Compute the following: (i) [ ]+[ ] (ii) [ ]+[ ] (iii) [ ]+[ ] (iv) [ ]+[ ]

Two matrices of the SAME order are added entrywise: if A =[a ij] and B =[b ij] are both m× n, then A + B =[a ij+b ij]. In every part below the two matrices have equal order, so each sum is defined. (i) Add corresponding entries: [ ]+[ ]=[ ]=[ ] (ii) Add entrywise first: [ ] Each entry is now a perfect square. Watch the signs in the second row: the added matrix contributes -2ac and -2ab there, so those entries are squares of DIFFERENCES: =[ ] (iii) Add the nine corresponding entries: [ ]+[ ]=[ ] (iv) Every entry of the sum pairs cos^2x with sin^2x, and sin^2x+ cos^2x=1: [ ]=[ ] Final answers: (i) [ ]; (ii) [ ]; (iii) [ ]; (iv) [ ].

Compute the indicated products. (i) [ ][ ] (ii) [ ][ ] (iii) [ ][ ] (iv) [ ][ ] (v) [ ][ ] (vi) [ ][ ]

The product AB exists only when the number of COLUMNS of A equals the number of ROWS of B. If A is m× n and B is n× p, then AB is m× p and ( AB ) ij=Σ k=1^na ikb kj, i.e. the i -th row of A is paired term-by-term with the j -th column of B. Check the order first in each part. (i) (2×2)(2×2)→ 2×2: [ ][ ]=[ ]=[ ] (ii) (3×1)(1×3)→ 3×3. The order of the factors is what makes this a 3×3 matrix rather than a single number: [ ][ ]=[ ] (iii) (2×2)(2×3)→ 2×3. For instance the (1,1) entry is 1(1)+(-2)(2)=-3 and the (1,2) entry is 1(2)+(-2)(3)=-4: [ ][ ]=[ ] (iv) (3×3)(3×3)→ 3×3. Row 1=(2,3,4) against the three columns gives 2(1)+3(0)+4(3)=14, 2(-3)+3(2)+4(0)=0, 2(5)+3(4)+4(5)=42; row 2=(3,4,5) gives 18,-1,56; row 3=(4,5,6) gives 22,-2,70: [ ][ ]=[ ] (v) (3×2)(2×3)→ 3×3. Row 3=(-1,1) gives (-1)(1)+1(-1)=-2, (-1)(0)+1(2)=2, (-1)(1)+1(1)=0: [ ][ ]=[ ] (vi) (2×3)(3×2)→ 2×2. Row 1=(3,-1,3) gives 3(2)+(-1)(1)+3(3)=14 and 3(-3)+(-1)(0)+3(1)=-6; row 2=(-1,0,2) gives (-1)(2)+0(1)+2(3)=4 and (-1)(-3)+0(0)+2(1)=5: [ ][ ]=[ ]

If A =[ ], B =[ ] and C =[ ], then compute ( A + B ) and ( B - C ). Also, verify that A +( B - C )=( A + B )- C.

A + B =, B - C =. All three matrices are of order 3×3, so every sum and difference below is defined, and both operations are carried out entrywise. A + B =[ ]=[ ] B - C =[ ]=[ ] Verification. Left side: A +( B - C )=[ ]+[ ]=[ ] Right side, using the A + B already found: ( A + B )- C =[ ]-[ ]=[ ] The two matrices are identical entry for entry, so A +( B - C )=( A + B )- C. This is exactly the associativity of matrix addition applied to A + B +(- C ).

If A =[ ] and B =[ ], then compute 3 A-5 B.

[ ]. Scalar multiplication multiplies EVERY entry: k A =[ka ij]. Compute the two scalar multiples separately before subtracting. 3 A =3[ ]=[ ] 5 B =5[ ]=[ ] The denominators are cleared by the very scalars chosen, and the two results are the same matrix. Hence, subtracting entrywise, 3 A -5 B =[ ]= O Final answer: 3 A -5 B is the null matrix of order 3×3.

Simplify cos θ[ ]+ sin θ[ ]

[ ]. Multiply each matrix by its scalar first, then add entrywise. cos θ[ ]=[ ] sin θ[ ]=[ ] Adding, the diagonal entries become cos^2θ+ sin^2θ=1, and the off-diagonal entries cancel in pairs, sin θ cos θ- sin θ cos θ=0: [ ]=[ ] Final answer: the expression simplifies to the identity matrix I of order 2, for every value of θ.

