Question
Find a 2 × 2 matrix A such that.
$\text{A}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
$\text{A}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
✓
Answer
Let $\text{A}=\begin{bmatrix}\text{w}&\text{x}\\\text{y}&\text{z}\end{bmatrix}$
Now,
$\begin{bmatrix}\text{w}&\text{x}\\\text{y}&\text{z}\end{bmatrix}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
$ \Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=6\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=\begin{bmatrix}6&0\\0&6\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ w + x = 6
⇒ w = 6 - x ...(1)
-2w + 4x = 0 ...(2)
Putting the value of w in eq. (2), we get
-2(6 - x) + 4x = 0
⇒ -12 + 2x + 4x = 0
⇒ -12 + 6x = 0
⇒ 6x = 12
⇒ x = 2
Putting the value of x in eq. (1), we get
w = 6 - 2
⇒ w = 4
Now,
y + z = 0
⇒ y = -z ...(3)
-2y + 4z = 6 ...(4)
Putting the value of y in eq. (4), we get
-2(-z) + 4z = 6
⇒ 2z + 4z = 6
⇒ 6z = 6
⇒ z = 1
Putting the value of z in eq. (3), we get
y = -1
$ \therefore\ \text{A}=\begin{bmatrix}4&2\\-1&1\end{bmatrix}$
Now,
$\begin{bmatrix}\text{w}&\text{x}\\\text{y}&\text{z}\end{bmatrix}\begin{bmatrix}1&-2\\1&4\end{bmatrix}=6\text{I}_2$
$ \Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=6\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{w}+\text{x}&-2\text{w}+4\text{x}\\\text{y}+\text{z}&-2\text{y}+4\text{z}\end{bmatrix}=\begin{bmatrix}6&0\\0&6\end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ w + x = 6
⇒ w = 6 - x ...(1)
-2w + 4x = 0 ...(2)
Putting the value of w in eq. (2), we get
-2(6 - x) + 4x = 0
⇒ -12 + 2x + 4x = 0
⇒ -12 + 6x = 0
⇒ 6x = 12
⇒ x = 2
Putting the value of x in eq. (1), we get
w = 6 - 2
⇒ w = 4
Now,
y + z = 0
⇒ y = -z ...(3)
-2y + 4z = 6 ...(4)
Putting the value of y in eq. (4), we get
-2(-z) + 4z = 6
⇒ 2z + 4z = 6
⇒ 6z = 6
⇒ z = 1
Putting the value of z in eq. (3), we get
y = -1
$ \therefore\ \text{A}=\begin{bmatrix}4&2\\-1&1\end{bmatrix}$
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