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Algebra — MATHS STD 10 — Question

Tamilnadu BoardEnglish MediumSTD 10MATHSAlgebra5 Marks
Question
$A=\left[\begin{array}{ll}3 & 0 \\ 4 & 5\end{array}\right], B=\left[\begin{array}{ll}6 & 3 \\ 8 & 5\end{array}\right], C=\left[\begin{array}{ll}3 & 6 \\ 1 & 1\end{array}\right]$ find the matrix $D$, such that $C D-A B=0$
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Answer

$\begin{aligned} & \text { Given } A=\left[\begin{array}{ll}3 & 0 \\ 4 & 5\end{array}\right], B=\left[\begin{array}{ll}6 & 3 \\ 8 & 5\end{array}\right], C=\left[\begin{array}{ll}3 & 6 \\ 1 & 1\end{array}\right] \\ & \text { Let } D \text { be }\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\ & \text { Given } C D- AB =0 \\ & AB = CD \\ & {\left[\begin{array}{cc}3 & 0 \\ 4 & 5\end{array}\right]\left[\begin{array}{ll}6 & 3 \\ 8 & 5\end{array}\right]=\left[\begin{array}{ll}3 & 6 \\ 1 & 1\end{array}\right]\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]} \\ & {\left[\begin{array}{cc}18+0 & 9+0 \\ 24+40 & 12+25\end{array}\right]=\left[\begin{array}{cc}3 a +6 c & 3 b +6 d \\ a + c & b + d \end{array}\right]} \\ & {\left[\begin{array}{cc}18 & 9 \\ 64 & 37\end{array}\right]=\left[\begin{array}{c}3 a +6 c \quad 3 b +6 d \\ a + c \quad b + d \end{array}\right]}\end{aligned}$
$\begin{array}{r}3 a+6 c=18 \\ (\div 3) \Rightarrow \quad a+2 c=6 \cdots(1)\\ a+c=64 \cdots(2)\\ \underline{(-) (-)(-)} \\ (1)-(2) \Rightarrow \quad c=-58\end{array}$
Substitute the value of $c=-58$ in (1)
$
\begin{aligned}
a-116 & =6 \\
a & =6+116 \\
& =122 \\
3 b+6 d & =9 \\
(\div 3) \Rightarrow b+2 d & =3 \\
b+d & =37 \quad \ldots .(3) \\
\ldots(-)(-) & (-)
\end{aligned}
$
$(3)-\overline{(4)} \Rightarrow d=-34$

Substitute the value of $d=-34$ in (4)
$
\begin{aligned}
& b-34=37 \\
& b=37+34 \\
& =71
\end{aligned}
$
Matrix $D=\left[\begin{array}{cc}122 & 71 \\ -58 & -34\end{array}\right]$

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