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Matrices (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)5 Marks
Question
If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]$, then find matrix $X$ such that $X A=B$.
✓

Answer

$
\begin{gathered}
X A=B \\
\therefore X\left(\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]=\left(\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right)
\end{gathered}
$
By $C_3-C_1$, we get
$
X\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 2 & 3 \\
1 & 2 & 0
\end{array}\right)=\left(\begin{array}{lll}
1 & 2 & 2 \\
1 & 1 & 4 \\
2 & 4 & 5
\end{array}\right)
$
By $\left(\frac{1}{2}\right) C_2$, we get
$
X\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 3 \\
1 & 1 & 0
\end{array}\right)=\left[\begin{array}{lll}
1 & 1 & 2 \\
1 & \frac{1}{2} & 4 \\
2 & 2 & 5
\end{array}\right]
$
By $C_3-3 C_2$, we get
$
X\left(\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 1 & -3
\end{array}\right]=\left[\begin{array}{rrr}
1 & 1 & -1 \\
1 & \frac{1}{2} & \frac{5}{2} \\
2 & 2 & -1
\end{array}\right]
$
By $\left(-\frac{1}{3}\right) C_3$, we get
$
X\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 1 & 1
\end{array}\right]=\left[\begin{array}{rrr}
1 & 1 & \frac{1}{3} \\
1 & \frac{1}{2} & -\frac{5}{6} \\
2 & 2 & \frac{1}{3}
\end{array}\right]
$
By $C _1- C _3$ and $C _2- C _3$, we get
$
\begin{aligned}
& X\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
\frac{2}{3} & \frac{2}{3} & \frac{1}{3} \\
\frac{11}{6} & \frac{4}{3} & -\frac{5}{6} \\
\frac{5}{3} & \frac{5}{3} & \frac{1}{3}
\end{array}\right] \\
& \therefore X=\frac{1}{6}\left[\begin{array}{rrr}
4 & 4 & 2 \\
11 & 8 & -5 \\
10 & 10 & 2
\end{array}\right] .
\end{aligned}
$

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