[1] 第一种方法:
由以下代码
A = ones(n); % n 阶全 1 矩阵
for k = 2 : n
A(k, k) = k; % 将主对角线上第 k 个元素赋值为 k
end
可得题中矩阵;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d4628535e5dde71112ad8188b7efce1b9d166155?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
[2] 第二种方法:
A = ones(n) + diag(0:n-1);
利用全 1 矩阵函数 ones 和对角矩阵函数 diag 可以直接得到题中矩阵;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/37d3d539b6003af36fc51e1e252ac65c1038b655?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
[3] 第三种方法:
A = ones(n);
b = 2 : n;
A(n * (b-1) + b) = b;
通过下标数组的方式给对角线元素赋值;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/94cad1c8a786c917cb7c9047d93d70cf3bc75755?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)