.

The quantity row is $[120\ 96\ 120]$ and the price column is $\begin{bmatrix}80\\60\\40\end{bmatrix}$. Their product is $120\cdot80+96\cdot60+120\cdot40=9600+5760+4800=20160$.

1. i. $k=3,\ p=n$
2. ii. $k$ is arbitrary, $p=2$
3. iii. $p$ is arbitrary, $k=3$
4. iv. $k=2,\ p=3$

$PY$ has order $(p\times k)(3\times k)$, so it is defined only if $k=3$, and then its order is $p\times k$. $WY$ has order $(n\times3)(3\times k)=n\times k$. To add $PY$ and $WY$, their row counts must match, so $p=n$.

1. i. $p\times2$
2. ii. $2\times n$
3. iii. $n\times3$
4. iv. $p\times n$

$X$ has order $2\times n$ and $Z$ has order $2\times p$. If $n=p$, then both have order $2\times n$, so $7X-5Z$ also has order $2\times n$.

Transpose is obtained by interchanging rows and columns. Applying this to each matrix gives the listed matrices.

Compute $A+B=\begin{bmatrix}-5&3&-2\\6&9&9\\-1&4&2\end{bmatrix}$ and transpose it to get the first matrix in the answer. Similarly, $A-B=\begin{bmatrix}3&1&8\\4&5&9\\-3&-2&0\end{bmatrix}$, whose transpose is the second matrix. These are exactly $A'+B'$ and $A'-B'$.

Since $A'=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}$, $A=\begin{bmatrix}3&-1&0\\4&2&1\end{bmatrix}$. Now compute $A+B$ and $A-B$, then transpose. The results match $A'+B'$ and $A'-B'$ as listed.

$A=\begin{bmatrix}-2&1\\3&2\end{bmatrix}$ and $2B=\begin{bmatrix}-2&0\\2&4\end{bmatrix}$. Thus $A+2B=\begin{bmatrix}-4&1\\5&6\end{bmatrix}$, so $(A+2B)'=\begin{bmatrix}-4&5\\1&6\end{bmatrix}$.

In (i), $AB=\begin{bmatrix}-1&2&1\\4&-8&-4\\-3&6&3\end{bmatrix}$, so transposing gives the stated matrix, equal to $B'A'$. In (ii), $AB=\begin{bmatrix}0&0&0\\1&5&7\\2&10&14\end{bmatrix}$, whose transpose equals $B'A'$.

(i) $A'=\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}$. Multiplication gives diagonal entries $\cos^2\alpha+\sin^2\alpha=1$ and off-diagonal entries 0. (ii) The same calculation gives diagonal entries $\sin^2\alpha+\cos^2\alpha=1$ and off-diagonal entries 0. Hence $A'A=I$ in both cases.

(i) The entries mirror across the main diagonal: $a_{12}=a_{21}=-1$, $a_{13}=a_{31}=5$, $a_{23}=a_{32}=1$, so $A'=A$. (ii) The diagonal entries are 0 and opposite entries satisfy $a_{ij}=-a_{ji}$, so $A'=-A$.

$A'=\begin{bmatrix}1&6\\5&7\end{bmatrix}$. Thus $A+A'=\begin{bmatrix}2&11\\11&14\end{bmatrix}$, equal to its transpose. Also $A-A'=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, whose transpose is its negative.

Here $A'=-A$, since $A$ is skew symmetric. Therefore $A+A'=O$ and $A-A'=2A$. Dividing by 2 gives $O$ and $A$, respectively.

Use $A=\dfrac12(A+A')+\dfrac12(A-A')$. The first term is symmetric and the second is skew symmetric. Applying this formula to each matrix gives the four decompositions in the answer.

1. i. Skew symmetric matrix
2. ii. Symmetric matrix
3. iii. Zero matrix
4. iv. Identity matrix

Since $A'=A$ and $B'=B$, $(AB-BA)'=(AB)'-(BA)'=B'A'-A'B'=BA-AB=-(AB-BA)$. Hence it is skew symmetric.

1. i. $\dfrac{\pi}{6}$
2. ii. $\dfrac{\pi}{3}$
3. iii. $\pi$
4. iv. $\dfrac{3\pi}{2}$

$A'=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$, so $A+A'=\begin{bmatrix}2\cos\alpha&0\\0&2\cos\alpha\end{bmatrix}$. For this to equal $I$, $2\cos\alpha=1$, hence $\cos\alpha=\dfrac12$ and the given option is $\alpha=\dfrac{\pi}{3}$.