Find X and Y, if (i) X + Y =[ ] and X - Y =[ ] (ii) 2 X +3 Y =[ ] and 3 X +2 Y =[ ]

(i) X =, Y = (ii) X =, Y =. Matrix equations in X and Y are solved exactly like simultaneous linear equations: add or subtract suitable multiples to eliminate one unknown matrix, then divide by the scalar (multiply by its reciprocal). (i) Adding the two given equations eliminates Y: ( X + Y )+( X - Y )=2 X =[ ]+[ ]=[ ] ⇒ X =1/2[ ]=[ ] Subtracting eliminates X: 2 Y =[ ]-[ ]=[ ] ⇒ Y =[ ] Check: X + Y =[ ], as required. (ii) To eliminate X, take 3(2 X +3 Y )-2(3 X +2 Y )=5 Y: 5 Y =3[ ]-2[ ]=[ ]-[ ]=[ ] ⇒ Y =1/5[ ]=[ ] To eliminate Y, take 3(3 X +2 Y )-2(2 X +3 Y )=5 X: 5 X =3[ ]-2[ ]=[ ]-[ ]=[ ] ⇒ X =1/5[ ]=[ ] Check: 2 X +3 Y =[ ]=[ ], as required.

Find X, if Y =[ ] and 2 X + Y =[ ]

X =[ ]. Treat the matrix equation like a scalar one: isolate 2 X by subtracting Y from both sides, then multiply by the scalar 1/2 (which halves every entry). 2 X =[ ]- Y =[ ]-[ ]=[ ]=[ ] X =1/2[ ]=[ ] Check: 2 X + Y =[ ]+[ ]=[ ]. Final answer: X =[ ].

Find x and y, if 2[ ]+[ ]=[ ]

x=3, y=3. Simplify the left side, then use the definition of equality of matrices: two matrices are equal only if they have the same order AND every pair of corresponding entries is equal. 2[ ]+[ ]=[ ]+[ ]=[ ] Equating this with [ ] position by position: 2+y=5, 6=6, 1=1, 2x+2=8 The (1,2) and (2,1) entries are automatically consistent; the other two give y=3, 2x=6 ⇒ x=3 Final answer: x=3, y=3.

Solve the equation for x, y, z and t, if 2[ ]+3[ ]=3[ ]

Carry out the scalar multiplications and the addition on the left, then equate corresponding entries. 2[ ]+3[ ]=[ ]+[ ]=[ ] 3[ ]=[ ] Note the placement of the unknowns in the given matrix: z sits in the (1,2) position and y in the (2,1) position, so y must be matched with 12 and z with 15, not the other way round. Equating entrywise: 2x+3=9 ⇒ x=3 2z-3=15 ⇒ 2z=18 ⇒ z=9 2y=12 ⇒ y=6 2t+6=18 ⇒ 2t=12 ⇒ t=6 Final answer: x=3, y=6, z=9, t=6.

EXERCISE 3.2 1–10 — NCERT Solutions for Class 12 Mathematics Chapter 3

EXERCISE 3.2 1–10 (part 2 of 6)

