If A = left[ {begin{array}{*{20}{c}}1&a0&1end{array}} right] , then {A^4} is equal to

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3 and 4 .Determinants and Matrices

easy

If $A = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]$, then ${A^4}$ is equal to

$\left[ {\begin{array}{*{20}{c}}1&{{a^4}}\\0&1\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}4&{4a}\\0&4\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}4&{{a^4}}\\0&4\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}1&{4a}\\0&1\end{array}} \right]$

Solution

(d) ${A^2} = A.\,A = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{2a}\\0&1\end{array}} \right]$

${A^3} = A.\,{A^2} = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&{2a}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{3a}\\0&1\end{array}} \right]$

${A^4} = A.\,{A^3} = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&{3a}\\0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&{4a}\\0&1\end{array}} \right]$.

Standard 12

Mathematics

In order that the matrix $\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}} \right]$ be non-singular, $\lambda $ should not be equal to

easy

A trust fund has Rs. $30,000$ that must be invested in two different types of bonds. The first bond pays $5 \%$ interest per year, and the second bond pays $7 \%$ interest per year. Using matrix multiplication, determine how to divide Rs. $30,000$ among the two types of bonds. If the trust fund must obtain an annual total interest of Rs. $1800$.

hard

If ${a_{ij}} = \frac{1}{2}(3i – 2j)$ and $A = {[{a_{ij}}]_{2 \times 2}}$, then $A$ is equal to

easy

Find the values of $a, \,b,\, c,$ and $d$ from the following equation:

$\left[\begin{array}{cc}
2 a+b & a-2 b \\
5 c-d & 4 c+3 d
\end{array}\right]=\left[\begin{array}{cc}
4 & -3 \\
11 & 24
\end{array}\right]$

easy

Compute the indicated products $\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$

easy

If $A = \left[ {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right]$, then ${A^4}$ is equal to

Solution

Similar Questions

In order that the matrix $\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\\3&\lambda &5\end{array}} \right]$ be non-singular, $\lambda $ should not be equal to

If ${a_{ij}} = \frac{1}{2}(3i – 2j)$ and $A = {[{a_{ij}}]_{2 \times 2}}$, then $A$ is equal to

Find the values of $a, \,b,\, c,$ and $d$ from the following equation: $\left[\begin{array}{cc} 2 a+b & a-2 b \\ 5 c-d & 4 c+3 d \end{array}\right]=\left[\begin{array}{cc} 4 & -3 \\ 11 & 24 \end{array}\right]$

Compute the indicated products $\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$

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Find the values of $a, \,b,\, c,$ and $d$ from the following equation:

$\left[\begin{array}{cc}
2 a+b & a-2 b \\
5 c-d & 4 c+3 d
\end{array}\right]=\left[\begin{array}{cc}
4 & -3 \\
11 & 24
\end{array}\right]$