Name: Finding Inverse by Elementary Transformations
Uploaded: 2019-07-19
Duration: 12 min 5 s
Description: In this video, we will learn to find the inverse of matrices using elementary transformations of matrices. We will learn how to find the inverse of a matrix,...

Question

If

A = [\begin{array}{cc} 0 & - x \\ x & 0 \end{array}], B = [\begin{array}{l} 0 & 1 \\ 1 & 0 \end{array}]

and

x^{2} = - 1

, then show that

(A + B)^{2} = A^{2} + B^{2}

.

Accepted Answer

We have $A = [\begin{array}{cc} 0 & - x \\ x & 0 \end{array}], B = [\begin{array}{l} 0 & 1 \\ 1 & 0 \end{array}]$ and $x^{2} = - 1$ .

$∴ (A + B) = [\begin{matrix} 0 & - x + 1 \\ x + 1 & 0 \end{matrix}]$

And $(A + B)^{2} = [\begin{matrix} 0 & - x + 1 \\ x + 1 & 0 \end{matrix}] [\begin{matrix} 0 & - x + 1 \\ x + 1 & 0 \end{matrix}]$

$= [\begin{matrix} 1 - x^{2} & 0 \\ 0 & 1 - x^{2} \end{matrix}]$ .......(i)

Also $A^{2} = A A$

$= [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}] [\begin{matrix} 0 & - x \\ x & 0 \end{matrix}]$

$= [\begin{matrix} - x^{2} & 0 \\ 0 & - x^{2} \end{matrix}]$

And $B^{2} = B B$

$= [\begin{array}{l} 0 & 1 \\ 1 & 0 \end{array}] [\begin{array}{l} 0 & 1 \\ 1 & 0 \end{array}]$

$= [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}]$

Now $A^{2} + B^{2} = [\begin{matrix} - x^{2} + 1 & 0 \\ 0 & - x^{2} + 1 \end{matrix}]$ [Using Eq (i) and (ii)]

$= [\begin{matrix} 1 - x^{2} & 0 \\ 0 & 1 - x^{2} \end{matrix}]$

$= {(A + B)}^{2}$

Hence, proved.

If $A = [\begin{array}{cc} 0 & - x \\ x & 0 \end{array}], B = [\begin{array}{l} 0 & 1 \\ 1 & 0 \end{array}]$ and $x^{2} = - 1$ , then show that $(A + B)^{2} = A^{2} + B^{2}$ .

Important Questions on Matrices

Learn and Prepare for any exam you want !

If A=0−xx0, B=0110 and x2=-1, then show that (A+B)2=A2+B2.

Important Questions on Matrices

Learn and Prepare for any exam you want !

If $A = [\begin{array}{cc} 0 & - x \\ x & 0 \end{array}], B = [\begin{array}{l} 0 & 1 \\ 1 & 0 \end{array}]$ and $x^{2} = - 1$ , then show that $(A + B)^{2} = A^{2} + B^{2}$ .