Exercise 1
Let $\displaystyle \mathrm{A}=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], \mathrm{B}=\left[\begin{array}{rr}1 & 3 \\ -2 & 5\end{array}\right], \mathrm{C}=\left[\begin{array}{rr}-2 & 5 \\ 3 & 4\end{array}\right]$
NCERT’s answer
(i)
$\displaystyle \mathrm{A}+\mathrm{B}=\begin{array}{rr}3 & 7 \\ 1 & 7\end{array}$ (ii) $\displaystyle \mathrm{A}-\mathrm{B}=\begin{array}{ll}1 & 1 \\ 5 & -3\end{array}$ (iii) $\displaystyle 3 \mathrm{~A}-\mathrm{C}=\begin{array}{ll}8 & 7 \\ 6 & 2\end{array}$ (iv) $\displaystyle \mathrm{AB}=\left[\begin{array}{ll}-6 & 26 \\ -1 & 19\end{array}\right]$ (v) $\displaystyle \mathrm{BA}=\begin{array}{lc}11 & 10 \\ 11 & 2\end{array}$
Working being prepared
Find each of the following:
Exercise 2
Compute the following:
(i)
$\displaystyle \left[\begin{array}{cc}a & b \\ -b & a\end{array}\right]+\left[\begin{array}{cc}a & b \\ b & a\end{array}\right]$
(ii)
$\displaystyle \left[\begin{array}{ll}a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2}\end{array}\right]+\left[\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right]$
(iii)
$\displaystyle \left[\begin{array}{rrr}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{array}\right]+\left[\begin{array}{ccc}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{array}\right]$
(iv)
$\displaystyle \left[\begin{array}{cc}\cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x\end{array}\right]+\left[\begin{array}{cc}\sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x\end{array}\right]$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
Two matrices of the SAME order are added entrywise: if $\displaystyle \mathrm{A}=[a_{ij}]$ and $\displaystyle \mathrm{B}=[b_{ij}]$ are both $\displaystyle m\times n$, then $\displaystyle \mathrm{A}+\mathrm{B}=[a_{ij}+b_{ij}]$. In every part below the two matrices have equal order, so each sum is defined.
(i)
Add corresponding entries:
\[\left[\begin{array}{rr}a & b \\ -b & a\end{array}\right]+\left[\begin{array}{rr}a & b \\ b & a\end{array}\right]=\left[\begin{array}{rr}a+a & b+b \\ -b+b & a+a\end{array}\right]=\left[\begin{array}{rr}2a & 2b \\ 0 & 2a\end{array}\right]\]
(ii)
Add entrywise first:
\[\left[\begin{array}{rr}a^{2}+b^{2}+2ab & b^{2}+c^{2}+2bc \\ a^{2}+c^{2}-2ac & a^{2}+b^{2}-2ab\end{array}\right]\]
Each entry is now a perfect square. Watch the signs in the second row: the added matrix contributes $\displaystyle -2ac$ and $\displaystyle -2ab$ there, so those entries are squares of DIFFERENCES:
\[=\left[\begin{array}{rr}(a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2}\end{array}\right]\]
(iii)
Add the nine corresponding entries:
\[\left[\begin{array}{rrr}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{array}\right]+\left[\begin{array}{rrr}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{array}\right]=\left[\begin{array}{rrr}11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9\end{array}\right]\]
(iv)
Every entry of the sum pairs $\displaystyle \cos^{2}x$ with $\displaystyle \sin^{2}x$, and $\displaystyle \sin^{2}x+\cos^{2}x=1$:
\[\left[\begin{array}{rr}\cos^{2}x+\sin^{2}x & \sin^{2}x+\cos^{2}x \\ \sin^{2}x+\cos^{2}x & \cos^{2}x+\sin^{2}x\end{array}\right]=\left[\begin{array}{rr}1 & 1 \\ 1 & 1\end{array}\right]\]
Final answers: (i) $\displaystyle \left[\begin{array}{rr}2a & 2b \\ 0 & 2a\end{array}\right]$; (ii) $\displaystyle \left[\begin{array}{rr}(a+b)^{2} & (b+c)^{2} \\ (a-c)^{2} & (a-b)^{2}\end{array}\right]$; (iii) $\displaystyle \left[\begin{array}{rrr}11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9\end{array}\right]$; (iv) $\displaystyle \left[\begin{array}{rr}1 & 1 \\ 1 & 1\end{array}\right]$.
Exercise 3
Compute the indicated products.
(i)
$\displaystyle \left[\begin{array}{rr}a & b \\ -b & a\end{array}\right]\left[\begin{array}{rr}a & -b \\ b & a\end{array}\right]$
(ii)
$\displaystyle \left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}2 & 3 & 4\end{array}\right]$
(iii)
