Matrices — ICSE Class 10 Maths Important Questions

Answer: (a) $3\times2$3×2.

Order is written as (number of rows) $\times$× (number of columns) $=3\times2$=3×2.

Answer: (a) $2\times4$2×4.

The inner dimensions match ($3=3$3=3), so $AB$AB is defined with order (rows of $A$A) $\times$× (columns of $B$B) $=2\times4$=2×4.

Answer: (a) $\begin{pmatrix}4&6\\2&0\end{pmatrix}$pmatrix4&6; 2&0pmatrix.

Scalar multiplication multiplies every element by $2$2: $\begin{pmatrix}2(2)&2(3)\\2(1)&2(0)\end{pmatrix}=\begin{pmatrix}4&6\\2&0\end{pmatrix}$pmatrix2(2)&2(3); 2(1)&2(0)pmatrix=pmatrix4&6; 2&0pmatrix.

Answer: (c) $BA$BA.

$B$B is of order $1\times2$1×2 and $A$A is of order $2\times2$2×2. For $BA$BA the inner orders match ($2=2$2=2), so $BA$BA is defined and has order $1\times2$1×2. $AB$AB would need $2=1$2=1, and $A+B$A+B needs both matrices to have the same order, so neither of these is defined.

Answer: (a) Non-commutativity means precisely that $AB\neq BA$AB≠ BA in general, which is what R states, so R is the correct explanation of A.

Subtract corresponding elements:

$A-B=\begin{pmatrix}3-1&1-2\\2-0&4-5\end{pmatrix}=\begin{pmatrix}2&-1\\2&-1\end{pmatrix}.$A-B=pmatrix3-1&1-2; 2-0&4-5pmatrix=pmatrix2&-1; 2&-1pmatrix.

Equating corresponding elements: $x+y=6$x+y=6 and $x-y=2$x-y=2.

Adding: $2x=8\Rightarrow x=\mathbf{4}$2x=8 x=4; then $y=6-4=\mathbf{2}$y=6-4=2.

$AB=\begin{pmatrix}2&1\\3&2\end{pmatrix}\begin{pmatrix}1&0\\-1&2\end{pmatrix}.$AB=pmatrix2&1; 3&2pmatrixpmatrix1&0; -1&2pmatrix.

Row 1: $(2\cdot1+1\cdot(-1),\ 2\cdot0+1\cdot2)=(1,2)$(2·1+1·(-1), 2·0+1·2)=(1,2).

Row 2: $(3\cdot1+2\cdot(-1),\ 3\cdot0+2\cdot2)=(1,4)$(3·1+2·(-1), 3·0+2·2)=(1,4).

$AB=\begin{pmatrix}1&2\\1&4\end{pmatrix}.$AB=pmatrix1&2; 1&4pmatrix.

$2X=\begin{pmatrix}5&6\\7&8\end{pmatrix}-\begin{pmatrix}1&2\\3&4\end{pmatrix}=\begin{pmatrix}4&4\\4&4\end{pmatrix}.$2X=pmatrix5&6; 7&8pmatrix-pmatrix1&2; 3&4pmatrix=pmatrix4&4; 4&4pmatrix.

Divide every element by $2$2:

$X=\begin{pmatrix}2&2\\2&2\end{pmatrix}.$X=pmatrix2&2; 2&2pmatrix.

$A^2=\begin{pmatrix}3&x\\0&1\end{pmatrix}\begin{pmatrix}3&x\\0&1\end{pmatrix}=\begin{pmatrix}9&3x+x\\0&1\end{pmatrix}=\begin{pmatrix}9&4x\\0&1\end{pmatrix}.$A^2=pmatrix3&x; 0&1pmatrixpmatrix3&x; 0&1pmatrix=pmatrix9&3x+x; 0&1pmatrix=pmatrix9&4x; 0&1pmatrix.

Comparing with $\begin{pmatrix}9&12\\0&1\end{pmatrix}$pmatrix9&12; 0&1pmatrix: $4x=12\Rightarrow x=\mathbf{3}$4x=12 x=3.

Order of $M$M: the right-hand side is of order $1\times2$1×2 and $\begin{pmatrix}1&2\\0&1\end{pmatrix}$pmatrix1&2; 0&1pmatrix is of order $2\times2$2×2, so $M$M must be of order $\mathbf{1\times2}$1×2.

Let $M=\begin{pmatrix}a&b\end{pmatrix}$M=pmatrixa&bpmatrix. Then
$\begin{pmatrix}a&b\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}a&2a+b\end{pmatrix}=\begin{pmatrix}3&10\end{pmatrix}.$pmatrixa&bpmatrixpmatrix1&2; 0&1pmatrix=pmatrixa&2a+bpmatrix=pmatrix3&10pmatrix.

So $a=3$a=3, and $2(3)+b=10\Rightarrow b=4$2(3)+b=10 b=4.

$M=\begin{pmatrix}3&4\end{pmatrix}.$M=pmatrix3&4pmatrix.

