Question

Solve the matrix and prove that it is equal 0​

Answer 1

Explanation:

\underline{ \underline{ \text{Given}}} :

\begin{gathered} \tt{ {A}^{T} = \begin{bmatrix} 2 & - 4 \\ 4 & 3 \\ \end{bmatrix}}\end{gathered}

\underline{ \underline { \text{To \: Find}}} :

\underline{ \underline{ \text{Solution}}} :

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by \sf{ {A}^{T}}

. Again , Interchange it's rows and columns in order to find ' A '.

\begin{gathered} \tt{A = \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix}}\end{gathered}

Now , LEFT HAND SIDE ( L.H.S )

\tt{ {A}^{2} - 5A+ 22I}

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼ \begin{gathered}\begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} - 5 \times \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} + 22 \times \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} 2 \times 2 + 4 \times ( - 4)& 2 \times 4 + 4 \times 3 \\ - 4 \times 2 + 3 \times ( - 4) & - 4 \times 4 + 3 \times 3 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} 4 + ( - 16) & 8 + 12 \\ - 8 + ( - 12) & - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered} \begin{bmatrix} - 12 & 20\\ - 20& - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} - 22 & 0 \\ 0& - 22 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} - 22 + 22 & 0 + 0 \\ 0 + 0 & - 22 + 22 \\ \end{bmatrix}\end{gathered}

⟼ \begin{gathered}\begin{bmatrix} 0 & 0\\ 0 & 0 \\ \end{bmatrix}\end{gathered}

⟼ \sf{0}0

RIGHT HAND SIDE ( R.H.S ) : 0

Answer 2

Explanation:

`\underline{ \underline{ \text{Given}}} :`

• 
  `\tt{ {A}^(T) = \begin{bmatrix} 2 & - 4 \\ 4 & 3 \\ \end{bmatrix}}`

`\underline{ \underline { \text{To \: Find}}} :`

• 
  `\sf{ {A}^(2) - 5A+ 22I= 0}`

`\underline{ \underline{ \text{Solution}}} :`

The new matrix obtained from a given matrix by interchanging it's rows and columns is called the transposition of matrix. It is denoted by
`\sf{ {A}^(T)}`. Again , Interchange it's rows and columns in order to find ' A '.

`\tt{A = \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix}}`

Now , LEFT HAND SIDE ( L.H.S )

`\tt{ {A}^(2) - 5A+ 22I}`

Here, I refers to identity matrix. A diagonal matrix in which all the elements of leading diagonal are 1 ( unit ) is called unit or identity matrix.

⟼
`\begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} * \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} - 5 * \begin{bmatrix} 2 & 4 \\ - 4 & 3 \\ \end{bmatrix} + 22 * \begin{bmatrix} 1 & 0 \\ 0 & 1\\ \end{bmatrix}`

⟼
`\begin{bmatrix} 2 * 2 + 4 * ( - 4)& 2 * 4 + 4 * 3 \\ - 4 * 2 + 3 * ( - 4) & - 4 * 4 + 3 * 3 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20& 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}`

⟼
`\begin{bmatrix} 4 + ( - 16) & 8 + 12 \\ - 8 + ( - 12) & - 16 + 9 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}`

⟼
`\begin{bmatrix} - 12 & 20\\ - 20& - 7 \\ \end{bmatrix} - \begin{bmatrix} 10 & 20 \\ - 20 & 15 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}`

⟼
`\begin{bmatrix} - 22 & 0 \\ 0& - 22 \\ \end{bmatrix} + \begin{bmatrix} 22 & 0 \\ 0 & 22 \\ \end{bmatrix}`

⟼
`\begin{bmatrix} - 22 + 22 & 0 + 0 \\ 0 + 0 & - 22 + 22 \\ \end{bmatrix}`

⟼
`\begin{bmatrix} 0 & 0\\ 0 & 0 \\ \end{bmatrix}`

⟼
`\sf{0}`

RIGHT HAND SIDE ( R.H.S ) : 0

L.H.S = R.H.S [ Hence , proved ! ]

Hope I helped ! ♡

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