$\displaystyle \left[\begin{array}{rr}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$
(iv)
$\displaystyle \left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{rrr}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$
(v)
$\displaystyle \left[\begin{array}{rr}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]$
(vi)
$\displaystyle \left[\begin{array}{rrr}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{rr}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
The product $\displaystyle \mathrm{AB}$ exists only when the number of COLUMNS of $\displaystyle \mathrm{A}$ equals the number of ROWS of $\displaystyle \mathrm{B}$. If $\displaystyle \mathrm{A}$ is $\displaystyle m\times n$ and $\displaystyle \mathrm{B}$ is $\displaystyle n\times p$, then $\displaystyle \mathrm{AB}$ is $\displaystyle m\times p$ and
\[(\mathrm{AB})_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj},\]
i.e. the $\displaystyle i$-th row of $\displaystyle \mathrm{A}$ is paired term-by-term with the $\displaystyle j$-th column of $\displaystyle \mathrm{B}$. Check the order first in each part.
(i)
$\displaystyle (2\times2)(2\times2)\to 2\times2$:
\[\left[\begin{array}{rr}a & b \\ -b & a\end{array}\right]\left[\begin{array}{rr}a & -b \\ b & a\end{array}\right]=\left[\begin{array}{rr}a\cdot a+b\cdot b & a(-b)+b\cdot a \\ (-b)a+a\cdot b & (-b)(-b)+a\cdot a\end{array}\right]=\left[\begin{array}{rr}a^{2}+b^{2} & 0 \\ 0 & a^{2}+b^{2}\end{array}\right]\]
(ii)
$\displaystyle (3\times1)(1\times3)\to 3\times3$. The order of the factors is what makes this a $\displaystyle 3\times3$ matrix rather than a single number:
\[\left[\begin{array}{r}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{rrr}2 & 3 & 4\end{array}\right]=\left[\begin{array}{rrr}2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12\end{array}\right]\]
(iii)
$\displaystyle (2\times2)(2\times3)\to 2\times3$. For instance the $\displaystyle (1,1)$ entry is $\displaystyle 1(1)+(-2)(2)=-3$ and the $\displaystyle (1,2)$ entry is $\displaystyle 1(2)+(-2)(3)=-4$:
\[\left[\begin{array}{rr}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{rrr}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]=\left[\begin{array}{rrr}-3 & -4 & 1 \\ 8 & 13 & 9\end{array}\right]\]
(iv)
$\displaystyle (3\times3)(3\times3)\to 3\times3$. Row $\displaystyle 1=(2,3,4)$ against the three columns gives $\displaystyle 2(1)+3(0)+4(3)=14$, $\displaystyle 2(-3)+3(2)+4(0)=0$, $\displaystyle 2(5)+3(4)+4(5)=42$; row $\displaystyle 2=(3,4,5)$ gives $\displaystyle 18,-1,56$; row $\displaystyle 3=(4,5,6)$ gives $\displaystyle 22,-2,70$:
\[\left[\begin{array}{rrr}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{rrr}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]=\left[\begin{array}{rrr}14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70\end{array}\right]\]
(v)
$\displaystyle (3\times2)(2\times3)\to 3\times3$. Row $\displaystyle 3=(-1,1)$ gives $\displaystyle (-1)(1)+1(-1)=-2$, $\displaystyle (-1)(0)+1(2)=2$, $\displaystyle (-1)(1)+1(1)=0$:
\[\left[\begin{array}{rr}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]=\left[\begin{array}{rrr}1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0\end{array}\right]\]
(vi)
$\displaystyle (2\times3)(3\times2)\to 2\times2$. Row $\displaystyle 1=(3,-1,3)$ gives $\displaystyle 3(2)+(-1)(1)+3(3)=14$ and $\displaystyle 3(-3)+(-1)(0)+3(1)=-6$; row $\displaystyle 2=(-1,0,2)$ gives $\displaystyle (-1)(2)+0(1)+2(3)=4$ and $\displaystyle (-1)(-3)+0(0)+2(1)=5$:
\[\left[\begin{array}{rrr}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{rr}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]=\left[\begin{array}{rr}14 & -6 \\ 4 & 5\end{array}\right]\]
Exercise 4
If $\displaystyle \mathrm{A}=\left[\begin{array}{rrr}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{rrr}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]$ and $\displaystyle \mathrm{C}=\left[\begin{array}{rrr}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]$ , then compute $\displaystyle (\mathrm{A}+\mathrm{B})$ and $\displaystyle (\mathrm{B}-\mathrm{C})$ . Also, verify that $\displaystyle \mathrm{A}+(\mathrm{B}-\mathrm{C})=(\mathrm{A}+\mathrm{B})-\mathrm{C}$ .
