Solve the matrix and prove that it is equal 0

Mathematics

1 answer:

Art [367]3 years ago

4 0

Step-by-step 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.

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

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

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

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