Name: The Eigenvectors and Eigenvalues
Uploaded: 2019-07-19
Duration: 17 min 15 s
Description: Studying linear algebra, we will inevitably stumble upon the concept of eigenvalues. These sound very exotic, but they are very important.

Question

Given

A = [\begin{array}{l} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 2 & 3 & 1 \end{array}], B = [\begin{array}{l} 2 & 3 \\ 3 & 4 \end{array}] .

Find P such that

B P A = [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}]

.

Accepted Answer

$B P A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$

$\Rightarrow B^{- 1} B P A = B^{- 1} [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$

$B^{- 1} = \frac{a d j \cdot B}{| B |}$

Now, $a d j . B = {[\begin{matrix} 4 & - 3 \\ - 3 & + 2 \end{matrix}]}^{T}$

$= [\begin{matrix} 4 & - 3 \\ - 3 & + 2 \end{matrix}]$

$| B | = 8 - 9 = - 1$

$B^{- 1} = \frac{1}{- 1} [\begin{matrix} 4 & - 3 \\ - 3 & + 2 \end{matrix}]$

$= [\begin{matrix} 4 & 3 \\ 3 & - 2 \end{matrix}]$

$B^{- 1} [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}] = [\begin{matrix} - 4 & + 3 \\ 3 & - 2 \end{matrix}] [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}]$

$B^{- 1} [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}] = [\begin{matrix} - 4 & + 3 & - 4 \\ 3 & - 2 & 3 \end{matrix}]$

$∴ P A = [\begin{matrix} - 4 & + 3 & - 4 \\ 3 & - 2 & 3 \end{matrix}]$

Now, $P A A^{- 1} = [\begin{matrix} - 4 & + 3 & - 4 \\ 3 & - 2 & 3 \end{matrix}] A^{- 1}$

$A^{- 1} = \frac{adj . A}{| A |}$

Now, $a d j . A = {[\begin{matrix} 1 & 0 & - 2 \\ 2 & - 1 & - 2 \\ - 3 & 1 & 2 \end{matrix}]}^{T}$

$= [\begin{matrix} 1 & 2 & - 3 \\ 0 & - 1 & 1 \\ - 2 & - 1 & 2 \end{matrix}]$

$| A | = 1 (4 - 3) - 1 (2 - 2) + 1 (6 - 8) = 1 - 2 = - 1$

$A^{- 1} = \frac{1}{- 1} [\begin{matrix} 1 & 2 & - 3 \\ 0 & - 1 & 1 \\ - 2 & - 1 & 2 \end{matrix}]$

$= [\begin{matrix} - 1 & - 2 & 3 \\ 0 & 1 & - 1 \\ 2 & 1 & - 2 \end{matrix}]$

$P I = [\begin{matrix} - 4 & + 3 & - 4 \\ 3 & - 2 & 3 \end{matrix}] [\begin{matrix} - 1 & - 2 & 3 \\ 0 & 1 & - 1 \\ 2 & 1 & - 2 \end{matrix}]$

$P = [\begin{matrix} 4 - 8 & 8 + 3 - 4 & - 12 - 3 + 8 \\ - 3 + 6 & - 6 - 2 + 3 & 9 + 2 - 6 \end{matrix}]$

$\Rightarrow P = [\begin{matrix} - 4 & 7 & - 7 \\ 3 & - 5 & 5 \end{matrix}]$

Given $A = [\begin{array}{l} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 2 & 3 & 1 \end{array}], B = [\begin{array}{l} 2 & 3 \\ 3 & 4 \end{array}] .$ Find P such that $B P A = [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}]$ .

Important Questions on Matrices and Determinants

Learn and Prepare for any exam you want !

Given A=111241231, B=2334. Find P such that BPA=101010.

Important Questions on Matrices and Determinants

Learn and Prepare for any exam you want !

Given $A = [\begin{array}{l} 1 & 1 & 1 \\ 2 & 4 & 1 \\ 2 & 3 & 1 \end{array}], B = [\begin{array}{l} 2 & 3 \\ 3 & 4 \end{array}] .$ Find P such that $B P A = [\begin{array}{l} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}]$ .