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle \mathrm{A}+\mathrm{B}=\begin{array}{ccc}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{array}, \mathrm{~B}-\mathrm{C}=\begin{array}{ccc}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{array}$
All three matrices are of order $\displaystyle 3\times3$, so every sum and difference below is defined, and both operations are carried out entrywise.\[\mathrm{A}+\mathrm{B}=\left[\begin{array}{rrr}1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3\end{array}\right]=\left[\begin{array}{rrr}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{array}\right]\]\[\mathrm{B}-\mathrm{C}=\left[\begin{array}{rrr}3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0+2 & 3-3\end{array}\right]=\left[\begin{array}{rrr}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{array}\right]\]Verification. Left side: \[\mathrm{A}+(\mathrm{B}-\mathrm{C})=\left[\begin{array}{rrr}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right]+\left[\begin{array}{rrr}-1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0\end{array}\right]=\left[\begin{array}{rrr}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1\end{array}\right]\]Right side, using the $\displaystyle \mathrm{A}+\mathrm{B}$ already found: \[(\mathrm{A}+\mathrm{B})-\mathrm{C}=\left[\begin{array}{rrr}4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4\end{array}\right]-\left[\begin{array}{rrr}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]=\left[\begin{array}{rrr}0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1\end{array}\right]\]The two matrices are identical entry for entry, so $\displaystyle \mathrm{A}+(\mathrm{B}-\mathrm{C})=(\mathrm{A}+\mathrm{B})-\mathrm{C}$. This is exactly the associativity of matrix addition applied to $\displaystyle \mathrm{A}+\mathrm{B}+(-\mathrm{C})$.
Exercise 5
If $\displaystyle \mathrm{A}=\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]$ and $\displaystyle \mathrm{B}=\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]$ , then compute $\displaystyle 3 A-5 B$ .
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle \left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
Scalar multiplication multiplies EVERY entry: $\displaystyle k\mathrm{A}=[ka_{ij}]$. Compute the two scalar multiples separately before subtracting.\[3\mathrm{A}=3\left[\begin{array}{rrr}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]=\left[\begin{array}{rrr}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]\]\[5\mathrm{B}=5\left[\begin{array}{rrr}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right]=\left[\begin{array}{rrr}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]\]The denominators are cleared by the very scalars chosen, and the two results are the same matrix. Hence, subtracting entrywise, \[3\mathrm{A}-5\mathrm{B}=\left[\begin{array}{rrr}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=\mathrm{O}\]Final answer: $\displaystyle 3\mathrm{A}-5\mathrm{B}$ is the null matrix of order $\displaystyle 3\times3$.
Exercise 6
Simplify $\displaystyle \cos \theta\left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\sin \theta\left[\begin{array}{rr}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right]$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle \left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Multiply each matrix by its scalar first, then add entrywise.