Solution (s) of equation left| {begin{array}{*{20}{c}}x&a&b a&x&ab&b&xend{ar

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3 and 4 .Determinants and Matrices

normal

The solution $(s)$ of the equation $\left| {\begin{array}{*{20}{c}}x&a&b\\ a&x&a\\b&b&x\end{array}} \right|$ $= 0$ is/are :

$x = - (a + b)$

$x = a$

$x = b$

All of the above

Solution

Use $c_1 \rightarrow c_1 -c_2$ and then $R_1 \rightarrow R_1 + R_2 $ to get $\left| {\,\begin{array}{*{20}{c}}0&{a + x}&{b + a}\\{ – (x – a)}&x&a\\0&b&x \end{array}\,} \right|$ $= 0$. Now open by $c_1$ and foctorize

Standard 12

Mathematics

Let $A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j} \neq 0$ for all $i, j$ and $A^2=I$. Let a be the sum of all diagonal elements of $A$ and $b =| A |$, then $3 a ^2+4 b ^2$ is equal to

hard

(JEE MAIN-2023)

$A$ is a $2 × 2$ matrix such that $A$ $\left[ {\begin{array}{{20}{c}}1\\{ – 1} \end{array}} \right]$ $=$$\left[ {\begin{array}{{20}{c}}{ – 1}\\2\end{array}} \right]$ and $A^2$ $\left[ {\begin{array}{{20}{c}}1\\{ – 1}\end{array}} \right]$ $=$ $\left[ {\begin{array}{{20}{c}}1\\0\end{array}} \right]$ . The sum of the elements of $A$, is

normal

Show that

$\left[ {\begin{array}{{20}{l}}
1&2&3 \\
0&1&0 \\
1&1&0
\end{array}} \right]\left[ {\begin{array}{{20}{c}}
{ – 1}&1&0 \\
0&{ – 1}&1 \\
2&3&4
\end{array}} \right]$ $ \ne \left[ {\begin{array}{{20}{c}}
{ – 1}&1&0 \\
0&{ – 1}&1 \\
2&3&4
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
1&2&3 \\
0&1&0 \\
1&1&0
\end{array}} \right]$

medium

Let $A=\left[a_{i j}\right]$ be a matrix of order $3 \times 3$, with $a_{i j}=(\sqrt{2})^{i+j}$. If the sum of all the elements in the third row of $A ^2$ is $\alpha+\beta \sqrt{2}, \alpha, \beta \in Z$, then $\alpha+\beta$ is equal to

hard

(JEE MAIN-2025)

Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right]$

Find $AB$

easy

The solution $(s)$ of the equation $\left| {\begin{array}{*{20}{c}}x&a&b\\ a&x&a\\b&b&x\end{array}} \right|$ $= 0$ is/are :

Solution

Similar Questions

Let $A=\left[a_{i j}\right]_{2 \times 2}$ where $a_{i j} \neq 0$ for all $i, j$ and $A^2=I$. Let a be the sum of all diagonal elements of $A$ and $b =| A |$, then $3 a ^2+4 b ^2$ is equal to

Let $A=\left[a_{i j}\right]$ be a matrix of order $3 \times 3$, with $a_{i j}=(\sqrt{2})^{i+j}$. If the sum of all the elements in the third row of $A ^2$ is $\alpha+\beta \sqrt{2}, \alpha, \beta \in Z$, then $\alpha+\beta$ is equal to

Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right]$ Find $AB$

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Let $A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right]$

Find $AB$