Question

Question 3
Plese show the steps

Answer 1

We have to find the matrix X such that,

$\longrightarrow\left[\begin{array}{cc}\sf{2}&\sf{-1}\\\sf{3}&\sf{1}\end{array}\right]\,\sf{X}=\left[\begin{array}{c}\sf{1}\\\sf{2}\end{array}\right]\quad\quad\dots\sf{(1)}$

We know that an $\sf{m\times n}$ matrix multiplied with an $\sf{n\times p}$ matrix gives an $\sf{m\times p}$ matrix as the product.

Here the matrix multiplied with X has 2 columns, hence X should have 2 rows. And the product has only 1 column, so should have X.

Hence the order of X is $\sf{2\times1.}$ So let,

$\longrightarrow\sf{X}=\left[\begin{array}{c}\sf{a}\\\sf{b}\end{array}\right]$

Then (1) becomes,

$\longrightarrow\left[\begin{array}{cc}\sf{2}&\sf{-1}\\\sf{3}&\sf{1}\end{array}\right]\,\left[\begin{array}{c}\sf{a}\\\sf{b}\end{array}\right]=\left[\begin{array}{c}\sf{1}\\\sf{2}\end{array}\right]$

$\longrightarrow\left[\begin{array}{c}\sf{2a-b}\\\sf{3a+b}\end{array}\right]=\left[\begin{array}{c}\sf{1}\\\sf{2}\end{array}\right]$

From this we obtain two equations,

$\longrightarrow\sf{2a-b=1\quad\quad\dots(2)}$

$\longrightarrow\sf{3a+b=2\quad\quad\dots(3)}$

On adding (2) and (3), we get,

$\longrightarrow\sf{5a=3}$

$\longrightarrow\sf{a=\dfrac{3}{5}}$

From (2),

$\longrightarrow\sf{\dfrac{6}{5}-b=1}$

$\longrightarrow\sf{b=\dfrac{1}{5}}$

Therefore,

$\longrightarrow\underline{\underline{\sf{X}=\left[\begin{array}{c}\sf{\dfrac{3}{5}}\\\\\sf{\dfrac{1}{5}}\end{array}\right]}}$

Answer 2

Answer:

We have to find the matrix X such that,

\begin{gathered}\longrightarrow\left[\begin{array}{cc}\sf{2}&\sf{-1}\\\sf{3}&\sf{1}\end{array}\right]\,\sf{X}=\left[\begin{array}{c}\sf{1}\\\sf{2}\end{array}\right]\quad\quad\dots\sf{(1)}\end{gathered}

⟶[

2

3

−1

1

]X=[

1

2

]…(1)

We know that an \sf{m\times n}m×n matrix multiplied with an \sf{n\times p}n×p matrix gives an \sf{m\times p}m×p matrix as the product.

Here the matrix multiplied with X has 2 columns, hence X should have 2 rows. And the product has only 1 column, so should have X.

Hence the order of X is \sf{2\times1.}2×1. So let,

\begin{gathered}\longrightarrow\sf{X}=\left[\begin{array}{c}\sf{a}\\\sf{b}\end{array}\right]\end{gathered}

⟶X=[

a

b

]

Then (1) becomes,

\begin{gathered}\longrightarrow\left[\begin{array}{cc}\sf{2}&\sf{-1}\\\sf{3}&\sf{1}\end{array}\right]\,\left[\begin{array}{c}\sf{a}\\\sf{b}\end{array}\right]=\left[\begin{array}{c}\sf{1}\\\sf{2}\end{array}\right]\end{gathered}

⟶[

2

3

−1

1

][

a

b

]=[

1

2

]

\begin{gathered}\longrightarrow\left[\begin{array}{c}\sf{2a-b}\\\sf{3a+b}\end{array}\right]=\left[\begin{array}{c}\sf{1}\\\sf{2}\end{array}\right]\end{gathered}

⟶[

2a−b

3a+b

]=[

1

2

]

From this we obtain two equations,

\longrightarrow\sf{2a-b=1\quad\quad\dots(2)}⟶2a−b=1…(2)

\longrightarrow\sf{3a+b=2\quad\quad\dots(3)}⟶3a+b=2…(3)

On adding (2) and (3), we get,

\longrightarrow\sf{5a=3}⟶5a=3

\longrightarrow\sf{a=\dfrac{3}{5}}⟶a=

5

3

From (2),

\longrightarrow\sf{\dfrac{6}{5}-b=1}⟶

5

6

−b=1

\longrightarrow\sf{b=\dfrac{1}{5}}⟶b=

5

1

Therefore,

\begin{gathered}\longrightarrow\underline{\underline{\sf{X}=\left[\begin{array}{c}\sf{\dfrac{3}{5}}\\\\\sf{\dfrac{1}{5}}\end{array}\right]}}\end{gathered}

⟶

X=

⎣

⎢

⎢

⎢

⎡

5

3

5

1

⎦

⎥

⎥

⎥

⎤