\[\cos\theta\left[\begin{array}{rr}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right]=\left[\begin{array}{rr}\cos^{2}\theta & \sin\theta\cos\theta \\ -\sin\theta\cos\theta & \cos^{2}\theta\end{array}\right]\]\[\sin\theta\left[\begin{array}{rr}\sin\theta & -\cos\theta \\ \cos\theta & \sin\theta\end{array}\right]=\left[\begin{array}{rr}\sin^{2}\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^{2}\theta\end{array}\right]\]Adding, the diagonal entries become $\displaystyle \cos^{2}\theta+\sin^{2}\theta=1$, and the off-diagonal entries cancel in pairs, $\displaystyle \sin\theta\cos\theta-\sin\theta\cos\theta=0$: \[\left[\begin{array}{rr}\cos^{2}\theta+\sin^{2}\theta & \sin\theta\cos\theta-\sin\theta\cos\theta \\ -\sin\theta\cos\theta+\sin\theta\cos\theta & \cos^{2}\theta+\sin^{2}\theta\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right]\]Final answer: the expression simplifies to the identity matrix $\displaystyle \mathrm{I}$ of order $\displaystyle 2$, for every value of $\displaystyle \theta$.
Exercise 7
Find X and Y , if
(i)
$\displaystyle \mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ and $\displaystyle \mathrm{X}-\mathrm{Y}=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$
(ii)
$\displaystyle 2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]$ and $\displaystyle 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
(i)
$\displaystyle \mathrm{X}=\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}, \mathrm{Y}=\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}$ (ii) $\displaystyle \mathrm{X}=\begin{array}{cc}\frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3\end{array}, \mathrm{Y}=\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}$
Matrix equations in $\displaystyle \mathrm{X}$ and $\displaystyle \mathrm{Y}$ are solved exactly like simultaneous linear equations: add or subtract suitable multiples to eliminate one unknown matrix, then divide by the scalar (multiply by its reciprocal).
(i)
Adding the two given equations eliminates $\displaystyle \mathrm{Y}$:
\[(\mathrm{X}+\mathrm{Y})+(\mathrm{X}-\mathrm{Y})=2\mathrm{X}=\left[\begin{array}{rr}7 & 0 \\ 2 & 5\end{array}\right]+\left[\begin{array}{rr}3 & 0 \\ 0 & 3\end{array}\right]=\left[\begin{array}{rr}10 & 0 \\ 2 & 8\end{array}\right]\]
\[\Rightarrow\ \mathrm{X}=\frac{1}{2}\left[\begin{array}{rr}10 & 0 \\ 2 & 8\end{array}\right]=\left[\begin{array}{rr}5 & 0 \\ 1 & 4\end{array}\right]\]
Subtracting eliminates $\displaystyle \mathrm{X}$:
\[2\mathrm{Y}=\left[\begin{array}{rr}7 & 0 \\ 2 & 5\end{array}\right]-\left[\begin{array}{rr}3 & 0 \\ 0 & 3\end{array}\right]=\left[\begin{array}{rr}4 & 0 \\ 2 & 2\end{array}\right]\ \Rightarrow\ \mathrm{Y}=\left[\begin{array}{rr}2 & 0 \\ 1 & 1\end{array}\right]\]
Check: $\displaystyle \mathrm{X}+\mathrm{Y}=\left[\begin{array}{rr}7 & 0 \\ 2 & 5\end{array}\right]$, as required.
(ii)
To eliminate $\displaystyle \mathrm{X}$, take $\displaystyle 3(2\mathrm{X}+3\mathrm{Y})-2(3\mathrm{X}+2\mathrm{Y})=5\mathrm{Y}$:
\[5\mathrm{Y}=3\left[\begin{array}{rr}2 & 3 \\ 4 & 0\end{array}\right]-2\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]=\left[\begin{array}{rr}6 & 9 \\ 12 & 0\end{array}\right]-\left[\begin{array}{rr}4 & -4 \\ -2 & 10\end{array}\right]=\left[\begin{array}{rr}2 & 13 \\ 14 & -10\end{array}\right]\]
\[\Rightarrow\ \mathrm{Y}=\frac{1}{5}\left[\begin{array}{rr}2 & 13 \\ 14 & -10\end{array}\right]=\left[\begin{array}{rr}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}\right]\]
To eliminate $\displaystyle \mathrm{Y}$, take $\displaystyle 3(3\mathrm{X}+2\mathrm{Y})-2(2\mathrm{X}+3\mathrm{Y})=5\mathrm{X}$:
\[5\mathrm{X}=3\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]-2\left[\begin{array}{rr}2 & 3 \\ 4 & 0\end{array}\right]=\left[\begin{array}{rr}6 & -6 \\ -3 & 15\end{array}\right]-\left[\begin{array}{rr}4 & 6 \\ 8 & 0\end{array}\right]=\left[\begin{array}{rr}2 & -12 \\ -11 & 15\end{array}\right]\]
\[\Rightarrow\ \mathrm{X}=\frac{1}{5}\left[\begin{array}{rr}2 & -12 \\ -11 & 15\end{array}\right]=\left[\begin{array}{rr}\frac{2}{5} & -\frac{12}{5} \\ -\frac{11}{5} & 3\end{array}\right]\]
Check: $\displaystyle 2\mathrm{X}+3\mathrm{Y}=\left[\begin{array}{rr}\frac{4}{5}+\frac{6}{5} & -\frac{24}{5}+\frac{39}{5} \\ -\frac{22}{5}+\frac{42}{5} & 6-6\end{array}\right]=\left[\begin{array}{rr}2 & 3 \\ 4 & 0\end{array}\right]$, as required.