1. i. $AB=BA$
2. ii. $AB=BA=0$
3. iii. $AB=0,\ BA=I$
4. iv. $AB=BA=I$

By definition, $B$ is the inverse of $A$ exactly when both products exist and $AB=BA=I$.

Since $A$ and $B$ are symmetric, $A'=A$ and $B'=B$. Therefore $(AB-BA)'=(AB)'-(BA)'=B'A'-A'B'=BA-AB=-(AB-BA)$. Hence $AB-BA$ is skew symmetric.

Transpose the expression: $(B'AB)'=B'A'B$. If $A$ is symmetric, then $A'=A$ and $(B'AB)'=B'AB$, so it is symmetric. If $A$ is skew symmetric, then $A'=-A$ and $(B'AB)'=-B'AB$, so it is skew symmetric.

The columns of $A$ are $C_1=(0,x,x)'$, $C_2=(2y,y,-y)'$, $C_3=(z,-z,z)'$. Since $A'A=I$, each column has length 1 and different columns are orthogonal. The orthogonality conditions are automatically satisfied, and the length equations are $2x^2=1$, $6y^2=1$, $3z^2=1$. Thus $x=\pm\dfrac1{\sqrt2}$, $y=\pm\dfrac1{\sqrt6}$ and $z=\pm\dfrac1{\sqrt3}$.

First $[1\ 2\ 1]\begin{bmatrix}1&2&0\\2&0&1\\1&0&2\end{bmatrix}=[6\ 2\ 4]$. Then $[6\ 2\ 4]\begin{bmatrix}0\\2\\x\end{bmatrix}=4+4x$. Setting this equal to 0 gives $x=-1$.

$A^2=\begin{bmatrix}8&5\\-5&3\end{bmatrix}$. Hence $A^2-5A+7I=\begin{bmatrix}8&5\\-5&3\end{bmatrix}-\begin{bmatrix}15&5\\-5&10\end{bmatrix}+\begin{bmatrix}7&0\\0&7\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$.

First $\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}x\\4\\1\end{bmatrix}=\begin{bmatrix}x+2\\9\\2x+3\end{bmatrix}$. Then $[x\ -5\ -1]\begin{bmatrix}x+2\\9\\2x+3\end{bmatrix}=x(x+2)-45-(2x+3)=x^2-48$. Thus $x^2=48$, so $x=\pm4\sqrt3$.

Sales matrix $S=\begin{bmatrix}10000&2000&18000\\6000&20000&8000\end{bmatrix}$. For prices $p=\begin{bmatrix}2.50\\1.50\\1.00\end{bmatrix}$, $Sp=\begin{bmatrix}46000\\53000\end{bmatrix}$. For costs $c=\begin{bmatrix}2.00\\1.00\\0.50\end{bmatrix}$, $Sc=\begin{bmatrix}31000\\36000\end{bmatrix}$. Gross profit is $Sp-Sc=\begin{bmatrix}15000\\17000\end{bmatrix}$.

Let $X=\begin{bmatrix}a&b\\c&d\end{bmatrix}$. Equating the first row of the product gives $a+4b=-7$ and $2a+5b=-8$, so $a=1,b=-2$. Equating the second row gives $c+4d=2$ and $2c+5d=4$, so $c=2,d=0$. Therefore $X=\begin{bmatrix}1&-2\\2&0\end{bmatrix}$.

1. i. $1+\alpha^2+\beta\gamma=0$
2. ii. $1-\alpha^2+\beta\gamma=0$
3. iii. $1-\alpha^2-\beta\gamma=0$
4. iv. $1+\alpha^2-\beta\gamma=0$

$A^2=\begin{bmatrix}\alpha^2+\beta\gamma&0\\0&\alpha^2+\beta\gamma\end{bmatrix}$. Since $A^2=I$, $\alpha^2+\beta\gamma=1$, i.e. $1-\alpha^2-\beta\gamma=0$.

1. i. $A$ is a diagonal matrix
2. ii. $A$ is a zero matrix
3. iii. $A$ is a square matrix
4. iv. None of these

If $A$ is symmetric, $A'=A$. If it is also skew symmetric, $A'=-A$. Hence $A=-A$, so $2A=O$ and therefore $A=O$.

1. i. $A$
2. ii. $I-A$
3. iii. $I$
4. iv. $3A$

Since $A^2=A$, also $A^3=A^2A=A^2=A$. Now $(I+A)^3=I+3A+3A^2+A^3=I+3A+3A+A=I+7A$. Therefore $(I+A)^3-7A=I$.