(i)
$AB=\begin{pmatrix}1&2\\0&3\end{pmatrix}\begin{pmatrix}2&0\\1&1\end{pmatrix}=\begin{pmatrix}1\cdot2+2\cdot1&1\cdot0+2\cdot1\\0\cdot2+3\cdot1&0\cdot0+3\cdot1\end{pmatrix}=\begin{pmatrix}4&2\\3&3\end{pmatrix}.$AB=pmatrix1&2; 0&3pmatrixpmatrix2&0; 1&1pmatrix=pmatrix1·2+2·1&1·0+2·1; 0·2+3·1&0·0+3·1pmatrix=pmatrix4&2; 3&3pmatrix.

$BA=\begin{pmatrix}2&0\\1&1\end{pmatrix}\begin{pmatrix}1&2\\0&3\end{pmatrix}=\begin{pmatrix}2\cdot1+0\cdot0&2\cdot2+0\cdot3\\1\cdot1+1\cdot0&1\cdot2+1\cdot3\end{pmatrix}=\begin{pmatrix}2&4\\1&5\end{pmatrix}.$BA=pmatrix2&0; 1&1pmatrixpmatrix1&2; 0&3pmatrix=pmatrix2·1+0·0&2·2+0·3; 1·1+1·0&1·2+1·3pmatrix=pmatrix2&4; 1&5pmatrix.

So $AB\neq BA$AB≠ BA: matrix multiplication is not commutative.

(ii) $A+B=\begin{pmatrix}3&2\\1&4\end{pmatrix}$A+B=pmatrix3&2; 1&4pmatrix and $A-B=\begin{pmatrix}-1&2\\-1&2\end{pmatrix}$A-B=pmatrix-1&2; -1&2pmatrix.

$(A+B)(A-B)=\begin{pmatrix}3(-1)+2(-1)&3(2)+2(2)\\1(-1)+4(-1)&1(2)+4(2)\end{pmatrix}=\begin{pmatrix}-5&10\\-5&10\end{pmatrix}.$(A+B)(A-B)=pmatrix3(-1)+2(-1)&3(2)+2(2); 1(-1)+4(-1)&1(2)+4(2)pmatrix=pmatrix-5&10; -5&10pmatrix.

$A^2=\begin{pmatrix}1&8\\0&9\end{pmatrix},\qquad B^2=\begin{pmatrix}4&0\\3&1\end{pmatrix},\qquad A^2-B^2=\begin{pmatrix}-3&8\\-3&8\end{pmatrix}.$A^2=pmatrix1&8; 0&9pmatrix, B^2=pmatrix4&0; 3&1pmatrix, A^2-B^2=pmatrix-3&8; -3&8pmatrix.

The two results are different, so $(A+B)(A-B)\neq A^2-B^2$(A+B)(A-B)≠ A^2-B^2. This is because $(A+B)(A-B)=A^2-AB+BA-B^2$(A+B)(A-B)=A^2-AB+BA-B^2, and $-AB+BA$-AB+BA is not the zero matrix when $AB\neq BA$AB≠ BA.

Compute $A^2$A^2:
$A^2=\begin{pmatrix}3&-2\\4&-2\end{pmatrix}\begin{pmatrix}3&-2\\4&-2\end{pmatrix}=\begin{pmatrix}9-8&-6+4\\12-8&-8+4\end{pmatrix}=\begin{pmatrix}1&-2\\4&-4\end{pmatrix}.$A^2=pmatrix3&-2; 4&-2pmatrixpmatrix3&-2; 4&-2pmatrix=pmatrix9-8&-6+4; 12-8&-8+4pmatrix=pmatrix1&-2; 4&-4pmatrix.

Write $kA-2I$kA-2I:
$kA-2I=\begin{pmatrix}3k-2&-2k\\4k&-2k-2\end{pmatrix}.$kA-2I=pmatrix3k-2&-2k; 4k&-2k-2pmatrix.

Equate corresponding elements (top-left): $3k-2=1\Rightarrow k=1$3k-2=1 k=1.

Check with the other entries: $-2k=-2$-2k=-2, $4k=4$4k=4, $-2k-2=-4$-2k-2=-4 — all hold for $k=1$k=1.

Hence $k=\mathbf{1}$k=1.

(i) $Q$Q is $2\times2$2×2 and $P$P is $2\times1$2×1, so $QP$QP has order $\mathbf{2\times1}$2×1.

(ii) $QP=\begin{pmatrix}2&3\\4&1\end{pmatrix}\begin{pmatrix}5\\20\end{pmatrix}=\begin{pmatrix}2(5)+3(20)\\4(5)+1(20)\end{pmatrix}=\begin{pmatrix}70\\40\end{pmatrix}.$QP=pmatrix2&3; 4&1pmatrixpmatrix5; 20pmatrix=pmatrix2(5)+3(20); 4(5)+1(20)pmatrix=pmatrix70; 40pmatrix.
Student 1 spends ₹$70$70 and Student 2 spends ₹$40$40.

(iii) Student 1 spends more, by $70-40=$70-40= ₹$\mathbf{30}$30.