First slide

Operations of matrix

Question

Let $A = [\begin{array}{l} 2 3 \\ - 1 2 \end{array}]$ and $f (x) = x^{2} - 4 x + 7$ Show that $f (A) = O$ Use this result to find $A^{5}$

Moderate

A

$A^{5} = [\begin{array}{l} - 118 - 93 \\ 93 - 118 \end{array}]$
B

$A^{5} = [\begin{array}{l} - 118 - 93 \\ 31 - 118 \end{array}]$
C

$A^{5} = [\begin{array}{l} - 118 - 31 \\ 31 - 118 \end{array}]$
D

$A^{5} = [\begin{array}{l} - 118 - 93 \\ - 31 - 118 \end{array}]$

Solution

we have,
$f (x) = x^{2} - 4 x + 7$ .Therefore, $f (A) = A^{2} - 4 A + 7 I_{2}$
$A^{2} = [\begin{array}{l} 2 3 \\ - 1 2 \end{array}] [\begin{array}{l} 2 3 \\ - 1 2 \end{array}]$ $= [\begin{array}{l} 4 - 3 6 + 6 \\ - 2 - 2 - 3 + 4 \end{array}]$ $= [\begin{array}{l} 1   12 \\ - 41 \end{array}]$
$- 4 A = [\begin{array}{l} -8   -12 \\ 4 -8 \end{array}]$ and $7 I_{2} = [\begin{array}{l} 7  0 \\ 07 \end{array}]$
$∴ f (A) = A^{2} - 4 A + 7 I_{2}$
$= [\begin{array}{l} 1   12 \\ - 41 \end{array}]$ $+ [\begin{array}{l} -8   -12 \\ 4 -8 \end{array}]$ $+ [\begin{array}{l} 70 \\ 07 \end{array}]$
$= [\begin{array}{l} 1-8+7      12-12+0 \\ - 4 + 4 + 0 1-8+7 \end{array}]$ $= [\begin{array}{l} 0  0 \\ 00 \end{array}] = O$
$f (A) = O$
$\Rightarrow A^{2} - 4 A + 7 I_{2} = O$
$\Rightarrow A^{2} = 4 A - 7 I_{2}$
$\Rightarrow A^{3} = A^{2} . A = (4 A - 7 I_{2}) A = 4 A^{2} - 7 I_{2} A$
$\begin{array}{l} \Rightarrow & A^{3} = 4 (4 A - 7 I_{2}) - 7 A \\ \Rightarrow & A^{3} = 9 A - 28 I_{2} \\ \Rightarrow & A^{4} = A^{3} A = (9 A - 28 I_{2}) A \\ \Rightarrow & A^{4} = 9 A^{2} - 28 A = 9 (4 A - 7 I_{2}) - 28 A \\ \Rightarrow & A^{4} = 36 A - 63 I_{2} - 28 A = 8 A - 63 I_{2} \\ \Rightarrow & A^{5} = A^{4} \cdot A = (8 A - 63 I_{2}) A = 8 A^{2} - 63 I_{2} A \\ \Rightarrow & A^{5} = 8 (4 A - 7 I_{2}) - 63 A = - 31 A - 56 I_{2} \\ \Rightarrow & A^{5} = - 31 [\begin{array}{r} 2 & 3 \\ - 1 & 2 \end{array}] - 56 [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] \\ \Rightarrow & A^{5} = [\begin{matrix} - 62 & - 93 \\ 31 & - 62 \end{matrix}] - [\begin{array}{lr} 56 & 0 \\ 0 & 56 \end{array}] \\ \Rightarrow & A^{5} = [\begin{array}{lr} - 118 & - 93 \\ 31 & - 118 \end{array}] \end{array}$

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