Exercise 8
Find X , if $\displaystyle \mathrm{Y}=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$ and $\displaystyle 2 \mathrm{X}+\mathrm{Y}=\left[\begin{array}{rr}1 & 0 \\ -3 & 2\end{array}\right]$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle \mathrm{X}=\left[\begin{array}{ll}-1 & -1 \\ -2 & -1\end{array}\right]$
Treat the matrix equation like a scalar one: isolate $\displaystyle 2\mathrm{X}$ by subtracting $\displaystyle \mathrm{Y}$ from both sides, then multiply by the scalar $\displaystyle \tfrac{1}{2}$ (which halves every entry).\[2\mathrm{X}=\left[\begin{array}{rr}1 & 0 \\ -3 & 2\end{array}\right]-\mathrm{Y}=\left[\begin{array}{rr}1 & 0 \\ -3 & 2\end{array}\right]-\left[\begin{array}{rr}3 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{rr}1-3 & 0-2 \\ -3-1 & 2-4\end{array}\right]=\left[\begin{array}{rr}-2 & -2 \\ -4 & -2\end{array}\right]\]\[\mathrm{X}=\frac{1}{2}\left[\begin{array}{rr}-2 & -2 \\ -4 & -2\end{array}\right]=\left[\begin{array}{rr}-1 & -1 \\ -2 & -1\end{array}\right]\]Check: $\displaystyle 2\mathrm{X}+\mathrm{Y}=\left[\begin{array}{rr}-2 & -2 \\ -4 & -2\end{array}\right]+\left[\begin{array}{rr}3 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{rr}1 & 0 \\ -3 & 2\end{array}\right]$.Final answer: $\displaystyle \mathrm{X}=\left[\begin{array}{rr}-1 & -1 \\ -2 & -1\end{array}\right]$.
Exercise 9
Find $\displaystyle x$ and $\displaystyle y$ , if $\displaystyle 2\left[\begin{array}{cc}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
$\displaystyle x=3, y=3$
Simplify the left side, then use the definition of equality of matrices: two matrices are equal only if they have the same order AND every pair of corresponding entries is equal.\[2\left[\begin{array}{rr}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{rr}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{rr}2 & 6 \\ 0 & 2x\end{array}\right]+\left[\begin{array}{rr}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{rr}2+y & 6 \\ 1 & 2x+2\end{array}\right]\]Equating this with $\displaystyle \left[\begin{array}{rr}5 & 6 \\ 1 & 8\end{array}\right]$ position by position: \[2+y=5,\qquad 6=6,\qquad 1=1,\qquad 2x+2=8\] The $\displaystyle (1,2)$ and $\displaystyle (2,1)$ entries are automatically consistent; the other two give \[y=3,\qquad 2x=6\ \Rightarrow\ x=3\]Final answer: $\displaystyle x=3,\ y=3$.
Exercise 10
Solve the equation for $\displaystyle x, y, z$ and $\displaystyle t$ , if $\displaystyle 2\left[\begin{array}{cc}x & z \\ y & t\end{array}\right]+3\left[\begin{array}{rr}1 & -1 \\ 0 & 2\end{array}\right]=3\left[\begin{array}{ll}3 & 5 \\ 4 & 6\end{array}\right]$
Not cross-checked
This solution has not been cross-checked against the answer printed in NCERT.
Carry out the scalar multiplications and the addition on the left, then equate corresponding entries.\[2\left[\begin{array}{rr}x & z \\ y & t\end{array}\right]+3\left[\begin{array}{rr}1 & -1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{rr}2x & 2z \\ 2y & 2t\end{array}\right]+\left[\begin{array}{rr}3 & -3 \\ 0 & 6\end{array}\right]=\left[\begin{array}{rr}2x+3 & 2z-3 \\ 2y & 2t+6\end{array}\right]\]\[3\left[\begin{array}{rr}3 & 5 \\ 4 & 6\end{array}\right]=\left[\begin{array}{rr}9 & 15 \\ 12 & 18\end{array}\right]\]Note the placement of the unknowns in the given matrix: $\displaystyle z$ sits in the $\displaystyle (1,2)$ position and $\displaystyle y$ in the $\displaystyle (2,1)$ position, so $\displaystyle y$ must be matched with $\displaystyle 12$ and $\displaystyle z$ with $\displaystyle 15$, not the other way round. Equating entrywise: \[2x+3=9\ \Rightarrow\ x=3\] \[2z-3=15\ \Rightarrow\ 2z=18\ \Rightarrow\ z=9\] \[2y=12\ \Rightarrow\ y=6\] \[2t+6=18\ \Rightarrow\ 2t=12\ \Rightarrow\ t=6\]Final answer: $\displaystyle x=3,\ y=6,\ z=9,\ t=6$.

EXERCISE 3.2 1–10 (part 2 of 6)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10